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  • Linking Post Title to Specific Page ID

    - by ThatMacLad
    I've created a form to update my websites homepage with content but I wanted to know how I could set it up so that a posts title links to a specific post ID. I'd also like to add a Read More link that directs anybody reading the blog to the correct post. Here is my PHP code: <html> <head> <title>Blog Name</title> </head> <body> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); $sql = "SELECT * FROM php_blog ORDER BY timestamp DESC LIMIT 5"; $result = mysql_query($sql) or print ("Can't select entries from table php_blog.<br />" . $sql . "<br />" . mysql_error()); while($row = mysql_fetch_array($result)) { $date = date("l F d Y", $row['timestamp']); $title = stripslashes($row['title']); $entry = stripslashes($row['entry']); $password = $row['password']; $id = $row['id']; if ($password == 1) { echo "<p><strong>" . $title . "</strong></p>"; printf("<p>This is a password protected entry. If you have a password, log in below.</p>"); printf("<form method=\"post\" action=\"post.php?id=%s\"><p><strong><label for=\"username\">Username:</label></strong><br /><input type=\"text\" name=\"username\" id=\"username\" /></p><p><strong><label for=\"pass\">Password:</label></strong><br /><input type=\"password\" name=\"pass\" id=\"pass\" /></p><p><input type=\"submit\" name=\"submit\" id=\"submit\" value=\"submit\" /></p></form>",$id); print "<hr />"; } else { ?> <p><strong><?php echo $title; ?></strong><br /><br /> <?php echo $entry; ?><br /><br /> Posted on <?php echo $date; ?> <hr /></p> <?php } } ?> </body> </html> Thanks for any help. I really appreciate any input!

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  • how to upload a audio file using REST webservice in Google App Engine for Java

    - by sathya
    Am using google app engine with eclipse IDE and trying to upload a audio file. I used the File Upload in Google App Engine For Java and can able to upload the file successfully. Now am planning to use REST web service for it. I had analyzed in developers.google but i failed. Can anyone suggest me how to implement REST Web services in google app engine using Eclipse. The code google provided is shown below, // file Upload.java public class Upload extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doPost(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException { Map<String, BlobKey> blobs = blobstoreService.getUploadedBlobs(req); BlobKey blobKey = blobs.get("myFile"); if (blobKey == null) { res.sendRedirect("/"); } else { res.sendRedirect("/serve?blob-key=" + blobKey.getKeyString()); }}} // file Serve.java public class Serve extends HttpServlet { private BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException { BlobKey blobKey = new BlobKey(req.getParameter("blob-key")); blobstoreService.serve(blobKey, res); }} // file index.jsp <%@ page import="com.google.appengine.api.blobstore.BlobstoreServiceFactory" %> <%@ page import="com.google.appengine.api.blobstore.BlobstoreService" %> <% BlobstoreService blobstoreService = BlobstoreServiceFactory.getBlobstoreService(); %> <form action="<%= blobstoreService.createUploadUrl("/upload") %>" method="post" enctype="multipart/form-data"> <input type="file" name="myFile"> <input type="submit" value="Submit"> </form> // web.xml <servlet> <servlet-name>Upload</servlet-name> <servlet-class>Upload</servlet-class> </servlet> <servlet> <servlet-name>Serve</servlet-name> <servlet-class>Serve</servlet-class> </servlet> <servlet-mapping> <servlet-name>Upload</servlet-name> <url-pattern>/upload</url-pattern> </servlet-mapping> <servlet-mapping> <servlet-name>Serve</servlet-name> <url-pattern>/serve</url-pattern> </servlet-mapping> Now how to provide a rest web service for the above code. Kindly suggest me an idea.

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  • why resubmit after refresh php page

    - by user2719452
    why resubmit after refresh php page? try it, go to: http://qass.im/message-envelope/ and upload any image now try click F5, after refresh page "resubmit" Why? I don't want resubmit after refresh page What is the solution? See this is my form code: <form id="uploadedfile" name="uploadedfile" enctype="multipart/form-data" action="upload.php" method="POST"> <input name="uploadedfile" type="file" /> <input type="submit" value="upload" /> </form> See this is php code upload.php file: <?php $allowedExts = array("gif", "jpeg", "jpg", "png", "zip", "pdf", "docx", "rar", "txt", "doc"); $temp = explode(".", $_FILES["uploadedfile"]["name"]); $extension = end($temp); $newname = $extension.'_'.substr(str_shuffle(str_repeat("0123456789abcdefghijklmnopqrstuvwxyz", 7)), 4, 7); $imglink = 'attachment/attachment_file_'; $uploaded = $imglink .$newname.'.'.$extension; if ((($_FILES["uploadedfile"]["type"] == "image/jpeg") || ($_FILES["uploadedfile"]["type"] == "image/jpeg") || ($_FILES["uploadedfile"]["type"] == "image/jpg") || ($_FILES["uploadedfile"]["type"] == "image/pjpeg") || ($_FILES["uploadedfile"]["type"] == "image/x-png") || ($_FILES["uploadedfile"]["type"] == "image/gif") || ($_FILES["uploadedfile"]["type"] == "image/png") || ($_FILES["uploadedfile"]["type"] == "application/msword") || ($_FILES["uploadedfile"]["type"] == "text/plain") || ($_FILES["uploadedfile"]["type"] == "application/vnd.openxmlformats-officedocument.wordprocessingml.document") || ($_FILES["uploadedfile"]["type"] == "application/pdf") || ($_FILES["uploadedfile"]["type"] == "application/x-rar-compressed") || ($_FILES["uploadedfile"]["type"] == "application/x-zip-compressed") || ($_FILES["uploadedfile"]["type"] == "application/zip") || ($_FILES["uploadedfile"]["type"] == "multipart/x-zip") || ($_FILES["uploadedfile"]["type"] == "application/x-compressed") || ($_FILES["uploadedfile"]["type"] == "application/octet-stream")) && ($_FILES["uploadedfile"]["size"] < 5242880) // Max size is 5MB && in_array($extension, $allowedExts)) { move_uploaded_file($_FILES["uploadedfile"]["tmp_name"], $uploaded ); echo '<a target="_blank" href="'.$uploaded.'">click</a>'; // If has been uploaded file echo '<h3>'.$uploaded.'</h3>'; } if($_FILES["uploadedfile"]["error"] > 0){ echo '<h3>Please choose file to upload it!</h3>'; // If you don't choose file } elseif(!in_array($extension, $allowedExts)){ echo '<h3>This extension is not allowed!</h3>'; // If you choose file not allowed } elseif($_FILES["uploadedfile"]["size"] > 5242880){ echo "Big size!"; // If you choose big file } ?> if you have solution, please edit my php code and paste your solution code! Thanks.

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  • Crystal Reports : How to add an external assembly class?

    - by Sunil
    I am using VS2010, CrystalReport13 & MVC3. My problem is unable to add an external assembly in Crystal Report using "Database Expert" Option. I have a class named WeeklyReportModel in an external assembly. In my web project, data retrieving from DB as IEnumerable collection of WeeklyReportModel. I tried ProjectData - .NetObjects in Crystal Report for adding the WeeklyReportModel. But this external assembly is not showing under ".NetObjects". Then I tried other option as Create New Connection - ADO.Net – Make New Connection and pointed this External Assembly. It has been added under Ado.Net node, but while expanding displays as "...no items found..." Totally frustrated. Please help. External Assembly Class: namespace SMS.Domain { public class WeeklyReportModel { public int StoreId { get; set; } public string StoreName{ get; set; } public decimal Saturday { get; set; } public decimal Sunday { get; set; } public decimal Monday { get; set; } public decimal Tuesday { get; set; } public decimal Wednesday { get; set; } public decimal Thurday { get; set; } public decimal Friday { get; set; } public decimal Average { get; set; } public string DateRange { get; set; } } } In Controller-action[Data retrieving as Collection Of WeeklyReportModel] namespace SMS.UI.Controllers { public class ReportController : Controller { public ActionResult StoreWeeklyReport(string id) { DateTime weekStart, weekClose; string[] dateArray = id.Split('_'); weekStart = Convert.ToDateTime(dateArray[0].ToString()); weekClose = Convert.ToDateTime(dateArray[1].ToString()); SMS.Infrastructure.Report.AuditReport weeklyReport = new SMS.Infrastructure.Report.AuditReport(); IEnumerable<SMS.Domain.WeeklyReportModel> weeklyRpt = weeklyReport.ReportByStore().WeeklyReport(weekStart, weekClose); Session["WeeklyData"] = weeklyRpt; Response.Redirect("~/Reports/Weekly/StoreWeekly.aspx"); return View(); } } } Thanks in advance.

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  • I am unsure of how to access a persistence entity from a JSP page?

    - by pharma_joe
    Hi, I am just learning Java EE, I have created a Persistence entity for a User object, which is stored in the database. I am now trying to create a JSP page that will allow a client to enter a new User object into the System. I am unsure of how the JSP page interacts with the User facade, the tutorials are confusing me a little. This is the code for the facade: <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>Add User to System</title> </head> <body> <h2>Add User</h2> <h3>Please fill out the details to add a user to the system</h3> <form action=""> <label>Email:</label> <input type="text" name="email"><br /> <label>Password:</label> <input type="password" name="name"><br /> <label>Name:</label> <input type="text" name="name"><br /> <label>Address:</label> <input type="text" name="address"><br /> <label>Type:</label> <select name="type"> <option>Administrator</option> <option>Member</option> </select><br /> <input type="submit" value="Add" name="add"/> <input type="reset" value="clear" /> </form> </body> This is the code I have to add a new User object within the User facade class: @Stateless public class CinemaUserFacade { @PersistenceContext(unitName = "MonashCinema-warPU") private EntityManager em; public void create(CinemaUser cinemaUser) { em.persist(cinemaUser); } I am finding it a little difficult to get my head around the whole MVC thing, getting there but would appreciate it if someone could turn the light on for me!

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  • Wordpress Search Results in Order

    - by Brad Houston
    One of my clients websites, www.kevinsplants.co.uk is not showing the search results in alphabetical order, how do I go about ordering the results in alphabetical order? We are using the Shopp plugin and I believe its that plugin that is generating the results! Cheers, Brad case "orderby-list": if (isset($Shopp->Category->controls)) return false; if (isset($Shopp->Category->smart)) return false; $menuoptions = Category::sortoptions(); $title = ""; $string = ""; $default = $Shopp->Settings->get('default_product_order'); if (empty($default)) $default = "title"; if (isset($options['default'])) $default = $options['default']; if (isset($options['title'])) $title = $options['title']; if (value_is_true($options['dropdown'])) { if (isset($Shopp->Cart->data->Category['orderby'])) $default = $Shopp->Cart->data->Category['orderby']; $string .= $title; $string .= '<form action="'.esc_url($_SERVER['REQUEST_URI']).'" method="get" id="shopp-'.$Shopp->Category->slug.'-orderby-menu">'; if (!SHOPP_PERMALINKS) { foreach ($_GET as $key => $value) if ($key != 'shopp_orderby') $string .= '<input type="hidden" name="'.$key.'" value="'.$value.'" />'; } $string .= '<select name="shopp_orderby" class="shopp-orderby-menu">'; $string .= menuoptions($menuoptions,$default,true); $string .= '</select>'; $string .= '</form>'; $string .= '<script type="text/javascript">'; $string .= "jQuery('#shopp-".$Shopp->Category->slug."-orderby-menu select.shopp-orderby-menu').change(function () { this.form.submit(); });"; $string .= '</script>'; } else { if (strpos($_SERVER['REQUEST_URI'],"?") !== false) list($link,$query) = explode("\?",$_SERVER['REQUEST_URI']); $query = $_GET; unset($query['shopp_orderby']); $query = http_build_query($query); if (!empty($query)) $query .= '&'; foreach($menuoptions as $value => $option) { $label = $option; $href = esc_url($link.'?'.$query.'shopp_orderby='.$value); $string .= '<li><a href="'.$href.'">'.$label.'</a></li>'; } } return $string; break;

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  • jquery ajax form success callback not being called

    - by Michael Merchant
    I'm trying to upload a file using "AJAX", process data in the file and then return some of that data to the UI so I can dynamically update the screen. I'm using the JQuery Ajax Form Plugin, jquery.form.js found at http://jquery.malsup.com/form/ for the javascript and using Django on the back end. The form is being submitted and the processing on the back end is going through without a problem, but when a response is received from the server, my Firefox browser prompts me to download/open a file of type "application/json". The file has the json content that I've been trying to send to the browser. I don't believe this is an issue with how I'm sending the json as I have a modularized json_wrapper() function that I'm using in multiple places in this same application. Here is what my form looks after Django templates are applied: <form method="POST" enctype="multipart/form-data" action="/test_suites/active/upload_results/805/"> <p> <label for="id_resultfile">Upload File:</label> <input type="file" id="id_resultfile" name="resultfile"> </p> </form> You won't see any submit buttons because I'm calling submit with a button else where and am using ajaxSubmit() from the jquery.form.js plugin. Here is the controlling javascript code: function upload_results($dialog_box){ $form = $dialog_box.find("form"); var options = { type: "POST", success: function(data){ alert("Hello!!"); }, dataType: "json", error: function(){ console.log("errors"); }, beforeSubmit: function(formData, jqForm, options){ console.log(formData, jqForm, options); }, } $form.submit(function(){ $(this).ajaxSubmit(options); return false; }); $form.ajaxSubmit(options); } As you can see, I've gotten desperate to see the success callback function work and simply have an alert message created on success. However, we never reach that call. Also, the error function is not called and the beforeSubmit function is executed. The file that I get back has the following contents: {"count": 18, "failed": 0, "completed": 18, "success": true, "trasaction_id": "SQEID0.231"} I use 'success' here to denote whether or not the server was able to run the post command adequately. If it failed the result would look something like: {"success": false, "message":"<error_message>"} Your time and help is greatly appreciated. I've spent a few days on this now and would love to move on.

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  • PHP & MySQL - saving and looping problems.

    - by R.I.P.coalMINERS
    I'm new to PHP and MySQL I want a user to be able to store multiple names and there meanings in a MySQL database tables named names using PHP I will dynamically create form fields with JQuery every time a user clicks on a link so a user can enter 1 to 1,000,000 different names and there meanings which will be stored in a table called names. Since I asked my last question I figured out how to store my values from my form using the for loop but every time I loop my values when I add one or more dynamic fields the second form field named meaning will not save the value entered also my dynamic form fields keep looping doubling, tripling and so on the entered values into the database it all depends on how many form fields are added dynamically. I was wondering how can I fix these problems? On a side note I replaced the query with echo's to see the values that are being entered. Here is the PHP code. <?php if(isset($_POST['submit'])) { $mysqli = mysqli_connect("localhost", "root", "", "site"); $dbc = mysqli_query($mysqli,"SELECT * FROM names WHERE userID='$userID'"); $name = $_POST['name']; $meaning = $_POST['meaning']; if(isset($name['0']) && mysqli_num_rows($dbc) == 0 && trim($name['0'])!=='' && trim($meaning['0'])!=='') { for($n = 0; $n < count($name); $n++) { for($m = 0; $m < count($meaning); $m++) { echo $name[$n] . '<br />'; echo $meaning[$m] . '<br /><br />'; break; } } } } ?> And here is the HTML code. <form method="post" action="index.php"> <ul> <li><label for="name">Name: </label><input type="text" name="name[]" id="name" /></li> <li><label for="meaning">Meaning: </label><input type="text" name="meaning[]" id="meaning" /></li> <li><input type="submit" name="submit" value="Save" /></li> </ul> </form> If needed I will place the JQuery code.

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  • EJB / JSF java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader

    - by Eric Sant'Anna
    I'm in my first time using EJB and JSF, and I can't resolve this: 20:23:12,457 Grave [javax.enterprise.resource.webcontainer.jsf.application] (http-localhost-127.0.0.1-8081-2) com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)]: java.lang.ClassNotFoundException: com.ericsantanna.jobFC.dao.DAOFactoryRemote from [Module "com.sun.jsf-impl:main" from local module loader @439db2b2 (roots: C:\jboss-as-7.1.1.Final\modules)] I'm getting this when I do an action like a selectOneMenu or a commandButton click. DAOFactory.class @Singleton @Remote(DAOFactoryRemote.class) public class DAOFactory implements DAOFactoryRemote { private static final long serialVersionUID = 6030538139815885895L; @PersistenceContext private EntityManager entityManager; @EJB private JobDAORemote jobDAORemote; /** * Default constructor. */ public DAOFactory() { // TODO Auto-generated constructor stub } @Override public JobDAORemote getJobDAO() { JobDAO jobDAO = (JobDAO) jobDAORemote; jobDAO.setEntityManager(entityManager); return jobDAO; } JobDAO.class @Stateless @Remote(JobDAORemote.class) public class JobDAO implements JobDAORemote { private static final long serialVersionUID = -5483992924812255349L; private EntityManager entityManager; /** * Default constructor. */ public JobDAO() { // TODO Auto-generated constructor stub } @Override public void insert(Job t) { entityManager.persist(t); } @Override public Job findById(Class<Job> classe, Long id) { return entityManager.getReference(classe, id); } @Override public Job findByName(Class<Job> clazz, String name) { return entityManager .createQuery("SELECT job FROM " + clazz.getName() + " job WHERE job.nome = :nome" , Job.class) .setParameter("name", name) .getSingleResult(); } ... TriggerFormBean.class @ManagedBean @ViewScoped @Stateless public class TriggerFormBean implements Serializable { private static final long serialVersionUID = -3293560384606586480L; @EJB private DAOFactoryRemote daoFactory; @EJB private TriggerManagerRemote triggerManagerRemote; ... triggerForm.xhtml (a portion with problem) </p:layoutUnit> <p:layoutUnit id="eastConditionPanel" position="center" size="50%"> <p:panel header="Conditions to Release" style="width:97%;height:97%;"> <h:panelGrid columns="2" cellpadding="3"> <h:outputLabel value="Condition Name:" for="conditionName" /> <p:inputText id="conditionName" value="#{triggerFormBean.newCondition.name}" /> </h:panelGrid> <p:commandButton value="Add Condition" update="conditionsToReleaseList" id="addConditionToRelease" actionListener="#{triggerFormBean.addNewCondition}" /> <p:orderList id="conditionsToReleaseList" value="#{triggerFormBean.trigger.conditionsToRelease}" var="condition" controlsLocation="none" itemLabel="#{condition.name}" itemValue="#{condition}" iconOnly="true" style="width:97%;heigth:97%;"/> </p:panel> </p:layoutUnit> In TriggerFormBean.class if comments daoFactory we get the same exception with triggerManagerRemote, both annotated with @EJB. I'm don't understand the relationship between my DAOFactory and the "Module com.sun.jsf-impl:main"... Thanks.

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  • mysql_fetch_array() expects parameter 1 to be resource problem

    - by user225269
    I don't get it, I see no mistakes in this code but there is this error, please help: mysql_fetch_array() expects parameter 1 to be resource problem <?php $con = mysql_connect("localhost","root","nitoryolai123$%^"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']); ?> <?php while ($row = mysql_fetch_array($result)) { ?> <table class="a" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3"> <tr> <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);"> <td> <table border="0" cellpadding="3" cellspacing="1" bgcolor=""> <tr> <td colspan="16" height="25" style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td> </tr> <tr> <td width="30" height="35"><font size="2">*Year:</td> <td width="30"><input name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td> <?php } ?> I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

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  • How can I run an android frame animation without it skewing?

    - by GameDev123
    I have a small state machine that runs a series of frame by frame animations in an ImageView, in a nested hierarchy of layouts. There is more than adequate space to display each frame of the animation. Each frame of the animation is cropped to fit the minimum amount of area, in order to save memory. If a frame only contains 50x50 worth of pixels then the png is 50x50. There is no transparent padding to keep them the same size. The ImageView is directly within a RelativeLayout, and is anchored to the bottom left with some padding. The general idea being that the character in the animation performs some action, which results in individual frames of the animation growing or shrinking. The issue is that individual frames of animation are skewed, and there does not appear to be any way of preventing this. If I set the source of the imageview directly to one of the frames of animation, it displays fine in the layout manager. I have tried this with Adjust View Bounds set to true, false, and undefined. I have tried using the background and the src attribute of imageview to set the animation drawable, I have tried every configuration of layout manager and setting minimum/maximum size that I can think of, and it still stretches the character on various frames depending on the size of the source png. In essence, all I want to do is say "I want this ImageView to anchor in the bottom left and then display any frame that happens to be in it without stretching or skewing it in any way aside from that which occurred when the frame png's were loaded." Seems simple, but I have yet to come across any way of doing it. Here is the layout of the imageview as of my last test, I had to remove bits of the XML to get it to display but nothing pertinent: RelativeLayout android:orientation="horizontal" android:layout_above="@+id/MenuOptions" android:layout_width="fill_parent" android:id="@+id/AnimationLayout" android:clipChildren="false" android:minHeight="180dp" android:layout_height="fill_parent" android:layout_below="@+id/GameBarLayout" ImageView android:id="@+id/animatedImg" android:layout_height="wrap_content" android:layout_width="wrap_content" android:visibility="visible" android:baselineAlignBottom="true" android:minHeight="180dp" android:minWidth="200dp" android:adjustViewBounds="true" android:layout_alignParentBottom="true" android:paddingLeft="30dp" android:paddingBottom="10dp" android:src="@drawable/idle01"/ImageView /RelativeLayout Here is how an animation is set up: animationDrawable = new AnimationDrawable(); animationDrawable.addFrame(res.getDrawable(R.drawable.idle01), 16); animationDrawable.addFrame(res.getDrawable(R.drawable.idle02), 16); animationDrawable.addFrame(res.getDrawable(R.drawable.idle03), 16);

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  • How do i know in the detail view what cell in tableview was selected?

    - by Daniel Rotaru
    how can i know what tableview cell was selected?(being in the detail view) The problem is that. I have an table view controller. Here are parsed from the internet entries to the table. So it's a dynamic tabe view that loads from internet. I will not know how many entries will be in the table so i will not know what details view to call when i click a row. So i have maked one view. This view contains an calendar. On this calendar(wich is the detail iew) i will parse data from internet depending on the selected row. For exemple: i have table: entry 1, entry 2,entry 3,entry 4 When i click entry 2 i need to call a php with the argument entry 2. The php will know what entry on the table i have selected and will generate me the correct xml that i will parse. Here is my tableview didSelectRow function: - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { // Navigation logic -- create and push a new view controller if(bdvController == nil) bdvController = [[BookDetailViewController alloc] initWithNibName:@"BookDetailView" bundle:[NSBundle mainBundle]]; Villa *aVilla = [appDelegate.villas objectAtIndex:indexPath.row]; [self.navigationController pushViewController:bdvController animated:YES] And here is my self view function on detailviewcontroller: -(void)loadView { [super loadView]; self.title=@"Month" UIBarButtonItem *addButtonItem = [[UIBarButtonItem alloc] initWithTitle:@"ListView" style:UIBarButtonItemStyleDone target:self action:@selector(add:)]; self.navigationItem.rightBarButtonItem = addButtonItem; calendarView = [[[KLCalendarView alloc] initWithFrame:CGRectMake(0.0f, 0.0f, 320.0f, 373.0f) delegate:self] autorelease]; appDelegate1 = (XMLAppDelegate *)[[UIApplication sharedApplication] delegate]; myTableView=[[UITableView alloc]initWithFrame:CGRectMake(0, 260, 320, 160)style:UITableViewStylePlain]; myTableView.dataSource=self; myTableView.delegate=self; UIView *myHeaderView=[[UIView alloc]initWithFrame:CGRectMake(0, 0, myTableView.frame.size.width,2)]; myHeaderView.backgroundColor=[UIColor grayColor]; [myTableView setTableHeaderView:myHeaderView]; [self.view addSubview:myTableView]; [self.view addSubview:calendarView]; [self.view bringSubviewToFront:myTableView]; } I think that here in self load i need to make the if procedure.. If indexPath.row=x parse fisier.php?variabila=title_of_rowx but the question is how i know the indexPath variable?

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  • Routing error when trying to use same view for update and create flows (Rails 3)

    - by Jamis Charles
    My overall use case: I have a Listing model that has many images. The Listing detail page lists all the fields that can be updated inline (through ajax). I want to be able to use the same view for both update listing and create new listing. My listing controller looks as follows: def detail @listing = Listing.find(params[:id]) @image = Image.new #should this link somewhere else? respond_to do |format| format.html # show.html.erb format.xml { render :xml => @listing } end end def create # create a new listing and save it immediately. Assign it to guest, with a status of "draft" @listing = Listing.new(:price_id => 1) # Default price id # save it to db # TODO add validation that it has to have a price ID, on record creation. So the view doesn't break. @listing.save @image = Image.new # redirect_to "/listings/detail/@listing.id" #this didn't work respond_to do |format| format.html # show.html.erb format.xml { render :xml => @listing } end end The PROBLEM I'm using a partial that shows the same form for the create view and the detail view. This works perfectly except for one thing: When I pull up http://0.0.0.0:3000/listings/detail/7, it works perfectly. When I pull up http://0.0.0.0:3000/listings/new, I get the following error: Showing /Applications/MAMP/htdocs/rails_testing/feedbackd/app/views/listings/_edit_form.html.erb where line #100 raised: No route matches {:action="show", :controller="images"} Extracted source (around line #100): 97: <!-- Form for new images --> 98: <div class="span-20 append-bottom"> 99: <!-- <%# form_for :image, @image, :url => image_path, :html => { :multipart => true } do |f| %> --> 100: <%= form_for @image, :url => image_path, :html => { :multipart => true } do |f| %> 101: <%= f.text_field :description %><br /> 102: <%= f.file_field :photo %> 103: <%= submit_tag "Upload" %> What I think the issue is: When I upload a new image (I'm using Paperclip), it requires the listing_id to create the image record. Since the listing_id isn't passed in with listings/new it can't find the listing_id. How can I pass in the id? Via a redirect? What's the best way to solve this? Thank you.

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  • Creating a mySQL query using PHP form dropdowns - If user ignores dropdown, do not filter by that pa

    - by user303043
    Hello, I am creating a simple MySQL query that will be built from the user selecting options from 2 dropdowns. The issue I am having is that I would like each drop down to default that if they do not actually choose an option, do not filter by that dropdown parameter. So, if they come in, and simply hit submit without choosing from a dropdown they should see everything. If they come in and pick from only one of the dropdowns, the query will basically ignore filtering by the other dropdown. I tried making <OPTION VALUE='any'>Choose but my query didn't know what to do with the 'any' and just shows no results. Here is my code. Thank you very much for whatever help you can provide. FORM <form method="POST" action="<?php echo $_SERVER['REQUEST_URI']; ?>"> <select name="GameType"> <OPTION VALUE='any'>Choose Game Type <option value="Game1">Option 1</option> <option value="Game2">Option 2</option> <option value="Game3">Option 3</option> </select> <select name="Instructor"> <OPTION VALUE='any'>Choose Instructor <option VALUE="InstructorA">Instructor A</option> <option value="InstructorB">Instructor B</option> <option value="InstructorC">Instructor C</option> </select> <input type='submit' value='Search Videos'> </form> MYSQL <?PHP connection stuff $db_handle = mysql_connect($server, $user_name, $password); $db_found = mysql_select_db($database, $db_handle); if ($db_found) { $SQL = "SELECT * FROM Videos WHERE GameType=\"{$_POST['GameType']}\" AND Instructor=\"{$_POST['Instructor']}\""; $result = mysql_query($SQL); while ($db_field = mysql_fetch_assoc($result)) { echo $db_field['ShortDescription'] . ", "; echo $db_field['LongDescription'] . ", "; echo $db_field['GameType'] . ", "; echo $db_field['NumberOfPlayers'] . ", "; echo $db_field['Instructor'] . ", "; echo $db_field['Stakes'] . ", "; echo $db_field['UserPermissionLevel'] . ", "; echo $db_field['DateCreated'] . "<BR>"; } mysql_close($db_handle); } else { print "Database NOT Found "; mysql_close($db_handle); } ?>

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  • jQuery slider onChange auto submit form

    - by bikey77
    I'm using a jQuery slider as a price range selector in a form. I'd like to have the form submit automatically when one of the values has been changed. I used a couple of examples I found on SO but they didn't work with my code. <form action="itemlist.php" method="post" enctype="application/x-www-form-urlencoded" name="priceform" id="priceform" target="_self"> <div id="slider-holder"> Prices: From <span id="pricefromlabel">100 &#8364;</span> To <span id="pricetolabel">500 &#8364;</span> <input type="hidden" id="pricefrom" name="pricefrom" value="100" /> <input type="hidden" id="priceto" name="priceto" value="500" /> <div id="slider-range"></div> <input name="Search" type="submit" value="Search" /> </div> </form> This is the code that displays the values of the slider and updates 2 hidden form fields I use to store the prices in order to submit: <script> $(function() { $("#slider-range" ).slider({ range: true, min: 0, max: 1000, values: [ <?=$minprice?>, <?=$maxprice?> ], start: function (event, ui) { event.stopPropagation(); }, slide: function( event, ui ) { $( "#pricefrom" ).val(ui.values[0]); $( "#priceto" ).val(ui.values[1]); $( "#pricefromlabel" ).html(ui.values[0] + ' &euro;'); $( "#pricetolabel" ).html(ui.values[1] + ' &euro;'); } }); return false; }); </script> I've tried adding this code as well as an data-autosubmit="true" attribute to the div but no result. $(function() { $('[data-autosubmit="true"]').change(function() { parentForm = $(this).('#priceform'); clearTimeout(submitTimeout); submitTimeout = setTimeout(function() { parentForm.submit() }, 100); }); I've also tried adding a $.post() event to the slider but I'm not very good with jQuery so I'm probably doing it wrong. Any help will be appreciated.

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  • Defining a select list through controller and view model

    - by Ibrar Hussain
    I have a View Model that looks like this: public class SomeViewModel { public SomeViewModel(IEnumerable<SelectListItem> orderTemplatesListItems) { OrderTemplateListItems = orderTemplatesListItems; } public IEnumerable<SelectListItem> OrderTemplateListItems { get; set; } } I then have an Action in my Controller that does this: public ActionResult Index() { var items = _repository.GetTemplates(); var selectList = items.Select(i => new SelectListItem { Text = i.Name, Value = i.Id.ToString() }).ToList(); var viewModel = new SomeViewModel { OrderTemplateListItems = selectList }; return View(viewModel); } Lastly my view: @Html.DropDownListFor(n => n.OrderTemplateListItems, new SelectList(Model.OrderTemplateListItems, "value", "text"), "Please select an order template") The code works fine and my select list populates wonderfully. Next thing I need to do is set the selected value that will come from a Session["orderTemplateId"] which is set when the user selects a particular option from the list. Now after looking online the fourth parameter should allow me to set a selected value, so if I do this: @Html.DropDownListFor(n => n.OrderTemplateListItems, new SelectList(Model.OrderTemplateListItems, "value", "text", 56), "Please select an order template") 56 is the Id of the item that I want selected, but to no avail. I then thought why not do it in the Controller? As a final attempt I tried building up my select list items in my Controller and then passing the items into the View: public ActionResult Index() { var items = _repository.GetTemplates(); var orderTemplatesList = new List<SelectListItem>(); foreach (var item in items) { if (Session["orderTemplateId"] != null) { if (item.Id.ToString() == Session["orderTemplateId"].ToString()) { orderTemplatesList.Add(new SelectListItem { Text = item.Name, Value = item.Id.ToString(), Selected = true }); } else { orderTemplatesList.Add(new SelectListItem { Text = item.Name, Value = item.Id.ToString() }); } } else { orderTemplatesList.Add(new SelectListItem { Text = item.Name, Value = item.Id.ToString() }); } } var viewModel = new SomeViewModel { OrderTemplateListItems = orderTemplatesList }; return View(viewModel); } Leaving my View like so: @Html.DropDownListFor(n => n.OrderTemplateListItems, new SelectList(Model.OrderTemplateListItems, "value", "text"), "Please select an order template") Nothing! Why isn't this working for me?

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  • Undefined index: email in C:\wamp\www\emailvalidate.php on line 4

    - by klari
    I am new to Ajax. In my Ajax I get the following error message : Notice: Undefined index: address in C:\wamp\www\test\sample.php on line 11 I googled but I didn't get a solution for my specified issue. Here is what I did. HTML Form with Ajax (test1.php) <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> <script> function loadXMLDoc() { var xmlhttp; if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("mydiv").innerHTML=xmlhttp.responseText; } } xmlhttp.open("POST","test2.php",true); xmlhttp.send(); } </script> </head> <body> <form id="form1" name="form1" method="post" action="sample.php"> <p> <label for="mail"></label> <input type="text" name="mail" id="mail" onblur="loadXMLDoc()" /> <input type="submit" name="submit" id="submit" value="Submit" /> </p> <p><div id = 'mydiv'></div></p> </form> </body> </html> test2.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <?php echo "Your Address is ".$_POST['address']; ?> </body> </html> I am sure it is very simple issue but I don't know how to solve it. Any help ?

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  • I am having trouble using jquery to submit a form. It was working before

    - by noah
    When a user clicks a link it uses jquery ajax to submit a form to go to paypal. Not working for some reason. Really appreciate any help... LINK TO CLICK I put this in an href for onClick: javascript:go_paypal(); CODE TO EXECUTE ON CLICK function go_paypal() { data = 'req_paypal=1'; $.blockUI({ message: '<h1> Going to Paypal...</h1>',css:{background:'#000'} }); $.ajax({ type: "POST", url: "index.php", data: data, success: function(data) { $("#paypal_form").html(data); $("#payPalForm").submit(); } , error: function() {$.unblockUI(); alert('Unable to communicate to server.'); } }); return false; } CODE TO GO ON SUBMIT if(isset($_POST['req_paypal']) && $_POST['req_paypal'] == 1 ) { $sql = 'INSERT INTO `transactions` (id,type,ip,time,ammount,status) VALUES (NULL,1,\''.$_SERVER['REMOTE_ADDR'].'\',\''.time().'\',\''.$global['paypal_prod_amount'].'\',0) '; echo $sql; mysql_query($sql); $id = mysql_insert_id(); $html = ' <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post" id="payPalForm"> <input type="hidden" name="item_number" value="One Year of Imgur Pro"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="no_note" value="1"> <input type="hidden" name="business" value="'.$global['paypal_email'].'"> <input type="hidden" name="custom" value="'.base64_encode($id).'"> <input type="hidden" name="currency_code" value="USD"> <input type="hidden" name="return" value="'.$global['paypal_return'].'"> <input name="item_name" type="hidden" id="item_name" value="One Year of Imgur Pro" > <input name="amount" type="hidden" id="amount" value="'.$global['paypal_prod_amount'].'" > </form> '; echo $html;exit; }

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  • How to update a TextView on ButtonClick with Spinner(s) values

    - by source.rar
    Hi, I am trying to populate a TextView based on the current selected options in 3 Spinner(s) but cant seem to figure out how to retrieve the selected values from the Spinners to invoke the update function with. Here is my current code (quite messy but I'm just learning Java :)), public class AgeFun extends Activity { private String[] dayNames; private String[] yearArray; private final static int START_YEAR = 1990; private static TextView textDisp; private Button calcButton; private static Spinner spinnerDay, spinnerYear, spinnerMonth; private static ArrayAdapter<?> monthAdapter, dayAdapter, yearAdapter; private int year, month, day; /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); year = 2000; month = 1; day = 1; textDisp = (TextView) findViewById(R.id.textView1); calcButton = (Button) findViewById(R.id.button); calcButton.setOnClickListener(new OnClickListener() { public void onClick(View v) { // Perform action on clicks AgeFun.updateAge(year, month, day); } }); // Month spinner spinnerMonth = (Spinner) findViewById(R.id.spinnerFirst); monthAdapter = ArrayAdapter.createFromResource( this, R.array.monthList, android.R.layout.simple_spinner_item); monthAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); spinnerMonth.setAdapter(monthAdapter); // Day spinner dayNames = new String[31]; for(int i =1; i <= 31; ++i) { dayNames[i-1] = Integer.toString(i); } spinnerDay = (Spinner) findViewById(R.id.spinnerSecond); dayAdapter = new ArrayAdapter<CharSequence>(this, android.R.layout.simple_spinner_item, dayNames); spinnerDay.setAdapter(dayAdapter); // Year spinner yearArray = new String[40]; for(int i =0; i < 40; ++i) { yearArray[i] = Integer.toString(START_YEAR+i); } spinnerYear = (Spinner) findViewById(R.id.spinnerThird); yearAdapter = new ArrayAdapter<CharSequence>(this, android.R.layout.simple_spinner_item, yearArray); spinnerYear.setAdapter(yearAdapter); updateAge(2000,1,1); } private static void updateAge(int year, int month, int day) { Date dob = new GregorianCalendar(year, month, day).getTime(); Date currDate = new Date(); long age = (currDate.getTime() - dob.getTime()) / (1000 * 60 * 60 * 24) / 365; textDisp.setText("Your are " + Long.toString(age) + " years old"); } } Any help with this would be great. TIA

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  • Parse error in PHP login form

    - by user225269
    I'm trying to have a login form in php. But my current code doesnt work. Here is the form: <form name="form1" method="post" action="loginverify.php"> <td><font size="3">Username:</td> <td></td> <td><input type="text" name="uname" value="" maxlength="15"/><br/></td> <td><font size="3">Password:</td> <td></td> <td><input type="text" name="pword" value="" maxlength="15"/><br/></td> <tr> <td>&nbsp;</td> <td>&nbsp;</td> <td><input type="submit" name="Submit" value="Login" /></td> </form> And the verify.php <?php session_start(); ?> <?php $host="localhost"; $username="root"; $password="nitoryolai123$%^"; $db_name="login"; $tbl="users"; $connection=mysql_connect($host, $username, $password) or die("cannot connect"); mysql_select_db($db_name, $connection) or die("cannot select db"); $user=$_POST['uname'] $pass=$_POST['pword'] $sql="SELECT Username, Password from users where Username='$user' and Password='$pass'"; $result=mysql_query[$sql]; $count=mysql_num_rows($result); if($count==1){ $SESSION['Username']=$user; echo"<a href='searchmain.php'> CONTINUE</a>"; } else{ echo"wrong username or password"; echo"<a href='loginform.php'>Back</a>"; } ?> Is there something wrong with my code. I get this parse error at line 15, which is this: $pass=$_POST['pword'] But when I try to remove it.It goes to line 16 or line 17 again. What do I do

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  • A script that writes errors and should create a event-error

    - by helmich
    this if it works should check the internet connection if there is a connection it does nothing. if there isn't a connection it should write a error in a txtfile if that happend 5 times it should create a error but it doesn't I will show you the whole code that i have now and the piece of code that i want in a loop. I can't get it in the way i want. I want it to creat 1 Event-error after 5 times writing to the file. this is the whole code i will put the code i want in a loop under it strDirectory = "Z:\text2" strFile = "\foutmelding.txt" strText = "De connectie is verbroken" strWebsite = "www.helmichbeens.com" If PingSite(strWebsite) Then WScript.Quit 'Website is pingable - no further action required Set objFSO = CreateObject("Scripting.FileSystemObject") RecordSingleEvent Dim fout For fout = 1 To 5 : Do If fout = 5 Then Exit Do Set WshShell = WScript.CreateObject("WScript.Shell") Call WshShell.LogEvent(1, "Test Event") Loop While False : next '------------------------------------ 'Record a single event in a text file '------------------------------------ Sub RecordSingleEvent If Not objFSO.FolderExists(strDirectory) Then objFSO.CreateFolder(strDirectory) Set objTextFile = objFSO.OpenTextFile(strDirectory & strFile, 8, True) objTextFile.WriteLine(Now & strText) objTextFile.Close End sub '---------------- 'Ping my web site '---------------- Function PingSite( myWebsite ) Set objHTTP = CreateObject( "WinHttp.WinHttpRequest.5.1" ) objHTTP.Open "GET", "http://" & myWebsite & "/", False objHTTP.SetRequestHeader "User-Agent", "Mozilla/4.0 (compatible; MyApp 1.0; Windows NT 5.1)" On Error Resume Next objHTTP.Send PingSite = (objHTTP.Status = 200) On Error Goto 0 End Function '----------------------------------------------- 'Counts the number of lines inside the text file '----------------------------------------------- Function EventCount(fout) strData = objFSO.OpenTextFile(strDirectory & strFile,ForReading).ReadAll arrLines = Split(strData,vbCrLf) EventCount = UBound(arrLines) End Function This is the whole code, and it doesnt work correctly becaus it creats a event-log rightaway and it should do that after the script has written 5 times to the textfile here is the code that writes to a textfile Sub RecordSingleEvent If Not objFSO.FolderExists(strDirectory) Then objFSO.CreateFolder(strDirectory) Set objTextFile = objFSO.OpenTextFile(strDirectory & strFile, 8, True) objTextFile.WriteLine(Now & strText) objTextFile.Close End sub and here is the code but this part doesnt not work or atleast i think it is this part Dim fout For fout = 1 To 5 : Do If fout = 5 Then Exit Do Set WshShell = WScript.CreateObject("WScript.Shell") Call WshShell.LogEvent(1, "Test Event") Loop While False : next Function EventCount(fout) strData = objFSO.OpenTextFile(strDirectory & strFile,ForReading).ReadAll arrLines = Split(strData,vbCrLf) EventCount = UBound(arrLines) End Function this is the not working part and I don't know what to do anymore so can you please take a look at it tank you very much. btw: this code can be very usefull for a network administrator

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  • Not sure what happens to my apps objects when using NSURLSession in background - what state is my app in?

    - by Avner Barr
    More of a general question - I don't understand the workings of NSURLSession when using it in "background session mode". I will supply some simple contrived example code. I have a database which holds objects - such that portions of this data can be uploaded to a remote server. It is important to know which data/objects were uploaded in order to accurately display information to the user. It is also important to be able to upload to the server in a background task because the app can be killed at any point. for instance a simple profile picture object: @interface ProfilePicture : NSObject @property int userId; @property UIImage *profilePicture; @property BOOL successfullyUploaded; // we want to know if the image was uploaded to out server - this could also be a property that is queryable but lets assume this is attached to this object @end Now Lets say I want to upload the profile picture to a remote server - I could do something like: @implementation ProfilePictureUploader -(void)uploadProfilePicture:(ProfilePicture *)profilePicture completion:(void(^)(BOOL successInUploading))completion { NSUrlSession *uploadImageSession = ..... // code to setup uploading the image - and calling the completion handler; [uploadImageSession resume]; } @end Now somewhere else in my code I want to upload the profile picture - and if it was successful update the UI and the database that this action happened: ProfilePicture *aNewProfilePicture = ...; aNewProfilePicture.profilePicture = aImage; aNewProfilePicture.userId = 123; aNewProfilePicture.successfullyUploaded = NO; // write the change to disk [MyDatabase write:aNewProfilePicture]; // upload the image to the server ProfilePictureUploader *uploader = [ProfilePictureUploader ....]; [uploader uploadProfilePicture:aNewProfilePicture completion:^(BOOL successInUploading) { if (successInUploading) { // persist the change to my db. aNewProfilePicture.successfullyUploaded = YES; [Mydase update:aNewProfilePicture]; // persist the change } }]; Now obviously if my app is running then this "ProfilePicture" object is successfully uploaded and all is well - the database object has its own internal workings with data structures/caches and what not. All callbacks that may exist are maintained and the app state is straightforward. But I'm not clear what happens if the app "dies" at some point during the upload. It seems that any callbacks/notifications are dead. According to the API documentation- the uploading is handled by a separate process. Therefor the upload will continue and my app will be awakened at some point in the future to handle completion. But the object "aNewProfilePicture" is non existant at that point and all callbacks/objects are gone. I don't understand what context exists at this point. How am I supposed to ensure consistency in my DB and UI (For instance update the "successfullyUploaded" property for that user)? Do I need to re-work everything touching the DB or UI to correspond with the new API and work in a context free environment?

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  • how to invoke an activity of a library project from an android apps

    - by Austin
    I have an open source android code that I need to use in my android apps. It has all the source code as well as resource files, manifest files and class path. It can be compiled as a separate android apps. I have constraints for using the open source. 1. I can't change a single line of code. 2. I can't use it as a separate apps. These constraints are non negotiable. What I have done is I have compiled the open source as class library(in Eclipse: Project Properties-Android- Tick check box Is Library). This has resulted in generation of .class files(in bin) for the java files and resource files. This open source has an android activity that i want to open from my application. So I have linked the directory of these sets of class files in the source section of my java build path( in .classpath). I have declared the activity in my manifest file with proper action intent filters. Now when I am trying to call activity from my code, its not working. Cleaning and rebuilding doesn't help. However, if I build the open source project and my apps in the same workspace of eclipse and link the open source in my apps in exact same manner it works fine. I am not able to identify the difference. All settings seems to be same(all files are identical in both the cases). But only in the second case it works. I have tried it as jar file also. I have build the open source as project library and exported it into a jar file(excluding manifest file). But in that case I am getting the following error UNEXPECTED TOP-LEVEL EXCEPTION: java.lang.IllegalArgumentException: already added: .... Conversion to Dalvik format failed with error 1 This I guess is coming because the android library(2.2) has been included twice in my apps( one for building my apps & another for building the open source). I dont know how to avoid this. Cleaning the project doesn't help. What i require is to use the open source and invoking it's activities in my apps without violating the constraints. If i can use the open source as bunch of .class files then great, or else any other way will do fine. Please look into it and let me know. Thanks

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  • Is there a way for a user to disable an AlertDialog completely?

    - by NewGuyChris
    In the app I'm making, I have an "if" statement where if two strings are saved to a certain string, an AlertDialog pops up. These strings will stay the same for some users, thus having this AlertDialog constantly pop up whenever they launch the activity where the ALertDialog is set to appear. Code (I have no setNegativeButton as of yet): private void SetWarning() { AlertDialog.Builder alert = new AlertDialog.Builder(this); alert.setTitle("Warning!"); alert.setMessage(R.string.Warning); alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int whichButton) { //No action needed; just close the AlertDialog. } }); alert.show(); } Here is a segment of my code that makes this AlertDialog appear: SharedPreferences sharedPreferences = getSharedPreferences("MY_PREF", MODE_PRIVATE); String s = sharedPreferences2.getString("MEM1", ""); String s2 = sharedPreferences2.getString("MEM2", ""); if(s.equals("String1") && s2.equals("String2")) SetWarning(); Is there a way to make an "alert.setNegativeButton" method where if the user clicks it, the AlertDialog will NEVER appear again? I'm thinking of maybe somehow implementing another SavedPreferences somehow so it saves the users selection and will then prevent the AlertDialog from ever appearing again. So far, to no luck. I've searched to find nothing, other than people asking how to disable buttons in an AlertDialog. Thank you! New updated code: alert.setNegativeButton("Cancel", new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int whichButton) { //set sharedpreferences boolean called DONTSHOWAGAIN to true; SharedPreferences sharedPreferences2 = getSharedPreferences("MY_PREF", MODE_PRIVATE); Boolean dontShowAgain = sharedPreferences2.getBoolean("dontShowAgain ", false); SharedPreferences.Editor ed = sharedPreferences2.edit(); ed.putBoolean("dontShowAgain", true); ed.commit(); } }); alert.show(); } private void StringWarning() { SharedPreferences sharedPreferences2 = getSharedPreferences("MY_PREF", MODE_PRIVATE); String s = sharedPreferences2.getString("MEM1", ""); String s2 = sharedPreferences2.getString("MEM2", ""); if(s.equals("String1") && s2.equals("String2")){ if(!dontShowAgain){ SetWarningExamConflict(); } }

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  • remote_form_for in index.html.erb file not working w/ AJAX...Ruby on Rails...

    - by bgadoci
    Just curious if I am overlooking something simple here. I have deployed the remote_form_for in the show.html.erb code before to render comments on a post (project in this case) without a problem. I have moved this code to the index view and seems to degrade to the normal form_for action (page refresh). I am not getting any javascript errors so not sure what is wrong here. Here is my code: index.html.erb <% remote_form_for [project, Comment.new] do |f| %> <p> <%= f.label :body, "New Comment" %><br/> <%= f.text_area (:body, :class => "textarea") %> </p> <p> <%= f.label :name, "Name" %> (Required)<br/> <%= f.text_field (:name, :class => "textfield") %> </p> <p> <%= f.label :email, "Email" %> (Required but will not be displayed)<br/> <%= f.text_field (:email, :class => "textfield") %> </p> <p><%= f.submit "Add Comment" %></p> <% end %> CommentsController#create def create @project = Project.find(params[:project_id]) @comment = @project.comments.create!(params[:comment]) respond_to do |format| format.html { redirect_to projects_path } format.js end end /views/comments/create.js.rjs page.insert_html :bottom, :commentwrapper, :partial => @comment page[@comment].visual_effect :highlight page[:new_comment].reset page.replace_html :notice, flash[:notice] flash.discard /views/comments/_comment.html.erb <% div_for comment do %> <div id="commentwrapper"> <% if admin? %> <%=link_to_remote "X", :url => [@project, comment], :method => :delete %> <% end %> <%= h(comment.body) %><br/><br/> Posted <%= time_ago_in_words(comment.created_at) %> ago by <%= h(comment.name) %> <% if admin? %> | <%= h(comment.email) %> <% end %></div> <% end %>

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