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  • how to get a text box with the value selected in select box

    - by udaya
    hi This is my view page here i have a select box <tr> <td>Chidren</td> <td>:</td> <td><select style="font-family: verdana; min-width: 52px;" id="ddlChildren" name="ddlChildren" class="required" onChange="return Check_Adult('dd1Age')" > <option value="">children</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option></select> </td> </tr> with the value from select box if the value is one then i need to create a text box if the value is two then i need to create two ,,, and respectively In my Check_Adult javascript function ,,I did this function Check_Adult() { var Child= document.getElementById('ddlChildren').value; if( Child==1) { <? echo '<input type="text" name="child" id="child">' ?> } } But the text box is not created how to create it?

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  • How to output multiple rows from an SQL query using the mysqli object

    - by Jonathan
    Assuming that the mysqli object is already instantiatied (and connected) with the global variable $mysql, here is the code I am trying to work with. class Listing { private $mysql; function getListingInfo($l_id = "", $category = "", $subcategory = "", $username = "", $status = "active") { $condition = "`status` = '$status'"; if (!empty($l_id)) $condition .= "AND `L_ID` = '$l_id'"; if (!empty($category)) $condition .= "AND `category` = '$category'"; if (!empty($subcategory)) $condition .= "AND `subcategory` = '$subcategory'"; if (!empty($username)) $condition .= "AND `username` = '$username'"; $result = $this->mysql->query("SELECT * FROM listing WHERE $condition") or die('Error fetching values'); $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; return $info; } } there are several hundred listings in the db and when I call $result-fetch_array() it places in an array the first row in the db. however when I try to call the object, I can't seem to access more than the first row. for instance: $listing_row = new Listing; while ($listing = $listing_row-getListingInfo()) { echo $listing[0]; } this outputs an infinite loop of the same row in the db. Why does it not advance to the next row? if I move the code: $this->listing = $result->fetch_array() or die('could not create object'); foreach ($this->listing as $key => $value) : $info[$key] = stripslashes(html_entity_decode($value)); endforeach; if I move this outside the class, it works exactly as expected outputting a row at a time while looping through the while statement. Is there a way to write this so that I can keep the fetch_array() call in the class and still loop through the records?

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  • Custom stream wrappers, what could they be useful for in web applications?

    - by michael
    I suppose the concept is language agnostic, but I don't know what it's called in other languages. In PHP they're Stream Wrappers. In short, a wrapper class that allows manipulation of a streamable resource (resource that can be read to/written to/seek into, such as a file, a db, an url). For example, in a template engine (a view), upon including a template file such as: include "view.wrapper://path/to/my/template/file.phtml"; my custom wrapper, declared elsewhere and associated with "view.wrapper", would first intercepts the file to replace such things as short tags (<?=) with a more verbose counterpart (<?php echo). This allows developers to use short tags in views, even if the server isn't set to allow it. It can also be applied to the preprocessing of views pseudo syntax such as {@myVar} (e.g. replacing it with $this->myVar). This is only one application of custom stream wrappers, but the feature seems powerful enough to make me think that there are others that could make life a lot simpler for developers. What have you built, or thought about building, custom stream wrappers for? where have you seen some interesting implementations? I'm particularly interested in their applications in web development.

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  • Memory fragmentation @ boost::asio ?

    - by Poni
    I'm pretty much stuck with a question I never got an answer for, a question which addresses an extremely important issue; memory fragmentation at boost::asio. Found nothing at the documentation nor here at SO. The functions at boost::asio, for example async_write() & async_read_some() always allocate something. (in my case it's 144 & 96 bytes respectively, in VC9 Debug build). How do I know about it? I connect a client to the "echo server" example provided with this library. I put a breakpoint at "new.cpp" at the code of "operator new(size_t size)". Then I send "123". Breakpoint is hit! Now using the stack trace I can clearly see that the root to the "new" call is coming from the async_write() & async_read_some() calls I make in the function handlers. So memory fragmentation will come sooner or later, thus I can't use ASIO, and I wish I could! Any idea? Any helpful code example?

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  • jQuery slider to load content one at a time

    - by Barrett
    I have a slider that load all of my content at once. Into a div. Like so: external page.php $get_users = mysql_query("SELECT * FROM user WHERE id!='$user_id'"); while ($rows = mysql_fetch_assoc($get_users)) { $id = $rows['id']; $firstname = $rows['firstname']; $display_info .= ' <div class="f_outer" id="' . $id . '"> <div class="f_name likeu">' . $firstname . '</div> </div>'; } echo $display_info; I call this page from my find.php page using bxslider Here is my find.php page below. <script type="text/javascript"> $(function() { var slider = $("#slider1").bxSlider(); $("#slider-like").live('click', function() { slider.goToNextSlide(); return false; }); }); </script> <div id="slider-like>Yes</div> <div id="slider1"> <?PHP include ("external.php"); ?> </div> So what I get is all of my .f_outer div on the find.php page. I have hundreds of user and they will all be loaded at once. I would like to only load one slide at a time. So when I click #slider-like it load one of my dive from my external page.

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  • PHP Multi-Domain Sessions; ini_set Not Working?

    - by SumWon
    Hello, I'm trying to set it up so if you log in to my website the session carries over to all sub-domains of my website. For example, if you go to domain.com and log in, then go to sub.domain.com, you'll already be logged in at sub.domain.com. To my understanding, you would want to use ini_set('session.cookie_domain','.domain.com') and then session_start(), then set your session variables, but this isn't working. Example of what I'm doing: Code for domain.com: <?php ini_set('session.cookie_domain','.domain.com'); session_start(); $_SESSION['variable'] = 1; ?> Code for sub.domain.com: <?php session_start(); echo $_SESSION['variable']; ?> But $_SESSION['variable'] isn't set. I've also tried using ini_set() in the sub.domain.com code, but it made no difference. I've verified that setting session.cookie_domain is working by using ini_get(). What am I doing wrong? Thanks!

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  • PHP include taking too long

    - by wxiiir
    I have a php file with around 100mb which is full of arrays (only arrays). I've made a script that includes this file (for processing), first it exhausted the default Xampp 128mb memory limit, i've raised it to 1024mb but it just takes forever and doesn't do anything. I'm sure the problem is being created by the sheer size of the file because i've tried removing all lines of code and just leaving the include and an echo for me to know when it finishes executing, and it does the same thing (which is taking forever), i've also tried to run the 100mb file in separate and same thing again. A 10mb file is taking forever as well but a similar 1mb file is almost instantly read and executed so the problem must be more than just the file size. I was avoiding using c++ for a simple project as this and would rather not to as php is easier for me and the task that will be executed doesn't need to benefit from the added speed that it would have if it had been done in c++ but if i have no luck in solving this problem i guess i'll have to. EDIT Reasons for not using a database: 1-Whoever made it didn't used a database and it will be pretty hard to store this in an organized database if i'm not able to do something with it first, like just reading it, copying parts from it or putting in memory or something. 2-I don't have experience working with databases as pretty much all stuff i've ever done in php didn't needed large amounts of stored data, 50kb at best, if i was thinking about a big project or huge chunks of data as this one, i definitely would, but i didn't made this mess to start with and now i have to undo it. 3-The logic for having to store a small portion of data like 10mb in hard drive when now every computer has pretty much enough ram to fit the whole OS in it is pretty much incomprehensible unless someone gives a good explanation about it, if i had to access a lot of said files simultaneously i would understand but like i said, this is a simple project, this is the only file that will be accessed at a given time this isn't even to make some kind of website, it's to run a few times and be done with it.

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  • PHP login, getting wrong count value from query / fetch array

    - by Chris
    Hello, *EDIT*Thanks to the comments below it has been figured out that the problem lies with the md5, without everything works as it should. But how do i implent the md5 then? I am having some troubles with the following code below to login. The database and register system are already working. The problem lies that it does not find any result at all in the query. IF the count is 0 it should redirect the user to a secured page. But this only works if i write count = 0, but this should be 0 , only if the user name and password is found he should be directed to the secure (startpage) of the site after login. For example root (username) root (password) already exists but i cannot seem to properly login with it. <?php session_start(); if (!empty($_POST["send"])) { $username = ($_POST["username"]); $password = (md5($_POST["password"])); $count = 0; $con = mysql_connect("localhost" , "root", ""); mysql_select_db("testdb", $con); $result = mysql_query("SELECT name, password FROM user WHERE name = '".$username."' AND password = '".$password."' ") or die("Error select statement"); $count = mysql_num_rows($result); if($count > 0) // always goes the to else, only works with >=0 but then the data is not found in the database, hence incorrect { $row = mysql_fetch_array($result); $_SESSION["username"] = $row["name"]; header("Location: StartPage.php"); } else { echo "Wrong login data, please try again"; } mysql_close($con); } ?>

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  • Logging in with sha1() encryption.

    - by Samir Ghobril
    Hey guys, I added this to my sign up code : $password=mysql_real_escape_string(sha1($_POST['password'])); and now it inserts the password into the database while its encrypted. But signing in doesn't seem to work anymore. Here is the login code. function checklogin($username, $password){ global $mysqli; $password=sha1($password); $result = $mysqli->prepare("SELECT * FROM users WHERE username = ? and password=?"); $result->bind_param("ss", $username, $password); $result->execute(); if($result != false){ $dbArray=$result->fetch(); if(!$dbArray){ echo '<p class="statusmsg">The username or password you entered is incorrect, or you haven\'t yet activated your account. Please try again.</p><br/><input class="submitButton" type="button" value="Retry" onClick="location.href='."'login.php'\">"; return; } $_SESSION['username']=$username; if(isset($_POST['remember'])){ setcookie("jmuser",$username,time()+60*60*24*356); setcookie("jmpass",$password ,time()+60*60*24*356); } redirect(); }

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  • Zend_Dojo_Form not rendering in layout

    - by Grant Collins
    Hi, I have a quick question about adding Zend_Dojo_Form into Zend_layouts. I have a Zend_Dojo_Form that I want to display in the layout that is used for a particular controller. I can add the form to the layout without any issue however the dojo elements fail to render, as they would do if I added the form to a standard view. Is there any reason why this would be the case? Do I need to do something to the layout so that it will enable the components for this embedded form in the layout. Any other dojo enabled forms that are added in the view using this layout work fine. My form is created in the usual way: class QuickAddJobForm extends Zend_Dojo_Form{ public function init(){ $this->setName('quickaddjobfrm') ->setMethod('post') ->setAction('/addjob/start/); /*We now create the elements*/ $jobTitle = new Zend_Dojo_Form_Element_TextBox('jobtitle', array( 'trim' => true ) ); $jobTitle->setAttrib('style', 'width:200px;') ->addFilter('StripTags') ->removeDecorator('DtDdWrapper') ->removeDecorator('HtmlTag') ->removeDecorator('Label'); .... $this->addElements(array($jobTitle, ....)); In the controller I declare the layout and the form in the init function: public function init(){ $this->_helper->layout->setLayout('add-layout'); $form = new QuickAddJobForm(); $form->setDecorators(array(array('ViewScript', array('viewScript' => 'quickAddJobFormDecorator.phtml')))); $this->_helper->layout()->quickaddjob = $form; In my layout Where I want the form I have: echo $this->layout()->quickaddjob; Why would adding this form in the layout fail to render/add the Dojo elements? All that is currently being displayed are text boxes, rather than some of the other components such as ComboBoxes/FilteringSelects etc... Thanks in advance.

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  • Trigger alert when database entries are added, not when they are removed

    - by Jeremy
    I have a jQuery script running that makes a periodic AJAX call using the following code. var a = moment(); var dayOfMonth = a.format("MMM Do"); var timeSubmitted = a.format("h:mm a"); var count_cases = -1; var count_claimed = -1; setInterval(function(){ //check if new lead was added to the db $.ajax({ type : "POST", url : "inc/new_lead_alerts_process.php", dataType: 'json', cache: false, success : function(response){ $.getJSON("inc/new_lead_alerts_process.php", function(data) { if (count_cases != -1 && count_cases != data.count) { window.location = "new_lead_alerts.php?id="+data.id; } count_cases = data.count; }); } }); This is the PHP that runs with each call: $count = mysql_fetch_array(mysql_query("SELECT count(*) as count FROM leads")); $client_id = mysql_fetch_array(mysql_query("SELECT id, client_id FROM leads ORDER BY id DESC LIMIT 1")); echo json_encode(array("count" => $count['count'], "id" => $client_id['id'], "client_id" => $client_id['client_id'])); I need to change the code so that the alert only triggers when a new entry is added to the database, not when an existing entry is removed. As it stands, the alert fires on both events. Any help is greatly appreciated.

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  • Eval string for wordpress "in_category" workaround outside the loop

    - by fimbim
    In Wordpress you can´t use "in_category" outside the loop, so I created a function that gives me all the categories my article is in and create a "is_category" if statement out of it. I put my function in my "functions.php": function inCatAlt($catID){ $allCats = get_categories('child_of='.$catID); $childCats = 'is_category('.$catID.') '; foreach($allCats as $childCat){ $childCats.= 'or is_category('.$childCat->cat_ID.') '; }; $allchildCats = trim(trim($childCats, 'or ')); return $allchildCats; } and call this in my sidebar, single and so on: echo inCatAlt(13); which gives me this as a string back: "is_category(13) or is_category(16) or is_category(15)" This is exactly what I needed, but now I want to evaluate the string to use it in a if function like this: if(eval(inCatAlt(13))){ do something } But it doesn´t work. Do I evaluate it wrong or what is the problem? If I copy paste the output into the if function it works fine… Thanks in advanced guys. Is my first time asking something here. I´m curious :)

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  • Query not being executed

    - by user2385241
    I'm trying to create a script that allows me to upload an image, grab the details sent through inputs (a description and chosen project number) and insert this information into a table. I currently have this function: public function NewEntry() { $connect = new dbconnect; $_SESSION['rnd'] = substr(number_format(time() * rand(),0,'',''),0,15); $allowedExts = array("gif", "jpeg", "jpg", "png"); $size = $_FILES["file"]["size"]; $path = $_FILES["file"]["name"]; $extension = pathinfo($path, PATHINFO_EXTENSION); $pr = $_POST['project']; $cl = $_POST['changelog']; $file = $_SESSION['rnd'] . "." . $extension; if (in_array($extension, $allowedExts) && $size < 200000000) { if ($_FILES["file"]["error"] == 0) { if (!file_exists("../uploads/" . $_SESSION['rnd'])) { move_uploaded_file($_FILES["file"]["tmp_name"], "../uploads/" . $_SESSION['rnd'] . "." . $extension); } } } else { echo "File validation failed."; } $row = $connect->queryExecute("INSERT INTO entries(project,file,changelog)VALUES($pr,$file,$cl)"); header('location:http://www.example.com/admin'); } When the form is posted the function runs, the image uploads but the query isn't executed. The dbconnect class isn't at fault as it's untampered and has been used in past projects. The error logs don't give any output and no MySQL errors show. Any ideas?

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  • PHP Error Form - Leave Contents of Form on Redirect

    - by user1371500
    I have a simple login form in which if an error occurs such as wrong password, I need it to be able to remember the username which was entered. Would I Go about doing this PHP or Javascript as I am not allowed to use JQuery. My current PHP - (Not Including the HTML Form) <?php //MySQl Connection mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("clubresults") or die(mysql_error()); //Initiates New Session - Cookie session_start(); // Start a new session // Get the data passed from the form $username = $_POST['username']; $password = md5($_POST['pass']); // Do some basic sanitizing $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); //Performs SQL Query to retrieve Login Details from DB $sql = "select * from admin_passwords where username = '$username' and password = '$password'"; $result = mysql_query($sql) or die ( mysql_error() ); //Assigns a Variable Count to 0 $count = 0; //Exectues a loop to increment on Successful Login while ($line = mysql_fetch_assoc($result)) { $count++; } //If count is equal to 1 Redirect user to the Members Page and Set Cookie if ($count == 1) { $_SESSION['loggedIn'] = "true"; header("Location: members.php"); // This is wherever you want to redirect the user to } else { //Else Echo that login was a failure. die('Login Failed. <a href=login.php>Click Here to Try Again</a>'); } ?> Any help would be appreciated. Cheers

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  • Display attribute from XML using PHP

    - by user560411
    Hello. I need to display the id attribute of CD from the following XML file. I display correctly everything except the id. Any help would be appreciate. display code <?php $doc = new DOMDocument(); $doc->load( 'insert.xml' ); $CATEGORIES = $doc->getElementsByTagName( "CD" ); foreach( $CATEGORIES as $CD ) { $TITLES = $CD->getElementsByTagName( "TITLE" ); $TITLE = $TITLES->item(0)->nodeValue; $BANDS= $CD->getElementsByTagName( "BAND" ); $BAND= $BANDS->item(0)->nodeValue; $YEARS = $CD->getElementsByTagName( "YEAR" ); $YEAR = $YEARS->item(0)->nodeValue; echo "<b>$TITLE - $BAND - $YEAR\n</b><br>"; } ?> XML <?xml version="1.0" encoding="utf-8"?> <MY_CD> <CATEGORIES> <CD id="3231"> <TITLE>NEVER MIND THE BOLLOCKS</TITLE> <BAND>SEX PISTOLS</BAND> <YEAR>1977</YEAR> </CD> <CD id="2453"> <TITLE>NEVERMIND</TITLE> <BAND>NIRVANA</BAND> <YEAR>1991</YEAR> </CD> </CATEGORIES> </MY_CD>

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  • Get Form Input via Ajax

    - by user3651491
    I have a jqgrid plugin which I call via Ajax. I have index.php and a getGridData.php. How will I pass form input in getGridData.php via ajax and use it in getGridData.php? I tried serialize but I can't pass or access it on getGridData.php. I need it as parameters for mysql. Here's my code. <script language="javascript" type="text/javascript"> function jgGrid() { $(document).ready(function () { $("#grid").jqGrid({ url: "inc/Controller/getGridData.php"+$("#thisForm").serialize(), data : formData, datatype: "json", mtype: "POST", colNames: ["SiteID", "TerminalID", "TransactionType", "Amount", "ServiceStatus"], colModel: [ { name: "SiteID"}, { name: "TerminalID"}, { name: "TransactionType"}, { name: "Amount"}, { name: "ServiceStatus"}, ], pager: "#pager", rowNum: 10, rowList: [10,20], sortname: "SiteID", sortorder: "asc", height: 'auto', viewrecords: true, gridview: true, caption: "" }); }); } </script> getGridData.php include('../Model/Queries.php'); $cardnumber = $_POST['cardnumber']; $transact_type = $_POST['transact_type']; $fromdate = $_POST['fromdate']; $todate = $_POST['todate']; $loyalty = new Queries(); $get_mid = $loyalty->loyaltyConn($cardnumber); $somedata = json_encode($loyalty->nposConn($get_mid, $transact_type, $fromdate, $todate)); echo $somedata;

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  • header location won't work in php

    - by Jayden Kelly
    I am making a login page for my website but the header location won't work. here is the code of login.php: <?php include ( './includes/header.php' ); if (isset($_POST['submit'])) { $username = $_POST['username']; $password = $_POST['password']; $check_username = mysql_query("SELECT username FROM users WHERE username='$username'"); $numrows = mysql_num_rows($check_username); if ($numrows != 1) { echo 'That User doesn\'t exist.'; } else { $check_password = mysql_query("SELECT password FROM users WHERE password='$password' && username='$username'"); while ($row = mysql_fetch_assoc($check_password)) { $password_db = $row['password']; if ($password_db == $password) { $_SESSION['username'] = $username; header("Location: members.php"); } } } } ?> <h2>Login to Your Account</h2> <form action='login.php' method='POST'> <input type='text' name='username' value='Username ...' onclick='value=""'/><p /> <input type='password' name='password' value='Password ...' onclick='value=""'/><p /> <input type='submit' name='submit' value='Login to my Account' /> </form> I would really appreciate it if someone could help me, thanks. P.S. If you need the php part of the header file it is here: <?php session_start(); include ( './includes/functions.php' ); include ( './includes/connect_to_mysql.php' ); ?>

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  • Error in inserting data into database

    - by Matthew
    When I try to run this code it will not insert the data into the database? <?php class Database { private $dsn; function __construct($dbname, $host, $user, $password, $enckey) { $this->dsn = "mysql:dbname=" . $dbname . ';host=' . $host; $this->user = $user; $this->password = $password; } private function createDSN() { return $this->dsn; } public function createConnection() { try { $dbh = new PDO(self::createDSN(), $this->user, $this->password); } catch (PDOException $e) { echo 'Connection failed: ' . $e->getMessage(); } return $dbh; } } $db = new Database('mytest', 'localhost', 'root', 'hashedpassword', null); $dbh = $db->createConnection(); $sql = $dbh->prepare("INSERT INTO contacts (firstname, lastname) VALUES (?,?)"); $sql->execute(array("abc", "xyz")); ?>

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  • PHP dynamic Page-level DocBlocks

    - by Obmerk Kronen
    I was wondering if there is a way to interact with the Page-level DocBlocks. My question is more specifically about wordpress plugin development, but this question has arised also in a non-wordpress environments. The reason is mainly the possibility to easily change VERSIONS and names throughout a large project with maybe a constant definition - but that will reflect also in the docblock.. The following example Docblock is from a wordpress plugin I write - /* Plugin Name: o99 Auxilary Functions v0.4.7 Plugin URI: http://www.myurl.com Description: some simple description that nobody reads. Version: 0.4.7 Author: my cool name Author URI: http://www.ok-alsouri.com */ Is there a way to transform it into : $ver = '0.4.7'; $uri = 'http://www.myurl.com'; $desc = 'some simple description that nobody reads.'; $mcn = 'my cool name'; etc.. etc.. /* Plugin Name: o99 Auxilary Functions ($ver) Plugin URI: ($uri) Description: ($desc) Version: ($ver) Author: ($mcn) Author URI: ($$uri) */ obviously for echo to work I would need to break the docblock itself, and I can not WRITE the docblock directly into it´s own file . In shorts : can I "generate" a docblock with php itself somehow (I would think that the answer is - "no" for the page itself.. But maybe I am wrong and someone has some neat hack :-) ) Is that even possible ?

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  • entering datas of php page to database

    - by Rahima farzana.S
    I have a php form with two text boxes and i want to enter the text box values into the database. I have created the table (with two columns namely webmeasurementsuite id and webmeasurements id) I used the following syntax for creating table: CREATE TABLE `radio` ( `webmeasurementsuite id` INT NOT NULL, `webmeasurements id` INT NOT NULL ); Utilising the tutorial in the following link, I wrote the php coding but unfortunately the datas are not getting entered into the database. I am getting an error in the insert sql syntax. I checked it but i am not able to trace out the error.Can anyone correct me? I got the coding from http://www.webune.com/forums/php-how-to-enter-data-into-database-with-php-scripts.html $sql = "INSERT INTO $db_table(webmeasurementsuite id,webmeasurements id) values ('".mysql_real_escape_string(stripslashes($_REQUEST['webmeasurementsuite id']))."','".mysql_real_escape_string(stripslashes($_REQUEST['webmeasurements id']))."')"; echo$sql; My error is as follows: INSERT INTO radio(webmeasurementsuite id,webmeasurements id) values ('','')ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'id,webmeasurements id) values ('','')' at line 1

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  • Pass Element value to $ajax->link in cakephp

    - by TwoThumbs
    I need to pass the value of an element to an $ajax-link without using a form/submit structure. (because I have a dynamically set number of clickable links through which I am triggering the action) I used to do this in Ruby using the Prototype javascript function $F like this: <%= link_to_remote "#{item.to_s}", :url => { :action => :add_mpc }, :with => "'category=' + $F('mpc_category')" -%> But this does not seem to work in Cakephp: <?php echo $ajax->link(substr($vehicle['vehicles']['year'], -2), array('action' => 'add_mpc', 'category' => '$F("mpc_category")'), array('update' => 'results', 'position' => 'top')); ?> PHP sees $F as a variable instead of a call to javascript. I'm not too familiar with Javascript, but is there another way to pass the value of the 'mpc_category' input element to the controller through this link? I have been looking for a couple days and can't find anyone dealing with this specific issue. Thanks for any assistance. Edit: fixed syntax in php statement.

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  • Acessing elements of this xml

    - by csU
    <wsdl:definitions targetNamespace="http://www.webserviceX.NET/"> <wsdl:types> <s:schema elementFormDefault="qualified" targetNamespace="http://www.webserviceX.NET/"> <s:element name="ConversionRate"> <s:complexType> <s:sequence> <s:element minOccurs="1" maxOccurs="1" name="FromCurrency" type="tns:Currency"/> <s:element minOccurs="1" maxOccurs="1" name="ToCurrency" type="tns:Currency"/> </s:sequence> </s:complexType> </s:element> <s:simpleType name="Currency"> <s:restriction base="s:string"> <s:enumeration value="AFA"/> <s:enumeration value="ALL"/> <s:enumeration value="DZD"/> <s:enumeration value="ARS"/> i am trying to get at all of the elements in enumeration but cant seem to get it right. This is homework so please no full solutions, just guidance if possible. $feed = simplexml_load_file('http://www.webservicex.net/CurrencyConvertor.asmx?WSDL'); foreach($feed->simpleType as $val){ $ns s = $val->children('http://www.webserviceX.NET/'); echo $ns_s -> enumeration; } what am i doing wrong? thanks

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  • Detect if Download is Complete

    - by user604138
    I have a very simple and standard PHP force download script. How do I check if/when the download has completed in order to notify the user on the clientside? I don't even need to show the progress in real time, I am only interested in the very specific event: "when the download completes". Based on my research, it seems like it would have to be determined from the serverside as there is noondownloadready event and I don't think it is possible to intercept browser events. So it seems that my best bet would be to compare bytes sent to total bytes with some sort of clientside/severside interaction. How would I go about checking the bytes sent from the server for a PHP forced download? is there some sort of global PHP variable that store these data that I can ping with AJAX? <?php header("Content-Type: video/x-msvideo"); header("Content-Disposition: attachment; filename=\"".basename($realpath)."\";"); ... $chunksize = 1 * (1024 * 1024); // how many bytes per chunk if ($size > $chunksize) { $handle = fopen($realpath, 'rb'); $buffer = ''; while (!feof($handle)) { $buffer = fread($handle, $chunksize); echo $buffer; ob_flush(); flush(); } fclose($handle); } else { readfile($realpath); } exit(); ?> The reason I need this: For the project I am working on, it is required that after the download starts, the page redirects to (or displays) a "please wait while the download completes" page. Then, once it is complete, it should redirect to (or display) a "Your download is complete, thank you" page. I am open to other ideas that would achieve the same result.

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  • Passing Javascript value to PHP Variable using ajax

    - by shels
    I am trying to use a Flash detection script to assess whether Flash Plugin is enabled in the user browser so that a different page loads. The Flash detection script is as follows, using jquery 1.8.2 and jquery.jqplugin 1.0.2 <script type="text/javascript" src="jquery-1.8.2.min.js"></script> <script type="text/javascript" src="jquery.jqplugin.1.0.2.min.js"></script> <script type="text/javascript"> $(document).ready(function() { $("#withflash").hide(); $("#noflash").hide(); if ($.browser.flash == true) $("#withflash").show (); else $("#noflash").show (); }); </script> <div id="withflash">Flash Supported</div> <div id="noflash">Flash Not Supported</div> I get the display that "Flash Supported" if Flash Plugin is present.. I need to capture the value whether flash plugin value is true in a php variable $hasFlashSupport as below: <?php echo " $hasFlashSupport"; exit; ?> I am aware that PHP is server based and Javascript is client based.. Hence Ajax would be a nice option to capture the javascript variable to my php variable. I am totally ignorant about Ajax syntax and how to achieve it. Request the experts here to help me out with the code on how this can be achieved... Thanking all of you in advance..

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  • [Zend Framework] Forms and success output

    - by rasouza
    Well, It's a beginer's question but I really don't know what is the best way. I have a basic CRUD (Create, Retrieve, Update and Delete) in my project and I'd like to output some message if succeded or not in a div inside the same page. So, basically, I have a form which action is setted to the same page and I have a div #statusDiv below this same form which I'd like to output something like Register included with success. What is the best way for doing this? Set a flag in the controller $this->view->flagStatus = 'message' then call it in the view? Just to make it more clear. It's my code: //IndexController.php indexAction() ... //Check if there's submitted data if ($this->getRequest()->isPost()) { ... $registries->insert($data); $this->view->flagStatus = 'message'; $this->_redirect('/'); } Then my view: .... <?php if ($this->flagStatus) { ?> <div id="divStatus" class="success span-5" style="display: none;"> <?php echo $this->flagStatus; ?> </div> <?php } ?> ....

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