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  • PHP JQuery: Where to specify uploadify destination folder

    - by Eamonn
    I have an uploadify script running, basic setup. Works fine when I hard code the destination folder for the images into uploadify.php - now I want to make that folder dynamic. How do I do this? I have a PHP variable $uploadify_path which contains the path to the folder I want. I have switched out my hard coded $targetPath = path/to/directory for $targetPath = $uploadify_path in both uploadify.php and check_exists.php, but it does not work. The file upload animation runs, says it is complete, yet the directory remains empty. The file is not hiding out somewhere else either. I see there is an option in the Javascript to specify a folder. I tried this also, but to no avail. If anyone could educate me on how to pass this variable destination to uploadify, I'd be very grateful. I include my current code for checking (basically default): The Javascript <script type="text/javascript"> $(function() { $('#file_upload').uploadify({ 'swf' : 'uploadify/uploadify.swf', 'uploader' : 'uploadify/uploadify.php', // Put your options here }); }); </script> uploadify.php $targetPath = $_SERVER['DOCUMENT_ROOT'] . $uploadify_path; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetFile = $targetPath . $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('jpg','jpeg','gif','png'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } }

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  • Executing system command in php, differs in using broswer and in using command line

    - by Amit
    Hi, I have to execute a Linux "more" command in php from a particular offset, format the result and display the result in Browser. My Code for the above is : <html> <head> <META HTTP-EQUIV=REFRESH CONTENT=10> <META HTTP-EQUIV=PRAGMA CONTENT=NO-CACHE> <title>Runtime Access log</title> </head> <body> <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "<br/>No of lines are : $accessDisplay[0] <br/>"; ?> The output at the browser is :: No of lines are : 3428 (This is wrong) While executing the same command using command line gives a different output. My code snippet for the same is : <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "No of lines are : $accessDisplay[0] \n"; ? The output at the command line is :: No of lines are : 279 (This is correct) While executing the same command directly in command line, gives me output as 279. I am unable to understand why the output of the same command is wrong in the browser. Its actually giving the word count of lines, ignoring the offset parameter. Please help !! Thanks, Amit

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  • jQuery Validate PHP Response

    - by Kurt
    Hello Everybody, my problem is, I want to validate an email adresse with jquery. Not only the syntax but rather if the email is already registrated. There are some tutorials but they are not working! At first the Jquery Code: <script id="demo" type="text/javascript"> $(document).ready(function() { // validate signup form on keyup and submit var validator = $("form#signupform").validate({ rules: { Vorname: { required: true, minlength: 3 }, Nachname:{ required: true, minlength: 4 }, password: { required: true, minlength: 5 }, password_confirm: { required: true, minlength: 5, equalTo: "#password" }, Email: { required: true, email: true, type: "POST", remote: "remotemail.php" }, dateformat: "required", ... </script> And now the PHP Code: <?php include('dbsettings.php'); $conn = mysql_connect($dbhost,$dbuser,$dbpw); mysql_select_db($dbdb,$conn); $auslesen1 = "SELECT Email FROM flo_user"; $auslesen2 = mysql_query($auslesen1,$conn); $registered_email = mysql_fetch_assoc($auslesen2); $requested_email = $_POST['Email']; if( in_array($requested_email, $registered_email) ){ echo "false"; } else{ echo "true"; } ?> I tried return TRUE/ return FALSE as well, but this displays "Email is registrated" all the time. json_encode didn't work as well. Thanks a lot!

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  • Run a command as cron would but from the command line.

    - by BCS
    I have a script that I'm trying to run from cron. When I run it from bash, it work just fine. However when I let cron do it's thing, I get a: myscript.sh: line 122: syntax error: unexpected end of file What I want is a way to run a command as if it was a cron job, but do it in my shell. As a side note: does anyone know what would be differnt under cron? (the script already has a #!/bin/sh line) To answer my own question: I added this to my crontab: * * * * * bcs for ((i=1; i <= 55; i++)) ; do find ~/.crontemp/ -name '*.run' -exec "{}" ";" ; sleep 1; done` and created this script: #!/bin/sh tmp=$(mktemp ~/.crontemp/cron.XXXXX) mknod $tmp.pipe p mv $tmp $tmp.pre echo $* '>' $tmp.pipe '1>&2' >> $tmp.pre echo rm $tmp.run >> $tmp.pre chmod 700 $tmp.pre mv $tmp.pre $tmp.run cat $tmp.pipe rm $tmp.pipe With that, I can run an arbitrary command with a delay of not more than one second. (And yes, I know there are all kinds of security issue involved in that)

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  • posting a variable with jquery and receiving it on other page

    - by Billa
    I want to post a varible "id" to a page. I'm trying following code, but I cant get the id value, it says "undefined". function box(){ var id=$(this).attr("id"); $("#votebox").slideDown("slow"); $("#flash").fadeIn("slow"); $.ajax({ type: "POST", //I want to post the "id" to the rating page. data: "id="+$(this).attr("id"), url: "rating.php", success: function(html){ $("#flash").fadeOut("slow"); $("#content").html(html); } }); } This function is called in following code. In the following code too, the id is posted to the page "votes.php", and it works fine, but in the above code when I'm trying to post the id to the rating.php page, it does not send. $(function(){ $("a.vote_up").click(function(){ the_id = $(this).attr('id'); $("span#votes_count"+the_id).fadeOut("fast"); $.ajax({ type: "POST", data: "action=vote_up&id="+$(this).attr("id"), url: "votes.php", success: function(msg) { $("span#votes_up"+the_id).fadeOut(); $("span#votes_up"+the_id).html(msg); $("span#votes_up"+the_id).fadeIn(); var that = this; box.call(that); } }); }); }); rating.php <? $id = $_POST['id']; echo $id; ?> The html part is: <a href='javascript:;' class='vote_up' id='<?php echo $row['id']; ?>'>Vote Up!</a> I'll appriciate any help.

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  • Variable not implementing correctly from if statement

    - by swiftsly
    I have an if statement that is supposed to set the variable $pc162v to a link specified in the MySQL table if content exists in the $vid column of the row. The problem is, the PHP is detecting that there's a link in the MySQL, but isn't setting the $pc162v variable correctly. Here's the variable declarations: $pc162v = ""; $vid162 = '<embed width="420" height="236" src="'.$pc162v.'" type="application/x-shockwave-flash"></embed>'; Here's the section of the if statement: if (empty($row[7])) { $vid162 = ''; } else { $pc162v = $row[7]; } In my web browsers, the part of the code where the variable $vid162 is used, shows up as the following: <embed width="420" height="236" src="" type="application/x-shockwave-flash"> I have also tried setting $vid162 to: <embed width="420" height="236" src="<?php echo $pc162v; ?>" type="application/x-shockwave-flash"></embed> and that just makes the code in my web browser: <embed width="420" height="236" src="<?php echo $pc162v; ?>" type="application/x-shockwave-flash"> Hope someone has a solution! Thanks in advance.

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  • How to control table width in code ignator?

    - by riad
    Dear Exparts, In codeIgnator frame work the below is my code working properly.But i cannot control the table width. So,when a long value come into the table then table going to extra large width.But i need to wrap the outcomes data.So,How i can fixed the table width? Pls see my code below.. ///controller code/// $config['base_url'] = base_url().'Search_Controller/songSearchPage/'; $config['total_rows'] = $this->db->count_all('tbl_rbt'); $config['per_page'] = '5'; $config['full_tag_open'] = '<p>'; $config['full_tag_close'] = '</p>'; $this->pagination->initialize($config); //load the model and get results $data[]=array(); $data['extraHeadContent'] = '<script type="text/javascript" src="' . base_url() . 'js/song_search.js"></script>'; $data['results'] = $this->search_model->getSongResult($config['per_page'],$this->uri->segment(3)); // load the HTML Table Class $this->table->set_heading('Song Name','Album Name','Artist Name'); // load the view $this->load->view('song_search_page',$data); /////view code///// <div class="song_element_output"> <?php echo $this->table->generate($results); ?> <?php echo $this->pagination->create_links(); ?> </div> Could anybody can help me to control the table??? Thanks Riad

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  • Yii Framework Tutorial - file not found

    - by Heshan Perera
    I am in the process of learning the Yii Framework. I have been following this Tutorial. I have followed all the steps and have by index page loading. In my index.php page I point to two other pages located in protected/views/message The following is the code in my index.php located in the message folder mentioned above. <html> <body> <h1>Welcome</h1> <p> To view our message go to... <?php echo CHtml::link('Here', '/message/show')?> </p> <p> To edit our message go to... <?php echo CHtml::link('Here', '/message/edit')?> </p> </body> </html> "message" is the ID I gave when generating the model and controller through the yii shell application. The problem is, after the above page loads, and I click on any one of the above URLs, it points to "localhost:8080/message/show" and "localhost:8080/message/edit" , whereas the real location of these files is "localhost:8080/test/protected/views/message/..." What could I be doing wrong ?

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  • Load PHP function with jQuery Ajax

    - by brandon14_99
    I have a file which is loaded at the top of my document, which is called Videos.php. Inside that file are several functions, such as getYoutubeVideos. On some pages, I need to call upon that function several times (up to 50), and it of course creates major lag on load times. So I have been trying to figure out how to call that function in, only when it is need (when someone clicks the show videos button). I have very little experience with jQuery's ajax abilities. I would like the ajax call to be made inside of something like this: jQuery('a[rel=VideoPreview1).click(function(){ jQuery ("a[rel=VideoPreview1]").hide(); jQuery ("a[rel=HideVideoPreview1]").show(); jQuery ("#VideoPreview1").show(); //AJAX STUFF HERE preventDefault(); }); Ok I have created this based on the responses, but it is still not working: jQuery Code: jQuery(document).ready(function(){ jQuery("a[rel=VideoPreview5]").click(function(){ jQuery("a[rel=VideoPreview5]").hide(); jQuery("a[rel=HideVideoPreview5]").show(); jQuery.post("/Classes/Video.php", {action: "getYoutubeVideos", artist: "Train", track: "Hey, Soul Sister"}, function(data){ jQuery("#VideoPreview5").html(data); }, 'json'); jQuery("#VideoPreview5").show(); preventDefault(); }); jQuery("a[rel=HideVideoPreview5]").click(function(){ jQuery("a[rel=VideoPreview5]").show(); jQuery("a[rel=HideVideoPreview5]").hide(); jQuery("#VideoPreview5").hide(); preventDefault(); }); }); And the PHP code: $Action = isset($_POST['action']); $Artist = isset($_POST['artist']); $Track = isset($_POST['track']); if($Action == 'getYoutubeVideos') { echo 'where are the videos'; echo json_encode(getYoutubeVideos($Artist.' '.$Track, 1, 5, 'relevance')); }

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  • PHP IF stops IE content from loading

    - by Rsmithy
    scenario I have a PHP if statement to test if a user is logging in for the first time ever. This then displays, a demo in a greybox popup box. Sadly when the box loads in IE, the content of the website doesn't. This means when I user closes the box, they are left the with background. Code - I'm now using PHP include. greybox.php <?php if ($fli == 0) {echo " <script type='text/javascript'> window.onload = function() { GB_showCenter('Your first login', '../video.php'); }; </script> ";} else echo "";?> <!-- GB scrip --> <script type="text/javascript"> var GB_ROOT_DIR = "greybox/"; </script> <script type="text/javascript" src="greybox/AJS.js"></script> <script type="text/javascript" src="greybox/AJS_fx.js"></script> <script type="text/javascript" src="greybox/gb_scripts.js"></script> <link href="greybox/gb_styles.css" rel="stylesheet" type="text/css" /> <!-- GB --> RELEVANT Script on main site <?php $fli = $_SESSION["USER"]["fli"]; ?> <?php include "greybox.php" ?> I would deeply appreicate any help at all please! :)

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  • CodeIgniter: problem using foreach in view

    - by krike
    I have a model and controller who gets some data from my database and returns the following array Array ( [2010] => Array ( [year] => 2010 [months] => Array ( [0] => stdClass Object ( [sales] => 2 [month] => Apr ) [1] => stdClass Object ( [sales] => 1 [month] => Nov ) ) ) [2011] => Array ( [year] => 2011 [months] => Array ( [0] => stdClass Object ( [sales] => 1 [month] => Nov ) ) ) ) It shows exactly what it should show but the key's have different names so I have no idea on how to loop through the years using foreach in my view. Arrays is something I'm not that good at yet :( this is the controller if you need to know: function analytics() { $this->load->model('admin_model'); $analytics = $this->admin_model->Analytics(); foreach ($analytics as $a): $data[$a->year]['year'] = $a->year; $data[$a->year]['months'] = $this->admin_model->AnalyticsMonth($a->year); endforeach; echo"<pre style='text-align:left;'>"; print_r($data); echo"</pre>"; $data['main_content'] = 'analytics'; $this->load->view('template_admin', $data); }//end of function categories()

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  • Draw a comparison between an integer in a specific row and a count

    - by XCoderX
    This is a follow-up question to this one: Check for specific integer in a row WHERE user = $name I want a user to be able to comment on my site for exactly five times a day. After this five times, the user has to wait 24 hours. In order to accomplish that I raise a counter in my MYSQL database, right next to the user. So where the name of the user is, there is where the counter gets raised. When it reaches 5 it should stop counting and reset after 24 hours. In order to check the time I use a timestamp. I check if the timestamp is older than 24 hours. If that is the case, the counter gets reseted (-5) and the user can comment again. In order to do that, I use the following code, however it never stops at five, my guess is that my comparison is wrong somehow: $counter = "SELECT FROM table VALUES CommentCounterReset WHERE Name = '$name'"; if(!isset($_SESSION['ts'])); { $_SESSION['ts'] = time(); } if ($counter >= 5) { if (time() - $_SESSION['ts'] <= 60*60*24){ echo "You already wrote five comments."; } else { $sql = "UPDATE table SET CommentCounterReset = CommentCounterReset-5 WHERE Name = '$name'"; } } else { $sql = "UPDATE table SET CommentCounterReset = CommentCounterReset+1 WHERE Name = '$name'"; echo "Your comment has been added."; }

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  • mysql query using global variables

    - by Carlos
    I am trying run a query to active the users account. I am not sure if I am having problem with the query itself or if there's something else that I dont know about. here is the code: if($_SESSION['lastid']&&$_SESSION['random']) { $check= mysql_query('SELECT * FROM members WHERE id= "$_SESSION[lastid]" AND random = " $_SESSION[random]"'); $checknum = mysql_num_rows($check); //$checknum = mysql_query($check) or die("Error: ". mysql_error(). " with query ". $check); if($checknum != 0) // run query to activate the account { $acti= mysql_query('UPDATE members SET activation = "1" WHERE id= "$_SESSION[lastid]"'); die('Your account has been activated. You may now log in!'); }else{ echo('Invalid id or activation code.') . ' lastid: ' .$_SESSION['lastid'] . ' random: ' .$_SESSION['random'] ; // die ('Invalid id or activation code.'); } }else{ die('Could not either find id or random number!'); } this is the warning I am getting from mysql: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /hermes/bosweb26b/b2501/servername/folder/file.php on line 30 but when I echo the variables out, I get the same values that are stored in the database.... Invalid id or activation code. lastid: 2 and random: 36308075 could someone please give me a hint? thank you.

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  • HTML relative links on various domains

    - by Adam Kiss
    I have quickie: When you code/develop themes, how do you link to various files in your html/css code? Example: We at our firm use mostly <base target="http://whatever"> in our main template and then just <img src="./images/file.png"> in our html, "/category/page" as links and something alike in our css. However, when testing on different machines, we use ip address rather than localhost on main dev station of coder, so all base links don't work (because localhost goes to viewing machine, not coder's, in our network). Same thing happens when updating pages - on dev server, we have to edit base target, so browsing site won't take us to live site - this part is actually rather simple PHP (if ... echo else echo something else), but it still not solve problem of more coding-testing problems. So, my question is, how do YOU solve it? How do you use relative links, which basically don't care for what domain is the page on and don't care for url rewrite? (because ../images/ is different for / and different for /something/somethingElse/page)?

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  • Problem when use compiled .A with simulator mode

    - by Paska
    Hi all, i have a lib .a that run only in device mode. My sdk vers is 4.2 with xcode 3.2.x. In simulator, compile correctle with no warning and no errors, but in run mode (simulator) it crash with this error: Detected an attempt to call a symbol in system libraries that is not present on the iPhone: strtod$UNIX2003 called from function pj_init in image MyAPP. If you are encountering this problem running a simulator binary within gdb, make sure you 'set start-with-shell off' first Program received signal: “SIGABRT”. I try to clean, rebuild and set "set start-with-shell off" from terminal in this way: cd ~ echo '' >> .gdbinit echo 'set start-with-shell 0' >> .gdbinit Restarted all, but nothing. The problem don't wont to resolve! Is there any tag or property that i forgotted to set in the options? In other linker flag is there only "-ObjC". It's very important to solve this issue... any idea please? thanks, A EDIT: It's my lib, compiled in simulator mode! EDIT: It run only with Simulator 4.1. Don't work with iphone 4.0, 4.2 and ipad 3.2 (all simulator).

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  • How to inject php code from database into php script ?

    - by luxquarta
    I want to store php code inside my database and then use it into my script. class A { public function getName() { return "lux"; } } // instantiates a new A $a = new A(); Inside my database there is data like "hello {$a->getName()}, how are you ?" In my php code I load the data into a variable $string $string = load_data_from_db(); echo $string; // echoes hello {$a->getName()}, how are you ? So now $string contains "hello {$a-getName()}, how are you ?" {$a-getName()} still being un-interpretated Question: I can't find how to write the rest of the code so that {$a-getName()} gets interpretated "into hello lux, how are you". Can someone help ? $new_string = ?????? echo $new_string; //echoes hello lux, how are you ? Is there a solution with eval() ? (please no debate about evil eval ;)) Or any other solution ?

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  • Send file FTP over SSL with custom port number

    - by JM4
    I have asked the question before but in a different manner. I am trying taking form data, compiling into a temporary CSV file and trying to send over to a client via FTP over SSL (this is the only route I am interested in hearing solutions for unless there is a workaround to doing this, I cannot make changes). I have tried the following: ftp_connect - nothing happens, the page just times out ftp_ssl_connect - nothing happens, the page just times out curl library - same thing, given URL it also gives error. I am given the following information: FTPS Server IP Address TCP Port (1234) Username Password Data Directory to dump file FTP Mode: Passive very, very basic code (which I believe should initiate a connection at minimum): Code: <?php $ftp_server = "00.000.00.000"; //masked for security $ftp_port = "1234"; // masked but not 990 $ftp_user_name = "username"; $ftp_user_pass = "password"; // set up basic ssl connection $conn_id = ftp_ssl_connect($ftp_server, $ftp_port, "20"); // login with username and password $login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); echo ftp_pwd($conn_id); // / echo "hello"; // close the ssl connection ftp_close($conn_id); ?> When I run this over a SmartFTP client, everything works just fine. I just can't get it to work using PHP (which is a necessity). Has anybody had success doing this in the past? I would be very interested to hear your approach.

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  • Getting the last User ID in Zend Framework

    - by Ryan Murphy
    Using MySQL query browser, I manually made a table called users and input some date in the fields. I set the primary key to id and set it to auto increment. There are 3 rows, the highest id is 3. I then made the following class in the method directory to call upon the data in the table etc. class Application_Model_DbTable_User extends Zend_Db_Table_Abstract { protected $_name = 'user'; public function getLatestUserId() { $id = $this->getAdapter()->lastInsertId(); return $id; } } In the controller I do the following which gets the value generated by the method and lets the view access it: $usersDbModel = new Application_Model_DbTable_User(); $lastUserId = $usersDbModel->getLatestUserId(); $this->view->lastUserId = $lastUserId; In the view I then echo it to display it to the user: echo $this->lastUserId; However, even though my last id in the users table is 3. It displays 0. I have also tried: public function getLatestUserId() { $sql = 'SELECT max(id) FROM user'; $query = $this->query($sql); $result = $query->fetchAll(); return $result; } But this just throws out a server error. What have I done wrong? Am I missing something? Is there another way of doing this?

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  • Increase and decrease row value by 1 in MySQL

    - by Elliott
    Hi I have a MySQL database table "points" the user can click a button and a point should be removed from their account, the button they pressed has an ID of another user, therefore their account must increase by one. I have it working in jQuery and checked the varibles/posts in Firebug, and it does send the correct data, such as: userid= 1 posterid = 4 I think the problem is with my PHP page: <?php include ('../functions.php'); $userid=mysql_real_escape_string($_POST['user_id']); $posterid=mysql_real_escape_string($_POST['poster_id']); if (loggedin()) { include ('../connection.php'); $query1 = "UPDATE `points` SET `points` = `points` - 1 WHERE `userID` = '$userid'"; $result1=mysql_query($query1); $query2 = "UPDATE `points` SET `points` = `points` + 1 WHERE `userID` = '$posterid'"; $result2=mysql_query($query2); if ($result1 && result2) { echo "Successful"; return 1; } else { echo mysql_error(); return 0; } } ?> Any ideas? Thanks :)

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  • gevent urllib is slow

    - by djay
    I've created a set of demos of a TCP server however my gevent examples are noticely slower. I'm sure must be how I compiled gevent but can't work out the problem. I'm using OSX leopard using fink compiled python 2.6 and 2.7. I've tried both the stable gevent and gevent 1.0b1 and it acts the same. The echo takes 5 seconds to respond, where the other examples take <1sec. If I remove the urllib call then the problem goes away. I put all the code in https://github.com/djay/geventechodemo To run the examples I'm using zc.buildout so to build $ python2.7 bootstrap.py $ bin/buildout To run the gevent example: $ bin/py geventecho3.py & [1] 80790 waiting for connection... $ telnet localhost 8080 Trying 127.0.0.1... ...connected from: ('127.0.0.1', 56588) Connected to localhost. Escape character is '^]'. hello echo: avast This will take 3-4 seconds to respond on my system. However the twisted example $ bin/py threadecho2.py or the twisted example $ bin/py twistedecho2.py Is less than 1s. Any idea what I'm doing wrong?

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  • Cache layer for MVC - Model or controller?

    - by Industrial
    Hi everyone, I am having some second thoughts about where to implement the caching part. Where is the most appropriate place to implement it, you think? Inside every model, or in the controller? Approach 1 (psuedo-code): // mycontroller.php MyController extends Controller_class { function index () { $data = $this->model->getData(); echo $data; } } // myModel.php MyModel extends Model_Class{ function getData() { $data = memcached->get('data'); if (!$data) { $query->SQL_QUERY("Do query!"); } return $data; } } Approach 2: // mycontroller.php MyController extends Controller_class { function index () { $dataArray = $this->memcached->getMulti('data','data2'); foreach ($dataArray as $key) { if (!$key) { $data = $this->model->getData(); $this->memcached->set($key, $data); } } echo $data; } } // myModel.php MyModel extends Model_Class{ function getData() { $query->SQL_QUERY("Do query!"); return $data; } } Thoughts: Approach 1: No multiget/multi-set. If a high number of keys would be returned, overhead would be caused. Easier to maintain, all database/cache handling is in each model Approach 2: Better performancewise - multiset/multiget is used More code required Harder to maintain Tell me what you think!

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  • merge my code with Ajax code>>> problem

    - by sandy
    I want to help me In the following link i found nice code in Ajax http://www.w3schools.com/php/php_ajax_livesearch.asp I want to link my code with the code you see in the link above and replace dropdown list How can I do it for I could not where is it change in code even my code work as Ajax ?? I wish .... I wish .... I wish any somebody can help me <?php include ("connect.php"); print_r($_POST['sector_list']); $member_id = intval($_POST['sector_list']); if($member_id == 0) { // Default choice was selected } else { $res = mysql_query("SELECT * FROM members WHERE MemberID = $member_id LIMIT 1"); if(mysql_num_rows($res) == 0) { // Not a valid member } else { // The member is in the database } } ?> <form method="POST" action=<?php echo $_SERVER["PHP_SELF"]; ?> > <input type="hidden" name="sector" value="sector_list"> <select name="sector_list[]" class="inputstandard" multiple="multiple"> <option size ="40" value="default">send to </option> <?php $result = mysql_query('SELECT * from members') or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo '<option value="' . $row['MemberName'] . '">' . $row['MemberName']. '</option>'; } ?> <input type ="submit" name ="go" value = "go" > </select> </form>

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  • Use Zip to Pre-Populate City/State Form with jQuery AJAX

    - by Paul
    I'm running into a problem that I can solve fine by just submitting a form and calling a db to retrieve/echo the information, but AJAX seems to be a bit different for doing this (and is what I need). Earlier in a form process I ask for the zip code like so: <input type="text" maxlength="5" size="5" id="zip" /> Then I have a button to continue, but this button just runs a javascript function that shows the rest of the form. When the rest of the form shows, I want to pre-populate the City input with their city, and pre-populate the State dropdown with their state. I figured I would have to find a way to set city/state to variables, and echo the variables into the form. But I can't figure out how to get/set those variables with AJAX as opposed to a form submit. Here's how I did it without ajax: $zip = mysql_real_escape_string($_POST['zip']); $q = " SELECT city FROM citystatezip WHERE zip = $zip"; $r = mysql_query($q); $row = mysql_fetch_assoc($r); $city = $row['city']; Can anybody help me out with using AJAX to set these variables? Thanks!

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  • Problem with checkboxes, sql select statements & php

    - by smokey20
    I am trying to display some rows from a database table based on choices submitted by the user. Here is my form code <form action="choice.php" method="POST" > <input type="checkbox" name="variable[]" value="Apple">Apple <input type="checkbox" name="variable[]" value="Banana">Banana <input type="checkbox" name="variable[]" value="Orange">Orange <input type="checkbox" name="variable[]" value="Melon">Melon <input type="checkbox" name="variable[]" value="Blackberry">Blackberry From what I understand I am placing the values of these into an array called variable. Two of my columns are called receipe name and ingredients(each field under ingredients can store a number of fruits). What I would like to do is, if a number of checkboxes are selected then the receipe name/s is displayed. Here is my php code. <?php // Make a MySQL Connection mysql_connect("localhost", "*****", "*****") or die(mysql_error()); mysql_select_db("****") or die(mysql_error()); $variable=$_POST['variable']; foreach ($variable as $variablename) { echo "$variablename is checked"; } $query = "SELECT receipename FROM fruit WHERE $variable like ingredients"; $row = mysql_fetch_assoc($result); foreach ($_POST['variabble'] as $ingredients) echo $row[$ingredients] . '<br/>'; ?> I am very new to php and just wish to display the data, I do not need to perform any actions on it. I have tried many select statements but I cannot get any results to display. My db connection is fine and it does print out what variables are checked. Many thanks in advance.

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  • Can't make HREF change based on PHP value

    - by Liso22
    I want to retrieve the user's location and then show a link that points to an URL that changes according to that location. I just want to place the user's city name at the end of the HREF. I need this work on my wordpress site, on a static page. I use a plugin called Exec-php which let's me run PHP in pages. I have a plugin that provides me the user's city through the shortcode "[mmjs-city]". I tried to make it work through different paths but I never get the link to work. Here I tried assigning that shortcode to a value, <?php $city= "[mmjs-city]"; echo $city; echo "<a href='?s=" . $city ."'>Search for your city</a>"; ?> I added the first two lines to check whether the shortcode is working or not and if it's correctly assinged to the value $city. That part works. Then it creates the link and put's the value $city at the end of it. But when trying it instead of taking me to: /?s=new+york It takes me to: /?s=%3Cscript%20language=%22javascript%22%3Edocument.write(geoip_city());%3C/script%3E I have no idea what to do. I would be really thankful for any info on how to make it work, it's really an important feature for my site. Please ask for any further info or idk anything. Also this is where I tested that code: http://chusmix.com/?page_id=1129 Thanks

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