Question:
I'm using jQuery and $.post(). My code snippet is as follows for chat .php :
<?php $msg=$_POST['msg']; mysql_connect("localhost","root"); mysql_select_db("user"); mysql_query("INSERT INTO space (name,msg,serial) VALUES('Test','$msg','1')"); ?> and here is the code for my HTML file :
<!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=iso-8859-1" /> <title>Shout!</title> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> var status=1; function action() { if(status==1) { $("#Layer1").hide("slow"); $("#Layer3").hide("fast"); $("#Layer4").hide("slow"); $("#close").attr("src","open.jpg"); status=0; } else if(status==0) { status=1; $("#Layer1").show("slow"); $("#Layer3").show("fast"); $("#Layer4").show("slow"); $("#close").attr("src","close.jpg"); } } function sendline() { var msg=$("#msg").val(); $.post("chat.php",{msg:msg}); $("#msg").val(" "); } function typeyo() { var text=$("#msg").val(); $("#Layer6").html(text); } </script> <style type="text/css"> <!-- body { background-color: #000000; } #Layer1 { position:absolute; width:200px; height:115px; z-index:1; left: 199px; top: 3px; } #Layer2 { position:absolute; width:69px; height:64px; z-index:2; left: 570px; top: 543px; } #Layer3 { position:absolute; width:131px; height:91px; z-index:3; left: 487px; top: 327px; } .style1 { color: #FFFFFF; font-family: "Segoe UI"; font-weight: bold; } #Layer4 { position:absolute; width:99px; height:38px; z-index:4; left: 744px; top: 485px; } #Layer5 { position:absolute; width:274px; height:70px; z-index:5; left: 422px; top: 62px; } #Layer6 { width:638px; height:356px; z-index:5; left: 352px; top: 105px; } --> </style></head> <body> <div class="style1" id="Layer3"> <textarea name="textarea" cols="30" rows="5" id="msg" ></textarea> </div> <div id="Layer1">Hello World!<img src="body.jpg" width="842" height="559" /></div> <div id="Layer2"><img src="close.jpg" alt="Go Online/Offline" name="close" width="63" height="64" id="close" OnClick="action()"/></div> <div id="Layer4"> <input type="button" value="Send Line" onclick="sendline()" /></div> <div id="Layer6" style="color:white;font-family:Segoe UI;font-size:16px;width:500px; height:400px; overflow:auto;"></div> </body> </html> Now,there seems to be some problem posting the variable msg.Im using chat.php for $.post() on the HTML code that i've provided.
There seems to be a problem with sending the "msg" here . The chat.php file is fine since if we run it directly ,and not thorugh a $.post() call it works perfectly
Kindly Help! thank you!
Solution:1
Your msg variable that you try to send via POST is not initialized, add the following line at the beginning of your sendline function:
var msg = $("#msg").val(); Note: you have a big/huge security issue inserting variables from POST in MySQL queries without prior treatment.
Solution:2
update: I suggest you use a javascript debugger, set a breakpoint at the beginning of function sendline() and step through the code. Which one to use depends on your browser(s).
Firefox -> e.g. Firebug
IE7 -> e.g. IE Developer Toolbar
IE8+ -> just press F12 to open the developer tools that are shipping with IE.
In addition to darma's answer: Your php script is prone to sql injections (intentional/malicious as well as unintentional ones). Either use prepared, parametrized statements or escape the data properly.
working example:
test.html:
<html> <head><title>test.html</title> <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js" type="text/javascript"></script> <script type="text/javascript"> function sendline() { var msg = $('#msg').val(); // --- add some tests on msg here, e.g. msg=="" --- $.post( "chat.php", {'msg':msg}, function(data, textStatus, req) { $('#reply').text('reply: ' + data); } ); } </script> </head> <body> <div> <textarea id="msg" rows="5" cols="30" name="textarea"></textarea> <button onclick="sendline()">send line</button> </div> <div id="reply"> </div> </body> </html> chat.php:
<?php // --- add some tests on $_POST['msg'] here // e.g. isset($_POST['msg']) and 0<strlen(trim($_POST['msg'])) --- // you might want to use a slightly more sophisticated error handling than "or die(mysql_error())" ...but this is only an example. $mysql = mysql_connect("localhost","localonly", "localonly") or die(mysql_error()); mysql_select_db("test", $mysql) or die(mysql_error()); $msg=mysql_real_escape_string($_POST['msg'], $mysql); $sql = "INSERT INTO space (name,msg,serial) VALUES('Test','$msg','1')"; // mysql_query($sql, $mysql) or die(mysql_error()); echo htmlspecialchars($sql); update2: You still don't have any error handling in your php script. Any of the mysql_* function can fail for various reasons; test the results. You need to "see" those errors, e.g. by writing them to a log file or something...
Try
<?php define('LOGERRORS', 1); function dbgLog($text) { if (LOGERRORS) { error_log(date('Y-m-d H:i:s : ').$text."\n", 3, 'error.log'); } } if ( !isset($_POST['msg']) ) { dbgLog('script called without post parameter "msg"'); die(); } $mysql = mysql_connect("localhost","root"); if ( !$mysql ) { dbgLog('database connection failed: '.mysql_error()); die(); } $result = mysql_select_db("user", $mysql); if ( !$result ) { dbgLog('database selection failed: '.mysql_error($mysql)); die(); } $msg=mysql_real_escape_string($_POST['msg'], $mysql); $sql = "INSERT INTO space (name,msg,serial) VALUES('Test','$msg','1')"; dbgLog('sending query: '.$sql); $result = mysql_query($sql, $mysql); if ( !$result ) { dbgLog('query failed: '.mysql_error($mysql)); die(); } Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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