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POhow to submit multiple data in mysql using php and jquery
 

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    <p>how to submit multiple values in a table please help me im using this code this code submit only 1 entry in database but i want to enter multiples value in database how can i do this</p> <p>live demo <a href="http://jsfiddle.net/eruZC/3/" rel="nofollow">http://jsfiddle.net/eruZC/3/</a></p> <p>this is entry.html</p> <pre><code> &lt;script type='text/javascript' src='http://code.jquery.com/jquery-1.10.1.js'&gt;&lt;/script&gt; &lt;link rel="stylesheet" type="text/css" href="http://fiddle.jshell.net/css/result-light.css"&gt; &lt;style type='text/css'&gt; &lt;/style&gt; &lt;script type='text/javascript'&gt;//&lt;![CDATA[ $(window).load(function(){ $(document).ready(function(){ var id=1; $("#butsend").click(function(){ $("#table1").append('&lt;tr valign="top"&gt;&lt;td width="100px"&gt;'+(id++)+'&lt;/td&gt;&lt;td width="100px"&gt;'+$("#sname").val()+'&lt;/td&gt;&lt;td width="100px"&gt;'+$("#age").val()+'&lt;/td&gt;&lt;td width="100px"&gt;&lt;a href="javascript:void(0);" class="remCF"&gt;Remove&lt;/a&gt;&lt;/td&gt;&lt;/tr&gt;'); }); $("#table1").on('click','.remCF',function(){ $(this).parent().parent().remove(); }); }); });//]]&gt; &lt;/script&gt; &lt;/head&gt; &lt;body&gt; &lt;form action="submit.php" d="form1" name="form1" method="post"&gt; &lt;label&gt;Student Name&lt;/label&gt;&lt;input type="text" name="sname" id="sname"&gt;&lt;/br&gt; &lt;label&gt;Student Age&lt;/label&gt;&lt;input type="text" name="age" id="age"&gt;&lt;/br&gt; &lt;input type="button" name="send" value="Add" id="butsend"&gt;&lt;/br&gt; &lt;input type="button" name="submit" value="Submit" &gt;&lt;/br&gt; &lt;/form&gt; &lt;table id="table1" name="table1" border=""&gt; &lt;tbody&gt; &lt;tr&gt;&lt;th width="100px"&gt;ID&lt;/th&gt;&lt;th width="100px"&gt;Name&lt;/th&gt;&lt;th width="100px"&gt;Age&lt;/th&gt;&lt;th width="100px"&gt;&lt;/th&gt;&lt;tr&gt; &lt;/tbody&gt; &lt;/table&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p>this is php page submit.php</p> <pre><code>&lt;?php /* NEW.PHP Allows user to create a new entry in the database */ // creates the new record form // since this form is used multiple times in this file, I have made it a function that is easily reusable function renderForm($sname, $age, $error) { ?&gt; &lt;?php // if there are any errors, display them if ($error != '') { echo '&lt;div style="padding:4px; border:1px solid red; color:red;"&gt;'.$error.'&lt;/div&gt;'; } ?&gt; &lt;?php } // connect to the database include('connect-db.php'); // check if the form has been submitted. If it has, start to process the form and save it to the database if (isset($_POST['submit'])) { // get form data, making sure it is valid $sname = mysql_real_escape_string(htmlspecialchars($_POST['sname'])); $age = mysql_real_escape_string(htmlspecialchars($_POST['age'])); // check to make sure both fields are entered if ($name == '') { // generate error message $error = 'ERROR: Please fill in all required fields!'; // if either field is blank, display the form again renderForm($sname, $age, $error); } else { // save the data to the database mysql_query("INSERT stock SET sname='$sname', age='$age'") or die(mysql_error()); // once saved, redirect back to the view page } } else // if the form hasn't been submitted, display the form { renderForm('','','','','','','',''); } ?&gt; </code></pre>
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    1. COis it true that you are actually only inserting the values of the labels? The values of the names and ages that you append in the table is not a field and not part of the form in order to get submitted, to your php code. Also take a look here it may help you http://stackoverflow.com/questions/10489046/how-to-pass-contents-of-an-html-table-as-form-data-on-a-post
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