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POHow to display MYSQL record using a combobox in PHP/AJAX?
 

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  1. POHow to display MYSQL record using a combobox in PHP/AJAX?
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    <p>Need help. I using 3 tables to store booking information: <code>eventinfo</code>, <code>customer</code> and <code>testbook</code> and I try to display records by using combo box to sort by username and eventID. </p> <p>Problem is the combobox does not load data from database.</p> <p>My code: </p> <p><strong>index.php</strong></p> <pre><code>&lt;html&gt; &lt;head&gt; &lt;script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4./jquery.min.js"&gt;&lt;/script&gt; &lt;script type="text/javascript"&gt; function showUser(str) { if (str=="") { document.getElementById("txtHint").innerHTML=""; return; } 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 &amp;&amp; xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","getuser.php?q="+str,true); xmlhttp.send(); } &lt;/script&gt; &lt;/head&gt; &lt;body&gt; &lt;form&gt; &lt;select name="users" onchange="showUser(this.value)"&gt; &lt;option selected="selected"&gt;--Select user--&lt;/option&gt; &lt;?php include('db.php'); $sql=mysql_query("select username, custNo from testbook"); while($row=mysql_fetch_array($sql)) { $id=$row['custNo']; $username=$row['username']; echo '&lt;option value="'.$id.'"&gt;'.$username.'&lt;/option&gt;'; } ?&gt; &lt;/select&gt; &lt;/form&gt; &lt;br /&gt; &lt;div id="txtHint"&gt;&lt;b&gt;Person info will be listed here.&lt;/b&gt;&lt;/div&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p><strong>getuser.php</strong></p> <pre><code>&lt;?php $q=$_GET["q"]; $con = mysql_connect('localhost', '', ''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("eventdb", $con); $sql="SELECT * FROM testbook WHERE custNo = '".$q."'"; $result = mysql_query($sql); echo "&lt;table border='1'&gt; &lt;tr&gt; &lt;th&gt;Booth&lt;/th&gt; &lt;th&gt;Event&lt;/th&gt; &lt;th&gt;Date&lt;/th&gt; &lt;th&gt;Status&lt;/th&gt; &lt;th&gt;Booking ID&lt;/th&gt; &lt;/tr&gt;"; while($row = mysql_fetch_array($result)) { echo "&lt;tr&gt;"; echo "&lt;td&gt;" . $row['boothAlias'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['eventID'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['date'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['bstatus'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['bookingID'] . "&lt;/td&gt;"; echo "&lt;/tr&gt;"; } echo "&lt;/table&gt;"; mysql_close($con); ?&gt; </code></pre> <p>Output:</p> <p><img src="https://i.stack.imgur.com/6beyv.jpg" alt="output"></p>
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    1. COWhen you view-source in the browser, do you see any PHP errors? Can you provide a link to a test version of the page?
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      1. POHow to display MYSQL record using a combobox in PHP/AJAX?
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    2. COhttp://i.imgur.com/F5CVP.jpg no php error, just do not load username from database
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      1. POHow to display MYSQL record using a combobox in PHP/AJAX?
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    3. COAre you certain that the query you make has any records? You should either be getting an error message, or the records should load. Make sure you have error reporting enabled - you can try turning on all errors for this page by putting the following at the top of the page `<?php error_reporting(E_ALL); ?>`. Errors may not necessarily appear on the screen if they are embedded inside of an HTML tag, so it's best to do view source (CTRL+U) and make sure there are no errors there.
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