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POUnable to display the data from mysql
 

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    <p>I am trying to display the data onselect method but the data is not getting displayed. This is the code.</p> <p>index.html</p> <pre><code>&lt;html&gt; &lt;head&gt; &lt;script&gt; function showUser(str) { alert(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 value=""&gt;Select a person:&lt;/option&gt; &lt;option value="1"&gt;MEGHALI&lt;/option&gt; &lt;option value="2"&gt;SONY&lt;/option&gt; &lt;option value="3"&gt;DHANU&lt;/option&gt; &lt;option value="4"&gt;ADITYA&lt;/option&gt; &lt;option value="5"&gt;MAHESH&lt;/option&gt; &lt;option value="6"&gt;NEERAJ&lt;/option&gt; &lt;option value="7"&gt;VINEET&lt;/option&gt; &lt;option value="8"&gt;ASHWINI&lt;/option&gt; &lt;option value="9"&gt;SANTOSH&lt;/option&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>and the getuser.php</p> <pre><code>&lt;?php $q = intval($_GET['q']); $con = mysqli_connect("localhost", "root", " ", "flowers"); if (!$con) { die('Could not connect: ' . mysqli_error($con)); } mysqli_select_db($con,"FLOWERS"); $sql="SELECT * FROM STUDENTS WHERE id = '".$q."'"; $result = mysqli_query($con,$sql); echo "&lt;table border='1'&gt; &lt;tr&gt; &lt;th&gt;ID&lt;/th&gt; &lt;th&gt;NAME&lt;/th&gt; &lt;th&gt;CITY&lt;/th&gt; &lt;th&gt;PHONE&lt;/th&gt; &lt;/tr&gt;"; while($row = mysqli_fetch_array($result)) { echo "&lt;tr&gt;"; echo "&lt;td&gt;" . $row['ID'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['NAME'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['CITY'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['PHONE'] . "&lt;/td&gt;"; echo "&lt;/tr&gt;"; } echo "&lt;/table&gt;"; mysqli_close($con); ?&gt; </code></pre> <p>I guess the <code>getuser.php?q="+str</code> is not getting passed.<br> Please help as I am newbie to ajax.</p>
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    1. USPasserby
    1. USSherry
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    1. COWhat is your database name? FLOWERS or flowers?
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      1. USRamesh
    2. COPlease narrow down your problem. If you don't even know yet whether it's the database connection/query that fails or the AJAX request, there's no reason yet to ask this question here.
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      1. USGolezTrol
    3. COSimplify by using j query instead [Click here](http://api.jquery.com/jQuery.ajax/)
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      1. USKrishnadas PC
 

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