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POPHP Ajax is not displaying
 

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    <p>I am trying to display from a MySQL Database using PHP and Ajax but the while loop is not displaying my data in the dropdown list do I have a syntax error? here is my code: </p> <pre><code>&lt;html&gt; &lt;head&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 value=""&gt;Select a person:&lt;/option&gt; &lt;option value="1"&gt;Peter Griffin&lt;/option&gt; &lt;option value="2"&gt;Lois Griffin&lt;/option&gt; &lt;option value="3"&gt;Glenn Quagmire&lt;/option&gt; &lt;option value="4"&gt;Joseph Swanson&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>the php file: </p> <pre><code>&lt;?php $q=$_GET["q"]; $con = mysql_connect('localhost', 'djronlove', 'Djronlove'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("djronlove", $con); $sql="SELECT * FROM user WHERE id = '".$q."'"; $result = mysql_query($sql); echo "&lt;table border='1'&gt; &lt;tr&gt; &lt;th&gt;Firstname&lt;/th&gt; &lt;th&gt;Lastname&lt;/th&gt; &lt;th&gt;Age&lt;/th&gt; &lt;th&gt;Hometown&lt;/th&gt; &lt;th&gt;Job&lt;/th&gt; &lt;/tr&gt;"; while($row = mysql_fetch_array($result)) { echo "&lt;tr&gt;"; echo "&lt;td&gt;" . $row['FirstName'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['LastName'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['Age'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['Hometown'] . "&lt;/td&gt;"; echo "&lt;td&gt;" . $row['Job'] . "&lt;/td&gt;"; echo "&lt;/tr&gt;"; } echo "&lt;/table&gt;"; mysql_close($con); ?&gt; </code></pre> <p>The url is <a href="http://imaginationeverywhere.info/mysql/ajax/ajaxdatabase.html" rel="nofollow">http://imaginationeverywhere.info/mysql/ajax/ajaxdatabase.html</a></p>
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    1. USAmen
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      1. VTUpMod
    1. COWell are you sure that your SQL code is returning values and you are asking for a SQL Injection Attack.
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      1. USepascarello
    2. COdo you get a response? try alert `(xmlhttp.responseText);` right after `xmlhttp.onreadystatechange`
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      1. USAleadam
    3. COAs a suggestion I will recommend you to go for jQuery ajax, though not the solution for this problem. :)
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      1. USShameer
 

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