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POhow to apply ajax on php code
 

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    <p>I am trying to apply ajax to my php code. However, when I click on the button I can't get any response. This means the ajax function I declare is not being called <code>onclick</code>.</p> <pre><code>&lt;?php $s_name= $_POST["submit"]; mysql_connect("localhost","root","");//database connection mysql_select_db("itcompanylist"); $query = "SELECT s_id FROM states WHERE `state_name` = '$s_name'"; $result1 = mysql_query($query); $row = mysql_fetch_array($result1); $result2 = mysql_query("SELECT city_name FROM `city` WHERE s_id ='".$row['s_id']."'"); $i = 0; echo "&lt;form method='post' name='myForm'&gt;&lt;table border='1' &gt;&lt;tr&gt;"; while ($row = mysql_fetch_row($result2)){ echo '&lt;td&gt;&lt;input type="submit" name="submit" onclick="ajaxFunction()" value="'.$row['0'].'"&gt;&lt;/td&gt;'; if ($i++ == 2) { echo "&lt;/tr&gt;&lt;tr&gt;"; $i=0; } } echo "&lt;/tr&gt;&lt;/table&gt;&lt;/form&gt;"; echo "&lt;div id='ajaxDiv'&gt;Your result will display here&lt;/div&gt;"; ?&gt; </code></pre> <p>ajax code:</p> <pre><code>&lt;script language="javascript" type="text/javascript"&gt; &lt;!-- //Browser Support Code function ajaxFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); }catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); }catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); }catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.value = ajaxRequest.responseText; } } var s1 = document.getElementById('submit').value; var queryString = "?submit=" + s1 ; ajaxRequest.open("GET", "" + queryString, true); ajaxRequest.send(null); } //--&gt; &lt;/script&gt; </code></pre>
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    1. USNikunj Jagad
    1. USNikunj Jagad
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    1. COAnd the code for `ajaxFunction()` is...?
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      1. USSeanWM
    2. COYou are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from.
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      1. USQuentin
 

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