php ajax - two drop down list with a submit button

Posted on 16th Feb 2014 07:03 pm by admin

i have created a form with 2 drop-down list and a submit button. I would like to know how to pass the selected values from the two drop down list to external javascript file and only will pass it after i clicked the submit button? Below is the example of my code:

<html>

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<head>
<script type="text/javascript" src="selecttidmth.js"></script>

<title>Report</title>
</head>

<body>
<form action="getID.php" method="get">
Magazines:
<select name="tid" onchange="showTid(this.value)">
<option value="">---Select---</option>
<option value="1">Magazine A</option>
<option value="2">Magazine B</option>
</select>         

Month:
<select name="mth" onchange="showMonth(this.value)">
<option value="">---Select---</option>
<option value="Jan">January</option>
<option value="Feb">February</option>
<option value="Mar">March</option>
<option value="Apr">April</option>
<option value="May">May</option>
<option value="Jun">June</option>
<option value="Jul">July</option>
<option value="Aug">August</option>
<option value="Sept">September</option>
<option value="Oct">October</option>
<option value="Nov">November</option>
<option value="Dec">December</option>
</select>    

<input type="submit" name="submit" Value="Submit">

</form>

<div id="txtHint"></div>
</body>
</html>

//javascript code

var xmlhttp;

function showTid(str)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}

function showMonth(str2)
{
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}

var url="getID.php";
url=url+"?q="+str+"?m="+str2;
url=url+"&sid="+Math.random();
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}

Hope someone here can help me out. Thanks in advance.

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php ajax - two drop down list with a submit button

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