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hi sir actually i use now day's ajax but i m having problem
my opinion i thing this code is right becuse i make code for
my side validation with ajax in php so but it is not run
complete i m posting my code pls confirm and clear my some
mistake and return send me with informatiom what was error
so next time i will be care fully so i m sending u but
replay so quick because tomorrow i have to present it some
where,frist one is index.php and second is show.php show.php
is database connectivity for testing
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Answer / rohit12
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0
Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>complete ajax validation</title>
<script type="text/javascript">
function ajaxFunction()
{
var httpxml;
try
{
// Firefox, Opera 8.0+, Safari
httpxml=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
httpxml=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
try
{
httpxml=new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e)
{
alert("Your browser does not support AJAX!");
return false;
}
}
}
function stateChanged()
{
if(httpxml.readyState==4)
{
document.getElementById("myDiv").innerHTML=httpxml.responseText;
myForm.reset();
}
}
function getFormData(myName) {
var myParameters = new Array();
//// Text field data collection //
var val=myName.fname.value;
val = "fname="+val;
myParameters.push(val);
var val=myName.lname.value;
val = "lname="+val;
myParameters.push(val);
// End of text field data collection //
////////////
return myParameters.join("&"); // return the string after
joining the array
}
var url="show.php";
var myForm = document.forms[0];
var parameters=getFormData(myName);
httpxml.onreadystatechange=stateChanged;
httpxml.open("POST", url, true)
httpxml.setRequestHeader("Content-type",
"application/x-www-form-urlencoded")
httpxml.send(parameters)
document.getElementById("myDiv").innerHTML="<img src=wait.gif>";
}
</script>
</head>
<body>
<form name="myName">
Name: <input type="text" name="fname" id="fname" />
<br /><br/>
LName: <input type="text" name="lname" id="lname" /><br />
<input type="button" onclick="ajaxFunction()" value="submit" />
</form>
<div id="myDiv"></div>
</body>
</html>
now show.php page start from here
<?Php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$status_form = "OK";
$msg="";
if(strlen($fname) < 3)
{
$status_form="NOTOK";
$msg .= "Your Name must be more than 3 char length<br>";
}
if($lname =="")
{
$status_form="NOTOK";
$msg .= "Your Last name is blank or invalid <br>";
}
//sleep(1);
if($status_form<>"OK"){
echo "<font color='red'>$msg $status</font>";
}else{
echo "<font color='green'><b>Form Validation is
passed.</b></font>";
}
?>
please replay me fast or imdetily
| Is This Answer Correct ? | 0 Yes | 0 No |
Hi I have a dynamic dropdown reading in data from a MQSql database, I need to refresh just the dropdown (not the whole page) for any new values in the database. I am trying to use the getElementById in javascript but I have no idea how to get it to refresh. Does any one have an examples or any ideas how this can be done? Regards Srilanka_3
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hi sir actually i use now day's ajax but i m having problem my opinion i thing this code is right becuse i make code for my side validation with ajax in php so but it is not run complete i m posting my code pls confirm and clear my some mistake and return send me with informatiom what was error so next time i will be care fully so i m sending u but replay so quick because tomorrow i have to present it some where,frist one is index.php and second is show.php show.php is database connectivity for testing
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how to validation ajax in php with database actually i use here it give some time erroe so pls tell how to do right and second think is that textfeild phone number it accept only int value not charecter so tell me both and my code is that so check it then send after correction
PHP Code 
