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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 |
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