I am using ajax and php ! I have three php files ,I want to display post using ajax and previous post display from database on same page !!

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Dear friends
I am using ajax and php ! I have three php files ,I want to display post using ajax and insert it in my databse,Want to display previous posts on same page !!but Iam getting output another page ! here is my three php file codes !! Help me to solve this friends !!

index.php

PHP

</div>!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Freak </title>
<link rel="stylesheet" href="root/Style/Style.css">
<link rel="stylesheet" href="root/Style/content.css">
<script language="javascript" type="text/javascript">

function ajax_post(){
// Creat out XMLHttpRequest Object

var hr = new XMLHttpRequest();
// Creat some variables we need to send to out php file
var url = "my_parse_file.php";
	
var fn = document.getElementById("first_name").value;
//var ln = document.getElementById("last_name").value;	
//var vars = "firstname="+fn+"&lastname="+ln;
var vars = "firstname="+fn;
hr.open("POST",url,true);
//set content type header information for sending url encoded variables in the request
hr.setRequestHeader("Content-type","application/x-www-form-urlencoded");
// Access the onreadystatechange event for the XMLHttpRequest object
hr.onreadystatechange = function(){
	if(hr.readyState==4 && hr.status==200) {
		var return_data = hr.responseText;
			document.getElementById("status").innerHTML = return_data;
		}
	}


// send the data to PHP now ... and wait for response to update the status div
hr.send(vars);// actually execute the request
document.getElementById("status").innerHTML = "processing...";
document.getElementById("first_name").value="";
}
</script>   
</head>
<body>
<?php include_once("Template_PageTop.php");>
<div id="pageMiddle">
<div id="sidebar">
<img src="root/Style/noimage.gif" alt="" id="profilepic" >
<div id="profilename">Jems Saiyed </div>
<a href=""><img src="root/Style/friend reqest_sidebar.png" alt=""></a> <a href=""> Friends </a>
<br>
<a href=""><img src="root/Style/classmates_sidebar.png" alt=""></a> <a href=""> Classmates </a>
<br>
<a href=""><img src="root/Style/photos_sidebar.png" alt=""></a> <a href="">Photos</a>
<br>
<a href=""><img src="root/Style/message_sidebar.png" alt=""></a> <a href="">Messages</a>
</br></br></br></div>
<div id="sidebar-b"> Status of the week </div>
<div id="content">
<br>
<a href=""><img src="root/Style/photos_sidebar.png" alt=""> Share Photo/Video</a>
<br>
<form action="inst.php" method="post">
<textarea  name="first_name"  style="width:50%;height:100px;"> </textarea>
<input name="myBtn" type="submit"  value="post" onClick="javascript:ajax_post();" style="margin-left:45%">
</form>
<div id="contener">
<div id="status" style="margin:10px;"></div>
</div>
</br></br></div>
</div>
<?php include_once("Template_PageBottom.php"); >
</body>
</html>

my_pase_file.php

<php <br mode="hold" />echo $_POST['firstname'];

>>inst.php

$con = mysqli_connect('localhost',"root","");
mysqli_select_db($con,"dbsocial");
$firstname = $_POST['first_name'];
echo $firstname;
$sql = "SELECT * FROM status";
mysqli_query($con,"INSERT INTO status (data) VALUES ('$firstname') ");

$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
	echo "<hr>".$row['data'];"<hr>";	
}
$output = mysqli_query($con,$sql);

mysqli_close($con);

Posted 8-Jul-13 7:01am

Md Jamaluddin Saiyed

Updated 8-Jul-13 17:39pm

Sudhakar Shinde

v4

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Comments

Zoltán Zörgő 8-Jul-13 14:03pm

Please, use proper formatting. Your post is looking ugly and it is unreadable.

Md Jamaluddin Saiyed 9-Jul-13 3:25am

now you can properly read the code ! Its edited !

Sudhakar Shinde 8-Jul-13 23:41pm

Hi Md Jamaluddin, Tried to format the code however please check if the code matches with your original post.

Md Jamaluddin Saiyed 9-Jul-13 3:24am

Thanks for editing the code

1 solution

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Md Jamaluddin Saiyed · Accepted Answer · 2013-07-12T23:07:00

Replace

Quote:
<textarea name="first_name" id = "first_name" style="width:50%;height:100px;"> </textarea>
<input name="myBtn" type="submit" value="POST" onClick="javascript:ajax_post();" style="margin-left:45%">

instead of -

Quote:
<form action="inst.php" method="post">
<textarea name="first_name" style="width:50%;height:100px;"> </textarea>
<input name="myBtn" type="submit" value="post" onClick="javascript:ajax_post();" style="margin-left:45%">
</form>

and other file inst.php

Quote:

$firstname = $_POST['firstname'];
$con = mysqli_connect('localhost',"root","");
mysqli_select_db($con,"dbsocial");
$sql = "SELECT * FROM status Order by Id DESC";
mysqli_query($con,"INSERT INTO status (data) VALUES ('$firstname')");
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{
echo "
".$row['data'];"
";
}
$output = mysqli_query($con,$sql);
mysqli_close($con);
?>