insert and fetch data with jquery post method in php

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I want to insert and fetch data with jquery post method in php. I need your help to correct my code friends
my code is

HTML

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Jquery ajax</title>
<script type="text/javascript" src="JQuery/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="JQuery/jquery.min.js"></script>
<script type="text/javascript" src="JQuery/jquery.form.js"></script>
</head>
<script>

$(document).ready(function() {
$("#btn1").click(function(){

$.post("testin.php",$("#frm").serialize(),function(data){

$("#div1").html(data);
	
});
	
});

});

</script>




<body>

<div id="div1">
<form id="frm" method="post">
First Name:<input type="text" name="firstname" />
Last Name:<input type="text" name="lastname" />
Age: <input type="text" name="age" />
Home Town:<input type="text" name="hometown" />
Job: <input type="text" name="job" />
<input type="submit" value="submit" id="btn1" />
</form>
</div>
</body>
</html>

my testin.php

PHP

 <?
$con = mysqli_connect("localhost","root");
mysqli_select_db($con,"ajaxdatabase");
$first = $_POST["firstname"];
$last = $_POST["lastname"];
$age = $_POST["age"];
$home = $_POST["hometown"];
$job = $_POST["job"];
$sql1 = "INSERT INTO user(FirstName,LastName,Age,HomeTown,Job)VALUES('$first','$last','$age','$home','$job')";
$query1 = mysqli_query($con,$sql1);

if (!mysqli_query($con,$sql1))
  {
  die('Error: ' . mysqli_error($con));
  }
echo "1 record added";
$sql = "SELECT * FROM user";
$query = mysqli_query($con,$sql);
echo "<table border='1'>
<tr>
<th>Id</th>
<th>FirstName</th>
<th>LastName</th>
<th>Age</th>
<th>HomeTown</th>
<th>Job</th>
</tr>";
while ($result = mysqli_fetch_array($query))
{
	echo '<tr>';
	echo '<td>'.$result['Id'];
	echo '<td>'.$result['FirstName'];
	echo '<td>'.$result['LastName'];
	echo '<td>'.$result['Age'];
	echo '<td>'.$result['HomeTown'];
	echo '<td>'.$result['Job'];
	echo '</tr>';
		
}
echo'</table>';

?>

Posted 4-Oct-13 7:08am

Md Jamaluddin Saiyed

Updated 4-Oct-13 13:14pm

Nelek

v2

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Emerald kwekowe · Answer 1 · 2013-10-04T09:06:00

What exactly is the error report?

here is a simple PHP/MySql code to insert/display user data..

PHP

<?php
								mysql_connect("localhost", "root","") or die (mysql_error());
mysql_select_db("ajaxdatabase") or die(mysql_error());
//This code runs if the form has been submitted

 if (isset($_POST['submit'])) { 



 	$insert = "INSERT INTO user (firstname, lastname, age, hometown, job)

 			VALUES ('".$_POST['firstname']."', '".$_POST['lastname']."', '".$_POST['age']."', '".$_POST['hometown']."', '".$_POST['job']."', )";

 	$add_member = mysql_query($insert);

 	?>
SUCCESSFUL INSERT

<?php 
 } 

 else 
 {	
 ?>
<form  method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
First Name:<input type="text" name="firstname" />
Last Name:<input type="text" name="lastname" />
Age: <input type="text" name="age" />
Home Town:<input type="text" name="hometown" />
Job: <input type="text" name="job" />
<input type="submit" value="submit" name="submit"/>
</form>

 <?php
 }
 ?>

And to Display the data

PHP

<?php
		// Connect to database server
	mysql_connect("localhost", "user", "") or die (mysql_error ());

	// Select database
	mysql_select_db("ajaxdatabase") or die(mysql_error());

	// SQL query
	$strSQL = "SELECT * FROM users ORDER BY Id DESC";

	// Execute the query (the recordset $rs contains the result)
	$rs = mysql_query($strSQL);
	
	// Loop the recordset $rs
	// Each row will be made into an array ($row) using mysql_fetch_array
	while($row = mysql_fetch_array($rs)) {

	   // Write the value of the column FirstName (which is now in the array $row)
	  echo $row['Id'] . " " . $row['firstame'] . " " . $row['lastname'] . " " . $row['age'] . " " . $row['hometown'] . " " . $row['job'] . " " . $row['BirthDate'] . "<br />";

	  }

	// Close the database connection
	mysql_close();
	?>