Reputation: 717
list.php
: A simple ajax code that I want to display only records of the Mysql table:
<html>
<head>
<script src="jquery-1.9.1.min.js">
</script>
<script>
$(document).ready(function() {
var response = '';
$.ajax({
type: "GET",
url: "Records.php",
async: false,
success: function(text) {
response = text;
}
});
alert(response);
});
</script>
</head>
<body>
<div id="div1">
<h2>Let jQuery AJAX Change This Text</h2>
</div>
<button>Get Records</button>
</body>
</html>
Records.php is the file to fetch records from Mysql.
In the Database are only two fields: 'Name', 'Address'.
<?php
//database name = "simple_ajax"
//table name = "users"
$con = mysql_connect("localhost","root","");
$dbs = mysql_select_db("simple_ajax",$con);
$result= mysql_query("select * from users");
$array = mysql_fetch_row($result);
?>
<tr>
<td>Name: </td>
<td>Address: </td>
</tr>
<?php
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>$row[1]</td>";
echo "<td>$row[2]</td>";
echo "</tr>";
}
?>
This code is not working.
Upvotes: 39
Views: 411342
Reputation: 2807
This answer was for @
Neha Gandhi but I modified it for people who use pdo and mysqli sing mysql functions are not supported. Here is the new answer
<html>
<!--Save this as index.php-->
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script src="//ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
<?php
// save this as display.php
// show errors
error_reporting(E_ALL);
ini_set('display_errors', 1);
//errors ends here
// call the page for connecting to the db
require_once('dbconnector.php');
?>
<?php
$get_member =" SELECT
empid, lastName, firstName, email, usercode, companyid, userid, jobTitle, cell, employeetype, address ,initials FROM employees";
$user_coder1 = $con->prepare($get_member);
$user_coder1 ->execute();
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($row =$user_coder1->fetch(PDO::FETCH_ASSOC)){
$firstName = $row['firstName'];
$empid = $row['empid'];
$lastName = $row['lastName'];
$cell = $row['cell'];
echo "<tr>";
echo "<td align=center>$firstName</td>";
echo "<td align=center>$empid</td>";
echo "<td align=center>$lastName </td>";
echo "<td align=center>$cell</td>";
echo "<td align=center>$cell</td>";
echo "</tr>";
}
echo "</table>";
?>
<?php
// save this as dbconnector.php
function connected_Db(){
$dsn = 'mysql:host=localhost;dbname=mydb;charset=utf8';
$opt = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);
#echo "Yes we are connected";
return new PDO($dsn,'username','password', $opt);
}
$con = connected_Db();
if($con){
//echo "me is connected ";
}
else {
//echo "Connection faid ";
exit();
}
?>
Upvotes: 0
Reputation: 851
For retrieving data using Ajax + jQuery, you should write the following code:
<html>
<script type="text/javascript" src="jquery-1.3.2.js"> </script>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to display.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
</script>
<body>
<h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
<tr>
<td> <input type="button" id="display" value="Display All Data" /> </td>
</tr>
</table>
<div id="responsecontainer" align="center">
</div>
</body>
</html>
For mysqli connection, write this:
<?php
$con=mysqli_connect("localhost","root","");
For displaying the data from database, you should write this :
<?php
include("connection.php");
mysqli_select_db("samples",$con);
$result=mysqli_query("select * from student",$con);
echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";
while($data = mysqli_fetch_row($result))
{
echo "<tr>";
echo "<td align=center>$data[0]</td>";
echo "<td align=center>$data[1]</td>";
echo "<td align=center>$data[2]</td>";
echo "<td align=center>$data[3]</td>";
echo "<td align=center>$data[4]</td>";
echo "</tr>";
}
echo "</table>";
?>
Upvotes: 68
Reputation: 2057
Please make sure your $row[1] , $row[2]
contains correct value, we do assume here that 1 = Name , and 2 here is your Address field
?
Assuming you have correctly fetched your records from your Records.php, You can do something like this:
$(document).ready(function()
{
$('#getRecords').click(function()
{
var response = '';
$.ajax({ type: 'POST',
url: "Records.php",
async: false,
success : function(text){
$('#table1').html(text);
}
});
});
}
In your HTML
<table id="table1">
//Let jQuery AJAX Change This Text
</table>
<button id='getRecords'>Get Records</button>
A little note:
Try learing PDO http://php.net/manual/en/class.pdo.php since mysql_* functions
are no longer encouraged..
Upvotes: 3
Reputation: 46270
$(document).ready(function(){
var response = '';
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
response = text;
}
});
alert(response);
});
needs to be:
$(document).ready(function(){
$.ajax({ type: "GET",
url: "Records.php",
async: false,
success : function(text)
{
alert(text);
}
});
});
Upvotes: 1
Reputation: 171
You can't return ajax return value. You stored global variable store your return values after return.
Or Change ur code like this one.
AjaxGet = function (url) {
var result = $.ajax({
type: "POST",
url: url,
param: '{}',
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (data) {
// nothing needed here
}
}) .responseText ;
return result;
}
Upvotes: 4