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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.
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Create HTML Button to display data on clickCreate an HTML button with id="showData" . Ajax script will execute on click this button. Create a div with id="table-container" . This div will use to show data while you click the button.
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>"; ?> 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; } If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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