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Home/ Questions/Q 8336393
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T03:58:34+00:00 2026-06-09T03:58:34+00:00

this is my html javascript $.ajax({ url: ‘post.php’, type: POST, data: ‘name=dan’, success: function(result){

  • 0

this is my html javascript

$.ajax({
        url: 'post.php',
        type: "POST",
        data: 'name="dan"',
        success: function(result){
            console.log(result);
        },
        error: function(){
            console.log('error');
        }   
    });

and here is the post.php

<?php
$host = "localhost";
$dbname = "test";
$user = "";
$pass = "";

$conn = mysql_connect($host, $user, $pass) or die (MYSQL_ERROR);
$select_db = mysql_select_db($dbname) or die("Could not select the database.");

$name = $_POST["name"];

$sql = "INSERT INTO user VALUES (null, '$name')";

mysql_query($sql) or die("Could not execute query.");


?>

Successful posted but when I do console.log of result, it shows (an empty string) in firebug. Can anyone help me with this issue please? thanks

I would like to have a return as a JSON type.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T03:58:35+00:00Added an answer on June 9, 2026 at 3:58 am

    Your code should be like this:

    $.ajax({
        url: 'post.php',
        type: "POST",
        data: {name: "dan"},
        success: function(result){
            console.log(result);
        },
        error: function(){
            console.log('error');
        }   
    });
    

    result = what is echoed in your post.php if nothing is echoed or printed, it returns nothing.

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