PHP : Return JSON content instead of String from database [duplicate]

Question

This question already has answers here:

JSON encode MySQL results (15 answers)

Closed 2 years ago.

I would like to return a JSON objet containing informations retrieved from my database. Today, i only check that the username/password couple is correct and return a string "success".

The structure of tbl_user table is the following :

================================
============ tbl_user ==========
================================
- id
- username
- password
- name
- surname
- city
================================
================================

If the username/password couple exist in database, i would like to return a JSON object like that :

{
   username : ..,
   name : ..,
   surname : ..,
   city : ..
}

How can i do this ?

Here is my PHP code :

<?php
$hostname_localhost ="localhost";
$database_localhost ="database";
$username_localhost ="myusername";
$password_localhost ="mypassword";
$localhost = mysql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
trigger_error(mysql_error(),E_USER_ERROR);

mysql_select_db($database_localhost, $localhost);

$username = $_POST['username'];
$password = $_POST['password'];
$query_search = "select * from tbl_user where username = '".$username."' AND password = '".$password. "'";
$query_exec = mysql_query($query_search) or die(mysql_error());
$rows = mysql_num_rows($query_exec);
//echo $rows;
 if($rows == 0) { 
 echo "fail"; 
 }
 else  {
    echo "success"; 
}
?>

Answer 1

First alter your select statement:

select field1, field2, ... from tbl_user where username = ....

then:

$rows = mysql_fetch_assoc($query_exec);

// you will get JSON data of all fields used in your selection
$jsonData = json_encode($rows);

Answer 2

Fetch your results in an array, then encode it to JSON as following:

echo json_encode($result);

Answer 3

Just use php json encode function for you output, like:

echo json_encode($rows);