Php outputing Json encoded data from mysql database

Question

I am just having an error in my output through my php file. I believe it is an easy fix. While it should be displaying data it is displaying this.

[{"first_name":"first_name","mobile_phone":"mobile_phone"},`

and here is my php file. Really hope this is a simple fix. Thanks

   <?php

    function connect() {


        // Connecting to mysql database
        $con = mysql_connect(private);

        // Selecing database
        $db = mysql_select_db("intraweb_db") or die(mysql_error()) or die(mysql_error());

        // returing connection cursor
        return $con;
    }




//end of connect to db
connect();

$sql=mysql_query("SELECT 'first_name', 'mobile_phone' FROM `admin_contacts`");

$output = array();
while($r = mysql_fetch_assoc($sql)){
    $output[]=$r;

}
echo json_encode($output);

?>

Awesome ;) thank you guys. got it working

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the pink box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — h2ooooooo
– h2ooooooo, Commented Jan 12, 2014 at 16:48
mysql_* is deprecated, consider using mysqli or PDO. Can you please use print_r($output) before the json_encode function and update us - what's the output. — Ofir Baruch
– Ofir Baruch, Commented Jan 12, 2014 at 16:50
Use backticks or no quotes for your column names, for one thing. — Funk Forty Niner
– Funk Forty Niner, Commented Jan 12, 2014 at 16:50
Funny how you think that if mysql_select_db() dies it can die again. — user555
– user555, Commented Jan 12, 2014 at 16:51

Awlad Liton · Accepted Answer · 2014-01-12 16:57:17Z

2

You are doing wrong in your query. don't use single or double quote in your column name.

Try this:

$sql=mysql_query("SELECT first_name,mobile_phone FROM `admin_contacts`");
$output = array();
while($r = mysql_fetch_assoc($sql)){
    $output[]=$r;    
}
echo json_encode($output);

?>

You also need to change in this line:

$db = mysql_select_db("intraweb_db") or die(mysql_error()) or die(mysql_error());

to

$db = mysql_select_db("intraweb_db") or die(mysql_error());

edited Jan 12, 2014 at 16:57

answered Jan 12, 2014 at 16:50

Awlad Liton

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2 Comments

towr Over a year ago

To elaborate, the single quotes around 'first_name' and 'mobile_phone' makes mysql return them as strings, rather than the values in the columns with those names. (You should use backticks for that)

Awlad Liton Over a year ago

@user3169485 : did you try this? if it worked then accept an answer.

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