how to convert multiple mysql result sets into json array

Question

i am executing multiple "select" statements in my php script. Then I am adding them to array by $Output=array($out, $out1); and then encoding it by json_encode()

the output is coming in format [ [{},{}] , [{}, {}] ]

how do i convert it in form - {"str1": [{},{}], "str2": [{},{}]}

php script:

<?php

require 'DbConnect.php';


$query = ("SELECT * FROM `cars`");
$query1 = ("SELECT * FROM `City`");

if ($query_run = mysql_query($query)){

while ($query_row = mysql_fetch_assoc($query_run)){

    $out [] = $query_row;

}

}else{
    echo 'Fail';
} 

//second query 

if ($query_run1 = mysql_query($query1)){

while ($query_row1 = mysql_fetch_assoc($query_run1)){

$out1 [] = $query_row1;

}

}else{
    echo 'Fail';
} 

$Output=array($out, $out1); 
echo(json_encode($Output));
?>

[ [{},{}] , [{}, {}] ] is ur output ...but before encoding put it in array — KOTIOS
– KOTIOS, Commented Jul 29, 2013 at 9:58

Rajeev Ranjan · Accepted Answer · 2013-07-29 10:03:23Z

1

if you are getting result in [ [{},{}] , [{}, {}] ]

then try

$Output=array('str1'=>$out,'str2'=>$out1); 
  echo(json_encode($Output,JSON_FORCE_OBJECT));

answered Jul 29, 2013 at 10:03

Rajeev Ranjan

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Comments

Padmanathan J · Accepted Answer · 2013-07-29 09:58:05Z

0

try this

    $query = "SELECT 
    post_title, 
    guid 
    FROM testwp_posts
    WHERE 
    post_author = $user_id 
    AND
    post_status = 'publish' 
    ORDER BY ID DESC LIMIT $number_of_posts";

        $result = mysql_query($query) or die('Errant query:  '.$query);

        /* create one master array of the records */
        $posts = array();
        if(mysql_num_rows($result)) {
            while($post = mysql_fetch_assoc($result)) {
                $posts[] = array('post'=>$post);
            }
        }


            header('Content-type: application/json');
            echo json_encode(array('posts'=>$posts));

answered Jul 29, 2013 at 9:58

Padmanathan J

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1 Comment

Carbon Over a year ago

wat is $link in your answer?

Muhammad Hamizi Jaminan · Accepted Answer · 2013-07-29 10:02:38Z

0

<?php

require 'DbConnect.php';


$query = ("SELECT * FROM `cars`");
$query1 = ("SELECT * FROM `City`");

if ($query_run = mysql_query($query))
{
    while ($query_row = mysql_fetch_assoc($query_run))
    {   
        $output['cars'][] = $query_row;
    }

}
else
{
    echo 'Fail';
} 

//second query 

if ($query_run1 = mysql_query($query1))
{
    while ($query_row1 = mysql_fetch_assoc($query_run1))
    {
        $output['city'][] = $query_row1;
    }
}
else
{
    echo 'Fail';
} 

echo(json_encode($output));
?>

may it helps

answered Jul 29, 2013 at 10:02

Muhammad Hamizi Jaminan

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Comments

Shehary · Accepted Answer · 2018-06-22 09:00:19Z

0

I am not sure if this helps your requirement, but try below things..

Instead of using

$out [] = $query_row;

Use

$out['str1'][] = $query_row;

AND Instead of

$out1 [] = $query_row1;

Use

$out['str2'][] = $query_row1;

And now try this

echo(json_encode($out));

edited Jun 22, 2018 at 9:00

Shehary

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answered Jul 29, 2013 at 10:18

Ravinder Reddy

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