correct way to return a json_object as an array

Question

I have an array populated using an sql statement in the following manner:

$index = 0;

while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
    $bookname[$index] = ($row['Bookname']);
    $subjectname[$index] = ($row['SubjectName']);
    $index++;
}

When I go to echo json encode the Arrays I get a blank [] when I know it has been populated which is really weird. Am I doing anything wrong in my context

echo json_encode($Bookname,$SubjectName);

Second param of json_encode() sets options on how to encode the data. Do not use some of your data in that place or you'll ge t really weired results. See php.net/manual/en/function.json-encode.php — maxhb
– maxhb, Commented Feb 10, 2016 at 11:32
@maxhb even with one param lets use $bookname it still gives [] when I test echo $Bookname[2]; it works perfectly fine and prints that element! — Eoin Ó Cribín
– Eoin Ó Cribín, Commented Feb 10, 2016 at 11:40
use var_dump(json_encode($Bookname)); or var_dump(json_encode($SubjectName)); you can not print an object or array using echo — Hirdesh Vishwdewa
– Hirdesh Vishwdewa, Commented Feb 10, 2016 at 11:45

devpro · Accepted Answer · 2016-02-10 11:33:12Z

2

You can use json_encode as like that:

<?php
$index = 0;
$data = array();
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
    $data[$index]['bookname'] = $row['Bookname'];
    $data[$index]['subjectname'] = $row['SubjectName'];
    $index++;
}
json_encode($data); // encode your array
?>

answered Feb 10, 2016 at 11:33

devpro

16.1k3 gold badges29 silver badges39 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Eoin Ó Cribín Over a year ago

Still getting [] I understand that two params shouldn't be used but even with one param nothing happens

devpro Over a year ago

@EoinÓCribín: please use print_r($row); inside the loop and share the result.

Eoin Ó Cribín Over a year ago

Nothing printed at all. but if I test using my original code and say for example: echo $bookname[0] after the loop I get a resut

devpro Over a year ago

@EoinÓCribín: u can not use $booksname[0] with my example... well, please use print_r($data); after loop (outside the loop)

Eoin Ó Cribín Over a year ago

just a blank array (Array ( )) prints I was sure data was going into it.

|

user5570620 · Accepted Answer · 2016-02-10 11:34:11Z

2

Try following:

       $index = 0;
       $data = array();
       while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
      {
        $data['Bookname'][$index] = $row['Bookname'] 
        $data['SubjectName'][$index] = $row['SubjectName'];
        $index++;

       }
       echo json_encode($data);

answered Feb 10, 2016 at 11:34

user5570620

Comments

Imran · Accepted Answer · 2016-02-11 05:08:47Z

1

You have passed two parameter while calling json_encode function. You batter combine two array in one. Then call the json_encode function like json_encode($combinedArray)

edited Feb 11, 2016 at 5:08

answered Feb 10, 2016 at 11:33

Imran

4,7802 gold badges23 silver badges40 bronze badges

3 Comments

devpro Over a year ago

second param of json_encode is for behaviour php.net/manual/en/function.json-encode.php

Hirdesh Vishwdewa Over a year ago

It takes only one parameter ??? I doubt on this line of your answer... see @devpro 's comment.

devpro Over a year ago

and third param is depth :) brother

Collectives™ on Stack Overflow

correct way to return a json_object as an array

3 Answers 3

8 Comments

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

8 Comments

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related