PHP/MySQL: SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data

Question

I have this MySQL table:

mysql> select * from  members;
+-------+-----------+-----------+
| memid | firstname | lastname  |
+-------+-----------+-----------+
|     1 | billal    | begueradj |
|     2 | bill      | gates     |
|     3 | steve     | jobs      |
+-------+-----------+-----------+
3 rows in set (0.00 sec)

I have this code:

<?php
$output = array('error' => false);
$members = array();

try {
    $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                   'root',
                   ''
    );
} catch(Exception $e) {
    die('Error in connecting to DB: <br/>'.$e->getMessage());
}

$response = $db->query('SELECT * FROM members');
while($row = $response->fetch()){
    echo $row['firstname'].' ';
    echo $row['lastname'].'<br/>';
    array_push($members, $row);
}

$output['members'] = $members;
$response->closeCursor();
$json = json_encode($out);
echo $json; // outputs correctly
header("Content-type: application/json");    // error here
die();
?>

And I am getting this error message when I run the PHP file containing the PHP code above:

SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data

Why this happens?

P.S. Of course, when I comment this line: //header("Content-type: application/json"); the error message disappears

EDIT: New code version following the comments below:

<?php
    header("Content-type: application/json");
    $output = array('error' => false);
    $members = array();

    try {
        $db = new PDO('mysql:host=localhost;dbname=bill;charset=utf8',
                       'root',
                       ''
        );
    } catch(Exception $e) {
        die('Error in connecting to DB: <br/>'.$e->getMessage());
    }

    $response = $db->query('SELECT * FROM members');
    while($row = $response->fetch()){       
        array_push($members, $row);
    }

    $output['members'] = $members;
    $response->closeCursor();
    $json = json_encode($out); 
    //echo $json;    

?>

Still getting the same error message

So after the header() line you've tried echo $json; and you get an error? — Nigel Ren
– Nigel Ren, Commented Aug 13, 2018 at 7:14
Oh ... yes, sorry, ... now it outputs correctly, but the error is still there — Billal BEGUERADJ
– Billal BEGUERADJ, Commented Aug 13, 2018 at 7:17

Anuga · Accepted Answer · 2018-08-13 07:56:15Z

header() should be placed above all output, so before you echo anything.

Plus, ~~since your not encoding the whole thing to JSON, instead only parts,~~ I'm guessing your missing start and end of the JSON.

Correct JSON outputs:

ex1:

{
  "key1": "value1",
  "key2": "value2",
  "key3": "value3"
}

ex2:

[
  {
    "key1": "value1",
    "key2": "value2",
    "key3": "value3"
  },
  {
    "key1": "value1",
    "key2": "value2",
    "key3": "value3"
  },
  {
    "key1": "value1",
    "key2": "value2",
    "key3": "value3"
  }
]

PHP:

<?php

$output = ["error" => false, "members" => []];

try {
  $db = new PDO("mysql:host=localhost;dbname=bill;charset=utf8", "root", "");
} catch(Exception $e) {
  die("Error in connecting to DB: <br/>{$e->getMessage()}");
}

$response = $db->query("SELECT * FROM members");

while($row = $response->fetch(PDO::FETCH_ASSOC)) {
  array_push($output["members"], $row);
}

$response->closeCursor();

$json = json_encode($output);

header("Content-type: application/json");

echo $json;

Vivekanand Panda · Accepted Answer · 2018-08-13 07:27:36Z

1

try thisit may help you out

if ($response->num_rows > 0) {
        while($row[] = $response -> fetch_assoc()) {
            $item = $row;
            $json = json_encode($item);
        }
    }

edited Aug 13, 2018 at 7:27

answered Aug 13, 2018 at 7:17

Vivekanand Panda

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1 Comment

Billal BEGUERADJ Over a year ago

It is the same error message I posted in my question

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