output error in php json-decode

Question

my php code generate a non-Valid json output error

my php code :

$questions = array();
while($question = mysql_fetch_array($result, MYSQL_ASSOC)) {
$questions[] = array('question'=> $question);
}
print_r ($questions);

$newQuestions = array('questions' => array());

foreach($questions as $key => $question){
    $newQuestion = array(
            'question' => $question['question']['question'],
            'correct' => $question['question']['correct'],
            'answers' => array(
                    $question['question']['answer1'],
                    $question['question']['answer2'],
                    $question['question']['answer3'],
                    $question['question']['answer4']
                )
    );

    $newQuestions['questions'][] = $newQuestion;

}

$output = json_encode(($newQuestions),JSON_UNESCAPED_UNICODE);

echo '<br/><br/>';
echo $output;

table fields :

Question :
correct  :
answer 1 :
answer 2 :
answer 3 :
answer 4 :

Example :

Question : is php a good language ?
correct  : 1
answer 1 : yes
answer 2 : no
answer 3 : maybe
answer 4 : good

the output is OK, and formated as I want.

output sample : http://pastebin.com/eefS7KYW

I am sure my php code is correct but I don't know where is exactly the issue !!

============== Fixed : it was just two echo $output !

json_encode works fine. The issue must be in your code, but I cannot see what exactly from handpainted screenshots. It would seem you're outputting the result twice. — Niels Keurentjes
– Niels Keurentjes, Commented Sep 4, 2014 at 23:17
@RicardoE as questions is the only key in the root object it's more likely a double print, otherwise it wouldn't appear right after the 'exception' line. — Niels Keurentjes
– Niels Keurentjes, Commented Sep 4, 2014 at 23:27
@RicardoE json output is missed up in one place only but the whole output is correct and formated properly . — saad
– saad, Commented Sep 4, 2014 at 23:31
Just FYI there are other issues with your code. We are long past using mysql_ functions, use mysqli or pdo instead. You also use variables needlessly. — php_nub_qq
– php_nub_qq, Commented Sep 4, 2014 at 23:32

nwolybug · Accepted Answer · 2014-09-05 03:17:55Z

1

Seems like you are doing a lot of variable passing which gets very confusing, very quickly. Especially when all the variables are iterations of 'question'. It appears that you are creating an array from information pulled from a database in the format [questions[question,correct,answers[1,2,3,4]]] This code format may work better?

   $newQuestions = array();
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
  $newQuestions['questions'][] = array(
    'question'  =>  $row['question'],
    'correct'   =>  $row['correct'],
    'answers'   =>  array(
        $row['answer1'],
        $row['answer2'],
        $row['answer3'],
        $row['answer4']
        )
    );
}

    $output = json_encode(($newQuestions),JSON_UNESCAPED_UNICODE);

    echo '<br/><br/>';
    echo $output;

Had a semicolon in the wrong place.Fixed the code above and test the code with the following:

<?php

$array = array(
    array('question'=>'Question1',
        'correct'=>3,
        'answer1' => 'Q1Answer1',
        'answer2' => 'Q1Answer2',
        'answer3' => 'Q1Answer3',
        'answer4' => 'Q1Answer4'
    ),
    array('question'=>'Question2',
        'correct'=>3,
        'answer1' => 'Q2Answer1',
        'answer2' => 'Q2Answer2',
        'answer3' => 'Q2Answer3',
        'answer4' => 'Q1Answer4'
    ),
    array('question'=>'Question3',
        'correct'=>3,
        'answer1' => 'Q3Answer1',
        'answer2' => 'Q3Answer2',
        'answer3' => 'Q3Answer3',
        'answer4' => 'Q1Answer4'
    )
);

$newQuestions = array();
//while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
foreach($array as $row){
    $newQuestions['questions'][] = array(
    'question'  =>  $row['question'],
    'correct'   =>  $row['correct'],
    'answers'   =>  array(
        $row['answer1'],
        $row['answer2'],
        $row['answer3'],
        $row['answer4']
        )
    );
}

print_r($newQuestions);
$output = json_encode(($newQuestions),JSON_UNESCAPED_UNICODE);

echo '<br/><br/>';
echo $output;

?>

edited Sep 5, 2014 at 3:17

answered Sep 4, 2014 at 23:53

nwolybug

4825 silver badges13 bronze badges

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

2 Comments

nwolybug Over a year ago

Had a semicolon in the wrong spot. Tested and fixed.

saad Over a year ago

it works ! thank you ! it was just twice echo that print 2 output :D

Collectives™ on Stack Overflow

output error in php json-decode

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related