How to encode multiple json elements using PHP?

Question

I'm trying to create a questionnaire app in Android and I can already get the questions from the database and show it as JSON Array but it will be displayed in this format:

{
    "result": [
        {
            "id_question": "1",
            "question_name": "Grade level",
            "choices": "Grade 11, Grade 12"
        },
        {
            "id_question": "2",
            "question_name": "Expected grade in this subject",
            "choices": "90-100, 75-89, 60-74, Below 60"
        }
    ]
}

But the library I'm using in Android is only accepting this kind of JSON format for showing the questions:

{
  "survey_properties": {
    "intro_message": "To get a reliable result for the evaluation, please respond to all questions.",
    "end_message": "Your answers have been recorded. <br>Thank you for taking the time to answer the evaluation."
  },
  "questions": [
    {
      "id_question": "1",
      "question_name": "Grade Level",
      "choices": [
        "Grade 11",
        "Grade 12"
      ]
    },
    {
      "id_question": "2",
      "question_name": "Expected Grade in this subject",
      "choices": [
        "90-100",
        "75-89",
        "60-74",
        "Below 60"
      ]
    }
  ]
}

How can I achieve this kind of output in PHP? This is the script I'm using:

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();

while($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        "choices"=>$row['choices']
        )
    );
}

echo json_encode(array("result"=>$result));

Loop through the array, explode the choices into an array et voila? — Martijn
– Martijn, Commented Sep 19, 2017 at 7:29
Thanks! Got the solution for the choices now from your and @tuan-duong 's answer. My problem now is how to add the survey_properties into the array — ppgdn
– ppgdn, Commented Sep 19, 2017 at 9:58

Tuan Duong · Accepted Answer · 2017-09-19 07:30:37Z

1

From your code, it could be modified like this:

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();

while ($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        "choices"=>explode(', ', $row['choices'])
    ));
}

echo json_encode(array("result"=>$result));

You have to add more information like survey_properties

answered Sep 19, 2017 at 7:30

Tuan Duong

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1 Comment

ppgdn Over a year ago

This worked! Thanks! :) how can I add the survey_properties? I tried to array_push the data but I can't seem to get the same output for the survey_properties like the format above.

David Witka · Accepted Answer · 2017-09-19 08:33:22Z

0

What I'm trying to do here is put the "choices" into an array. Hope that works.

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();
$choices = array();

while($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        array_push($choices,array($row['choices'])
        )
    );
}

echo json_encode(array("result"=>$result));

edited Sep 19, 2017 at 8:33

answered Sep 19, 2017 at 8:10

David Witka

945 bronze badges

1 Comment

ppgdn Over a year ago

This showed a Parse error: syntax error, unexpected ';', expecting ',' or ')'

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