How do I make my php code correctly output a JSON Array containing objects and arrays

Question

At present my JSON looks like this:

[{
    "user_review_ids": {
        "category": "cat1",
        "name": "name1",
        "phone": "phone1",
        "comment": "com1",
        "reviewid": 32
    }
}, {
    "user_review_ids": {
        "category": "cat2",
        "name": "name2",
        "phone": "phone2",
        "comment": "com2",
        "reviewid": 76
    }
}], [{
    "private_review_ids": {
        "category": "cat1",
        "name": "name1",
        "phone": "phone1",
        "comment": "com1",
        "reviewid": 240
    }
}, {
    "private_review_ids": {
        "category": "cat2",
        "name": "name2",
        "phone": "phone2",
        "comment": "com2",
        "reviewid": 241
    }
}]

Which...is not correct JSON. Jsonlint.com will tell you that.

The way it should be is:

[{
    "user_review_ids": [{
        "category": "cat1",
        "name": "name1",
        "phone": "phone1",
        "comment": "com1",
        "reviewid": 32
    }, {
        "category": "cat2",
        "name": "name2",
        "phone": "phone2",
        "comment": "com2",
        "reviewid": 76
    }],
    "private_review_ids": [{
        "category": "cat1",
        "name": "name1",
        "phone": "phone1",
        "comment": "com1",
        "reviewid": 240
    }, {
        "category": "cat2",
        "name": "name2",
        "phone": "phone2",
        "comment": "com2",
        "reviewid": 241
    }]
}]

Can you tell me what I need to do with my php code so it outputs the correct JSON?

The code right now is:

<?php
    require('file.php');

    $UserReviewID = ('32,76');
    $UserReviewID = explode(",",$UserReviewID);

    $PrivateReviewID = ('240,241');
    $PrivateReviewID = explode(",",$PrivateReviewID);

    //for user_review_ids
    $results = array();

    //for private_review_ids
    $results2 = array();

    foreach($UserReviewID as $UserReviewID) {
        $sql2 = "SELECT * FROM review WHERE review_id = ?";
        $stmt2 = $con->prepare($sql2) or die(mysqli_error($con));
        $stmt2->bind_param('i', $UserReviewID) or die ("MySQLi-stmt binding failed ".$stmt2->error);
        $stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);
        $result2 = $stmt2->get_result();

        while($row = mysqli_fetch_array($result2)) {//make an array called $results
            $results[] = array('user_review_ids' => array(
                'category' => $row['cat_name'],
                'name' => $row['name'],
                'phone' => $row['phone'],
                'comment' => $row['comment'],
                'reviewid' => $row['review_id'],
            ));
        }
    }

        foreach($PrivateReviewID as $PrivateReviewID) {
        $sql2 = "SELECT * FROM review WHERE review_id = ?";
        $stmt2 = $con->prepare($sql2) or die(mysqli_error($con));
        $stmt2->bind_param('i', $PrivateReviewID) or die ("MySQLi-stmt binding failed ".$stmt2->error);
        $stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);
        $result2 = $stmt2->get_result();

        while($row = mysqli_fetch_array($result2)) {//make an array called $results2
            $results2[] = array('private_review_ids' => array(
                'category' => $row['cat_name'],
                'name' => $row['name'],
                'phone' => $row['phone'],
                'comment' => $row['comment'],
                'reviewid' => $row['review_id'],
            ));
        }
    }

        echo json_encode($results) .",";
        echo json_encode($results2);
?>

You cant json_encode twice, instead join the arrays and do it once. — Lawrence Cherone
– Lawrence Cherone, Commented Apr 18, 2018 at 21:31

ADyson · Accepted Answer · 2018-04-18 21:46:05Z

Concatenating the outputs of multiple json_encode operations does not produce valid JSON. You need to create a data structure such than you can run json_encode once and include everything in it.

Also, to get the structure you showed, you need to change how you create the arrays resulting from the database rows - you're creating multiple items with the same index.

This should work:

<?php
    require('file.php');

    $UserReviewID = ('32,76');
    $UserReviewID = explode(",",$UserReviewID);

    $PrivateReviewID = ('240,241');
    $PrivateReviewID = explode(",",$PrivateReviewID);

    //for user_review_ids
    $results = array();

    //for private_review_ids
    $results2 = array();

    foreach($UserReviewID as $UserReviewID) {
        $sql2 = "SELECT * FROM review WHERE review_id = ?";
        $stmt2 = $con->prepare($sql2) or die(mysqli_error($con));
        $stmt2->bind_param('i', $UserReviewID) or die ("MySQLi-stmt binding failed ".$stmt2->error);
        $stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);
        $result2 = $stmt2->get_result();

        while($row = mysqli_fetch_array($result2)) {//make an array called $results
            $results[] = array(
                'category' => $row['cat_name'],
                'name' => $row['name'],
                'phone' => $row['phone'],
                'comment' => $row['comment'],
                'reviewid' => $row['review_id'],
            ));
        }
    }

        foreach($PrivateReviewID as $PrivateReviewID) {
        $sql2 = "SELECT * FROM review WHERE review_id = ?";
        $stmt2 = $con->prepare($sql2) or die(mysqli_error($con));
        $stmt2->bind_param('i', $PrivateReviewID) or die ("MySQLi-stmt binding failed ".$stmt2->error);
        $stmt2->execute() or die ("MySQLi-stmt execute failed ".$stmt2->error);
        $result2 = $stmt2->get_result();

        while($row = mysqli_fetch_array($result2)) {//make an array called $results2
            $results2[] = array(
                'category' => $row['cat_name'],
                'name' => $row['name'],
                'phone' => $row['phone'],
                'comment' => $row['comment'],
                'reviewid' => $row['review_id'],
            ));
        }
    }

    $combinedResults = array(array('user_review_ids' => $results, 'private_review_ids' => $results2));
    echo json_encode($combinedResults);
?>

Having said that, I'm not convinced you actually need the very outer array in your sample result, it doesn't seem to serve much purpose since it only has one - fixed - item in it. If you wanted to ditch that, then the last- but-one line would be

    $combinedResults = array('user_review_ids' => $results, 'private_review_ids' => $results2);

instead, which will just return a single object with those field names.

Andrew Chart · Accepted Answer · 2018-04-18 21:33:39Z

1

I think you just need to json_encode a single array. Perhaps something like:

echo json_encode( array_merge($results, $results2) );

answered Apr 18, 2018 at 21:33

Andrew Chart

6556 silver badges11 bronze badges

5 Comments

CHarris Over a year ago

Thanks, however I am getting the keys user_review_ids and private_review_ids repeated in each object whereas I just want them once, let me investigate...

ADyson Over a year ago

@CHarris this is because you create those key sagain every time you output a row from the respective DB tables. See my answer for a solution which avoids that

Andrew Chart Over a year ago

I was wondering whether actually you should just be pushing the array to a variable rather than an array. I.e. $results = array(...); instead of $results[] = array(...)

ADyson Over a year ago

@AndrewChart since that code is in a loop, that would simply overwrite one variable lots of times, and result in only the last row of data ever being returned from each table.

Andrew Chart Over a year ago

Ah yes I see. Looking too quickly by eye, sorry.

Apps-n-Add-Ons · Accepted Answer · 2018-04-18 21:34:15Z

1

instead of

 echo json_encode($results) .",";
 echo json_encode($results2);

use

echo json_encode(array_merge($results,$results2));

answered Apr 18, 2018 at 21:34

Apps-n-Add-Ons

2,1651 gold badge19 silver badges31 bronze badges

Comments

Diego Avila · Accepted Answer · 2018-04-18 21:30:51Z

-2

You can use the inbuilt php function json_encode() example:

<?php
$pdo = new PDO("mysql:dbname=database;host=127.0.0.1", "user", "password");
$statement = $pdo->prepare("SELECT * FROM table");
$statement->execute();
$results = $statement->fetchAll(PDO::FETCH_ASSOC);
$json = json_encode($results);

good luck !

answered Apr 18, 2018 at 21:30

Diego Avila

7387 silver badges25 bronze badges

2 Comments

Lawrence Cherone Over a year ago

OP is using mysqli

ADyson Over a year ago

this is a generic example of no relevance or direct benefit to the question. It uses the wrong DB library and entirely misses the point that OP already knows about json_encode, but is having an issue with the specific format of the data being output

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