PHP to JSON encode data not working

Question

I want to select all the recipes in my food database and encode them into a simple JSON response. I've looked at solutions around stack overflow and have implemented a sensible looking script.

No JSON array is returned / echoed.

Could someone tell me where I'm going wrong?

<?php
$type = "recipe";
$conn = mysqli_connect("localhost", "root", "", "food"); // SAMPLE CREDENTIALS
$sql = "SELECT * FROM wp_posts WHERE post_type = '$type' ";
$query = mysqli_query($conn, $sql);

$i = 0; // INCREMENT
$arr = [];

if($query) {
    while($row = mysqli_fetch_assoc($query)) {
        $jsonArrayObject = (array('title' => $row['post_title'] , 'excerpt' => $row['post_content']));
        $arr[$i] = $jsonArrayObject;
        $i++;
    }
    $json_array = json_encode($arr, JSON_PRETTY_PRINT);
    echo $json_array;
}
else {
    echo "Error in database";
}
?>

ARR VARIABLE PRINT OUT

I can't see the error but php isn't a language I know well. However standard debugging should solve this quite quickly... add an echo inside the loop for each jsonArrayObject. The results should point you straight at the problem (or at least give you more information for the question here) — PeteB
– PeteB, Commented Jan 26, 2016 at 11:39
@pritesh I have added a screenshot of the print out of the $arr variable. — WebDevDanno
– WebDevDanno, Commented Jan 26, 2016 at 11:45

Andreas Schrammel · Accepted Answer · 2016-01-26 11:49:29Z

1

Check if your PHP version is 5.4.0 or above.

The option JSON_PRETTY_PRINT was introduced in that version according to http://php.net/manual/en/json.constants.php

If your PHP version is below 5.4.0 leave the option.

answered Jan 26, 2016 at 11:49

Andreas Schrammel

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