Output JSON array append additional element PHP

Question

I'm pretty sure I'm quite close so hopefully quick answer.

I'm able to generate this JSON:

apps: [
    {
        0: {
            PublisherCount: "7"
    },
        Id: "87",
        AppName: "Productivity, Focus, Habits & Life Success by Audiojoy",
        AppBundle: "productivitymind"
    }
]

But I'm trying to get to:

apps: [
    {
        Id: "87",
        AppName: "Productivity, Focus, Habits & Life Success by Audiojoy",
        AppBundle: "productivitymind",
        PublisherCount: "7"
    }
]

Here is my output loop (I think the issue is in the 5th row where I array_push the new value for PublisherCount. It creates an additional node instead of adding it to the end.

$temp_array = array();
$i = 0;
while ($row = mysqli_fetch_assoc($publisher_apps)) {
$temp_array[] = $row;
$temp_array[$i][] = fetch_all(get_publisher_count_by_app_id($row['Id']))[0];
$i++;
}

$publisher_apps = $temp_array;

$result = array("apps"=>$publisher_apps);

output_json($result);

Thanks.

hum, try $temp_array[$i]['PublisherCount'] = fetch_all(get_publisher_count_by_app_id($row['Id']))[0]['PublisherCount']; (and indent your code) — nicolast
– nicolast, Commented Feb 13, 2017 at 23:33

Don't Panic · Accepted Answer · 2017-02-13 23:49:19Z

1

You have rows like this:

['Id' => "87",
 'AppName' => "Productivity, Focus, Habits & Life Success by Audiojoy",
 'AppBundle' => "productivitymind"]

and fetch_all(get_publisher_count_by_app_id($row['Id']))[0] returns an array like this:

['PublisherCount' => 7]

so when you append it with $temp_array[$i][], the entire array gets assigned to the 0 key of $temp_array[$i].

There are various different ways you could get just the PublisherCount value. One way is to use array_merge to combine the result of get_publisher_count_by_app_id with $row, and then add the modified $row to your main array.

while ($row = mysqli_fetch_assoc($publisher_apps)) {
    $count = fetch_all(get_publisher_count_by_app_id($row['Id']))[0];
    $temp_array[] = array_merge($row, $count);
}

If you do it this way, $i should become unneccessary.

answered Feb 13, 2017 at 23:49

Don't Panic

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1 Comment

Jeff Solomon Over a year ago

Yup! @Dont' Panic that worked. Thank you. Makes sense, i kept getting a separate key for it.

ablopez · Accepted Answer · 2017-02-13 23:49:59Z

0

Change it to this:

$temp_array = array();

while ($row = mysqli_fetch_assoc($publisher_apps)) {
  $row['PublisherCount'] = fetch_all(get_publisher_count_by_app_id($row['Id']))[0]['Pub‌lisherCount'];
  $temp_array[] = $row;
}

$publisher_apps = $temp_array;

$result = array("apps"=>$publisher_apps);

output_json($result);

edited Feb 13, 2017 at 23:49

answered Feb 13, 2017 at 23:33

ablopez

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2 Comments

Jeff Solomon Over a year ago

this results in this: apps: [ { Id: "87", AppName: "Productivity, Focus, Habits & Life Success by Audiojoy", AppBundle: "productivitymind", }, { PublisherCount: { PublisherCount: "12" } } ]

ablopez Over a year ago

My bad, I've updated my answer. Please the updated code instead. However, after you get it working, rather than using a "fetch_all" function you should be looking to use a "fetch_one" or "fetch_first" kind of function, if one exists in your codebase, for performance purposes.

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