Parse Nested JSON with PHP

Question

I've look high and low for a solution for this, but haven't found anything that is exactly what I am running into. I am using php to parse a JSON response that is nested in a way that makes it difficult to extract specific fields.

Here is a bit of the JSON:

{
    "collection":{
    "items":[
        {
            "href":"https://api.teamsnap.com/v3/members/MEMBERID",
            "data":[
                "name":"id",
                "value": 0000000 // MEMBERID
            },
            {
                "name":"type",
                "value":"member"
            },
            {
               "name":"address_city",
               "value":""
            },
            {
               "name":"address_state",
               "value":""
            },

Here is my php:

$json = file_get_contents($url, false, $context);

$result = json_decode($json);
foreach ($result->collection->items as $items=>$val) {
    foreach ($val->data as $data=>$datasets) {
        foreach ($datasets as $dataset=>$val) {
            echo $dataset.': '.$val;
            echo "<br>";  
        }
    }
};

and this is a bit of the current output:

name: id
value: MEMBERID
name: first_name
value: MEMBERNAME
name: last_name
value: MEMBERNAME

what I want to do is be able to list all of the members in this response with their first and last name. I have tried using $val->first_name, $val['first_name']; and such but all result in a foreach error.

That JSON structure you posted is invalid. You cannot close an array opened with a square brace with a curly brace as done below the MEMBERID value. — arkascha
– arkascha, Commented Mar 13, 2017 at 6:44
That is just a sample and I changed the actual member id to protect security. I have updated the json in my code above — RiotAct
– RiotAct, Commented Mar 13, 2017 at 6:46
Certainly that is just an example. But still it is invalid. We cannot help to fix your code if the structure of the data is invalid, thus unclear. — arkascha
– arkascha, Commented Mar 13, 2017 at 6:48
You're probably missing an opening curly brace ({) before your first data member. — Kep
– Kep, Commented Mar 13, 2017 at 6:49
I am not writing the json, it is coming from another server. I am trying to parse the json with php. — RiotAct
– RiotAct, Commented Mar 13, 2017 at 6:50

Kep · Accepted Answer · 2017-03-13 06:58:26Z

3

If you can ensure every data member has a name and value field you can use something like this and skip a for-loop:

$json = file_get_contents($url, false, $context);

$result = json_decode($json);
foreach ($result->collection->items as $itemKey => $val) {
    $data = array();
    foreach ($val->data as $dataKey => $dataset) {
        $data[$dataset->name] = $dataset->value;
    }
    var_dump($data);
};

Although you should probably check for $result->collection and $result->collection->items first before looping over them to avoid errors.

edited Mar 13, 2017 at 6:58

answered Mar 13, 2017 at 6:52

Kep

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5 Comments

RiotAct Over a year ago

That resulted in the following error: Cannot use a scalar value as an array

Kep Over a year ago

Be sure to check the variable names, I renamed the key in the foreach loop from $data to $dataKey - make sure you don't overwrite the array there.

RiotAct Over a year ago

Ok so that resulted in an array output of the entire string. How do I access a single variable like first_name?

Kep Over a year ago

It's just a normal array, just use $data["first_name"] after the loop, if it was part of the dataset it'll be in the $data array afterwards.

RiotAct Over a year ago

You sir, are the bomb!

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