Converting JSON from URL to CSV with PHP

Question

When I run the code below it results in the first row in a long list of columns containing only the text Array.

I've used $decoded = json_decode($json_file, true); to indicate that I want arrays instead of objects so I'm not sure what the issue is.

Should I be using bracket or arrow notation somewhere to specify that I want the array values?

$json_file = file_get_contents('https://fakeurl.com&contentType=json', false);

$decoded = json_decode($json_file, true);

$fp = fopen('output.csv', 'w');
foreach($decoded as $comment) {
    fputcsv($fp, $comment);
}
fclose($fp);

Here is a small snippet of the JSON. It looks like this all the way through:

{
    "rows": [{
        "columns": [{
            "name": "clientId",
            "value": "1839",
            "type": "xs:int",
            "format": ""
        }, {
            "name": "campaignId",
            "value": "25646",
            "type": "xs:int",
            "format": ""
        }, {
            "name": "campaignStatus",
            "value": "Live",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "campaignName",
            "value": "Template Donation Litle",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "campaignExportName",
            "value": "Template Donation Litle",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "description",
            "value": "/donate/template/Litle",
            "type": "xs:string",
            "format": ""
        }]
    }, {
        "columns": [{
            "name": "clientId",
            "value": "1839",
            "type": "xs:int",
            "format": ""
        }, {
            "name": "campaignId",
            "value": "25812",
            "type": "xs:int",
            "format": ""
        }, {
            "name": "campaignStatus",
            "value": "Live",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "campaignName",
            "value": "Monthly Only",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "campaignExportName",
            "value": "Monthly Only",
            "type": "xs:string",
            "format": ""
        }, {
            "name": "description",
            "value": "A donation receipt will be emailed to the address you submitted. This donation is tax-deductible to the fullest extent of the law.",
            "type": "xs:string",
            "format": ""
        }]
    }]
}

Do a print_r($decoded); so we can see what $decoded actually contains. We can't guess. — M. Eriksson
– M. Eriksson, Commented Aug 14, 2017 at 20:26
Hey @War10ck and Magnus I've added a small portion of the JSON — gabed123
– gabed123, Commented Aug 14, 2017 at 20:37

AbraCadaver · Accepted Answer · 2017-08-14 20:38:12Z

4

Your array is much deeper and the values need to be extracted:

foreach($decoded['rows'] as $row) {
    $values = array_column($row['columns'], 'value');
    fputcsv($fp, $values);
}

If you want the names/headers in the first row then you need to do this before the loop:

$headers = array_column($decoded['rows'][0]['columns'], 'name');
fputcsv($fp, $headers);

answered Aug 14, 2017 at 20:38

AbraCadaver

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4 Comments

gabed123 Over a year ago

array_column expects parameter 1 to be an array but null is given

AbraCadaver Over a year ago

3v4l.org/GiGmH Then the JSON didn't decode, see php.net/manual/en/function.json-last-error.php

gabed123 Over a year ago

But I used this decode function: $decoded = json_decode($json_file, true);

gabed123 Over a year ago

Hey thanks man I play around with what was in the foreach loop but I managed to figure it out. Thanks!

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