PHP | Error when sending JSON POST to API URL

Question

I am trying to send JSON to an API url. I am sending the json from test.php and I am receiving the data from api.php. I want to log the data into a text file, but nothing is logged, I assume that $data is returning null? In my test.php file it keeps returning "Sent BAD, request terminated." What am I doing wrong? How would I fix this?

test.php

<?php

$ch = curl_init();

$array = array('foo' => 'bar');

curl_setopt($ch, CURLOPT_URL,"https://dropchat.net/api.php");
curl_setopt($ch, CURLOPT_POST, 1);

curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($array));

curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

$server_output = curl_exec($ch);

curl_close ($ch);

if ($server_output == "OK") { 
    echo "Sent OK";
} else { 
    echo "Sent BAD, request terminated.";
}

?>

api.php

<?php

$data = json_decode(file_get_contents('php://input'), true);

$fp = fopen("myText.txt","wb");
fwrite($fp, $data);
fclose($fp);

?>

I assume...don't assume, test. And configure PHP to log errors and warnings to a file. — ADyson
– ADyson, Commented Feb 2, 2021 at 20:38
if ($server_output == "OK") { will never be true because api.php doesn't echo "OK" (or anything else, for that matter). So that doesn't prove anything one way or another. — ADyson
– ADyson, Commented Feb 2, 2021 at 20:38
The 2nd argument to fwrite() should be a string. $data is an array. — Barmar
– Barmar, Commented Feb 2, 2021 at 20:41

Barmar · Accepted Answer · 2021-02-02 20:49:43Z

1

Add error checking to api.php so you'll know what happened.

You can't write an array to a file, you can only write a string. So write the foo element of the array.

<?php
$data = json_decode(file_get_contents('php://input'), true);
if ($data === null) {
    echo json_last_error_msg();
} else {
    file_put_contents("myText.txt", $data['foo']);
    echo "OK";
?>

Then change the caller to print the error message:

$server_output = curl_exec($ch);

curl_close ($ch);

if ($server_output == "OK") { 
    echo "Sent OK";
} else { 
    echo "Sent BAD, reason: $server_output";
}

answered Feb 2, 2021 at 20:49

Barmar

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