Get location from a stream url in PHP - php

I'm trying to get the redirect url from a stream using php.
Here's the code I have right now:
<?php
$stream = 'https://api.soundcloud.com/tracks/178525956/stream?client_id=XXXXXX';
$ch = curl_init($stream);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$url = curl_exec($ch);
echo $url;
curl_close($ch);
?>
Which outputs as a string:
{"status":"302 - Found","location":"THE_URL_I_WANT"}
So how ould I go about getting the url I want as a variable?
Thanks

It's simple use json_decode
$data = json_decode($url);
$your_url = $data->location;

How about
$data = json_decode($url);
$location = $data["location"];

Related

Roblox API ( Avatar Equipment )

I am currently working on Roblox API. I am stuck on one question. Reason :
I have this link https://avatar.roblox.com/v1/users/2/currently-wearing.
This shows what specified users have equipped on them. this link right here shows this:
{"assetIds":[382537569,607702162,607785314]}
My goal is to get the assetIds to string.
I tried this:
<?php
$id = 2;
$ch = file_get_contents("https://avatar.roblox.com/v1/users/$id/currently-wearing");
$ch = curl_init($ch);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$data = curl_exec($ch);
$data = json_decode($data);
$id = "$data->assetIds";
echo $id;
But it Shows Array. I need some help.
The cURL is not necessary as the file_get_contents() get you the data you appear to require
$id = 2;
$ch = file_get_contents("https://avatar.roblox.com/v1/users/2/currently-wearing");
$data = json_decode($ch);
foreach ($data->assetIds as $id){
echo $id . PHP_EOL;
}
RESULT
382537569
607702162
607785314

get userid from username roblox php

$user = "linkmon99";
$page = file_get_contents("https://api.roblox.com/users/get-by-username?username=$user");
$msg = "> Rolimons : "https://www.rolimons.com/player/$page"
The expected output should be:
https://www.rolimons.com/player/2207291
But it is:
https://www.rolimons.com/player/%7B%22Id%22:2207291,%22Username%22:%224MHF%22,%22AvatarUri%22:null,%22AvatarFinal%22:false,%22IsOnline%22:false%7D
I hope someone can help! Thank you!
The file_get_contents() function reads a file into a string. You should be using cURL instead.
$user = "linkmon99";
$ch = curl_init("https://api.roblox.com/users/get-by-username?username=$user");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$data = curl_exec($ch);
$data = json_decode($data);
echo "> Rolimons : https://rolimons.com/player/$data->Id";
// Output: > Rolimons : https://rolimons.com/player/2207291

How to use json to pass data from a php file to another php file

As mentioned earlier , i am working on the local server as of now in xampp. I have created 2 files index.php and test.php. What i want to achieve is that , index.php will send json data to test.php , with the json data received , the test.php is able to use that json data to turn the statistics into graph.
I am working with the first step , but however , nothing seems to display on my test.php when i tried to do a var_dump($data) , and what i get is NULL. Tried alot of solutions online but none to seems to fix it. I am relatively new to this , so really thanks and appreciate of your help. First php is index.php , second php is test.php.
Do i require a live server as of now in order to see the results in test.php or local server unable to display the result?
<?php
$array = array();
$product = array();
$product[0]['id_product'] = 'A01';
$product[0]['name_product'] = 'Sandal';
$product[0]['price_product'] = '500';
$product[1]['id_product'] = 'A02';
$product[1]['name_product'] = 'Shoes';
$product[1]['price_product'] = '2500';
$array['id'] = '123';
$array['note'] = 'this is my short example';
$array['data'] = $product;
$data = json_encode($array);
$ch = curl_init('http://localhost:8080/practice3/test.php');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_POST, count($data));
$result = curl_exec($ch);
var_dump($result);
?>
```````````````````````````````````
<?php
$fp = fopen('php://input', 'r');
$raw = stream_get_contents($fp);
$data = json_decode($raw,true);
echo "hello";
echo $data['id'];
echo $data['note'];
foreach ($data['data'] as $key) {
echo 'id_product : '.$key['id_product'].'<br/>';
echo 'name_product : '.$key['name_product'].'<br/>';
echo 'price_product : '.$key['price_product'].'<br/>';
}
?>
````````````````````````````````````

How to send JSON and retrieve it using cURL?

This link = http://localhost/api_v2/url?key=***
will return this list of JSON :
I've already tested by :
Navigate to http://localhost/api_v2/url?key=*** through a browser
And make a curl request via command line $ curl http://localhost/api_v2/url?key=***
either way will give return the JSON, and give me the same result.
Well, what I can tell by that is my $array is storing something in it.
Even if I did dd($array) - I still get the same result.
Here is what I've
Here is how I establish my JSON
public function index_2(){
$file_name = 'inventory.csv';
$file_path = 'C:\\QuickBooks\\'.$file_name;
$csv= file_get_contents($file_path);
$utf8_csv = utf8_encode($csv);
$array = array_map("str_getcsv", explode("\n", $utf8_csv));
return Response::json($array);
}
Here is how I make a cURL request and trying to retrieve that JSON
<?php
$ch = curl_init("http://localhost/api_v2/url?key=***");
curl_setopt($ch, CURLOPT_USERPWD, "admin:*****");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$body = curl_exec($ch);
curl_close($ch);
$json_decode = json_decode($array, TRUE);
I keep getting complaint that $array variable is not define, but in fact I did define and send it over like this return Response::json($array); .
I am not sure what I did wrong here.
Spot the difference:
$body = curl_exec($ch);
^^^^^
$json_decode = json_decode($array, TRUE);
^^^^^^

How get content from specific link?

I have link in specific variable eg.
$link = 'http://google.com'
and I try to get content from this link with function fopen.
Eg. : $var = fopen("'".$link."'", "rb");
echo stream_get_contents($var); ,
but without success. Error is
Warning: file_get_contents('http://google.com'): failed to open stream: No such file or directory in /var/www/...
If I use directly
$var = fopen('http://google.com', "rb");
echo stream_get_contents($var)
this work perfectly?
How do I fix this or what method to use if I link is a variable?
Based on your posted code, this worked for me. Try it using this method:
<?php
$link = "http://www.google.com";
$var = fopen($link, "rb");
echo stream_get_contents($var)
?>
This always worked for me.
$url = 'http://google.com';
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$data = curl_exec($ch);
curl_close($ch);

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