How to load URL in json result response from API server with PHP?

Question

<?
$password = substr(md5(microtime()) , rand(0, 26) , 8);
$username = 'SAG' . $password;
$params = ['username' => $username, 'mobile' => 'yes', 'lang' => 'en', 'game_code' => 'lobby', 'page_site' => 'live'];
$curl = curl_init();
curl_setopt_array($curl, array(
    CURLOPT_URL => "https://api.gmaster8.com/BBIN/game/open",
    CURLOPT_CUSTOMREQUEST => "POST",
    CURLOPT_POSTFIELDS => $params,
    CURLOPT_RETURNTRANSFER => true,
    CURLOPT_HTTPHEADER => array(
        "Authorization: Basic U0FHQVBJOjEyMzRxd2Vy"
    ) ,
    CURLOPT_USERPWD => "$username:$password",
));
$result = curl_exec($curl);
echo $result;
?>

I got the $result like this:

{"url":"https:\/\/888.gsoftbb.com\/app\/WebService\/JSON\/display.php\/Login?website=gamingsoft&uppername=dgmaster8rmb&username=C71SAGabfbbcc9&lang=en-us&page_site=live&page_present=live&key=t7ksrtefa9e54610003dfb2b05986de7fcfa6btsmtpho&"}

How can I post the API and then load the url in response from API server? What code should I replace the line

echo $result;

to load the url in json response from API server?

Nicholas Byfleet · Accepted Answer · 2020-06-16 02:42:35Z

1

Try this where you have echo $result

$result = json_decode($result, true);
header("Location: {$result['url']}");

answered Jun 16, 2020 at 2:42

Nicholas Byfleet

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Comments

Juliano Bez · Accepted Answer · 2020-06-16 14:14:37Z

0

You can do it this way:

[...]
$ json = json_decode ($ result);
print_r ($ json);

Or that way if you want to print in a more friendly way:

[...]
$ result = curl_exec ($ curl);
header ("Content-type: application / json");
$ json = json_decode ($ result, JSON_PRETTY_PRINT);
print_r ($ json);

json_decode transforms your string into JSON and json_encode transforms your JSON into a string.

json_decode accepts flags, such as JSON_PRETTY_PRINT. See more at https://www.php.net/manual/pt_BR/function.json-encode.php

EDIT

That said, if you want to redirect to the URL returned in JSON, instead of printing the returned value, just do the following:

[...]
$ result = curl_exec ($curl);
$ json = json_decode ($result);
header("Location: {$result['url']}");

Hope this helps.

edited Jun 16, 2020 at 14:14

answered Jun 16, 2020 at 1:20

Juliano Bez

244 bronze badges

3 Comments

Learn Code Over a year ago

I follow your code but it still only print the result in array with beatiful format: but the page not forward or reload to open URL in json.

Learn Code Over a year ago

I follow your code but it still only print the result in array with beatiful format

Array ( [url] => https://888.gsoftbb.com/app/WebService/JSON/display.php/Login?website=gamingsoft&uppername=dgmaster8rmb&username=C71SAGbc8b93ff&lang=en-us&page_site=live&page_present=live&key=ieiviz7c6ba0db174d9d828db5bbec2c23acc0aglzbdb& )

but not open that URL.

Learn Code Over a year ago

I have try with replace function and I can handle it now

<? ........ $result = str_replace('\/','/',curl_exec($curl)); $result = str_replace('{"url":"','',$result); $result = str_replace('"}','',$result); ?> <a href="<? echo $result; ?>" target="_blank">OPEN URL</a>

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