Php api via array

Question

Am trying to read a script api which returns an array:

$api_key = "Uez4TWYH6OAQQHoUcICWJ8UUFYmwQ";
$file_id = "LNzcOp2b1352047884";
$ch=curl_init(); 
curl_setopt($ch,CURLOPT_URL,'http://ads.ngsms.tk/api/info.php?api_key='.$api_key.'&file_id='.$file_id); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
$api = curl_exec($ch); $api = print_r($api); 
Echo $api['file_id'];

The above displays:

Array ( 
   [file_id] => LNzcOp2b1352047884 
   [file_name] => sic_ftp_3rda.sisx 
   [file_type] => application/octet-stream 
   [file_size] => 65144
)

How can I get the file_name and other values?

$api["file_name"]... but im seeing some more problems with your script. you are overriding the $api variable.. why? — mr.alex
– mr.alex, Commented Nov 4, 2012 at 19:09
@Lordcash The last block of code on the answer copy and paste it all and it should work. — 1321941
– 1321941, Commented Nov 4, 2012 at 19:27

1321941 · Accepted Answer · 2012-11-04 19:55:40Z

1

To get any value within the array reference it in the same way you have done for file_id but instead replace the key with the value you are after e.g:

$api['file_id'];
$api['file_name'];
$api['file_type'];
$api['file_size'];

Then to simply show this on the page, you can use echo:

echo    $api['file_id'];
echo    $api['file_name'];
echo    $api['file_type'];
echo    $api['file_size'];

So your full code would be:

    <?php
$api_key = "Uez4TWYH6OAQQHoUcICWJ8UUFYmwQ";
$file_id = "LNzcOp2b1352047884";
$ch=curl_init(); 
curl_setopt($ch,CURLOPT_URL,'http://ads.ngsms.tk/api/info.php?api_key='.$api_key.'&file_id='.$file_id); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
$api = curl_exec($ch); 
//print_r($api);
$fileid = get_string_between($api, "[file_id] => ", "[file_name] ");
$filename = get_string_between($api, " [file_name] => ", " [file_type] => ");
$filetype = get_string_between($api, " [file_type] => ", " [file_size] =>");
$filesize = get_string_between($api, " [file_size] => ", ")");
echo "ID: $fileid <br /> name: $filename <br />type: $filetype   <br /> size: " .$filesize;
function get_string_between($string, $start, $end){
    $string = " ".$string;
    $ini = strpos($string,$start);
    if ($ini == 0) return "";
    $ini += strlen($start);
    $len = strpos($string,$end,$ini) - $ini;
    return substr($string,$ini,$len);
}

edited Nov 4, 2012 at 19:55

answered Nov 4, 2012 at 19:11

1321941

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6 Comments

Trapcode Over a year ago

it displayed Filename: A, pls if you have php installed on your system, pls try it out and help me.. The api_key and file_id is working

1321941 Over a year ago

@Lordcash I've narrowed it down, to the fact that inside $api is not actually an array, but a string of an array, which is why it is not being parsed, but give me a sec and I will get a solution.

1321941 Over a year ago

@Lordcash most eloquent way I could think of but it works see edit

Trapcode Over a year ago

ok letme try it. But i wrote the api my self, and am sure i return printed array, any ideas on where i went wrong in the info.php

Trapcode Over a year ago

it worked thanks, buh can you tell me how i would make the code in d api print out array i.e correct the error? Because i wrote them both, and i am testing them before realeasing the script

|

mabako · Accepted Answer · 2012-11-04 19:10:58Z

0

The simplest solution just popping to my mind would be the exact same one already being the last command.

echo $api['file_name'] // to print it
$file_name = $api['file_name'];

Remove the $api = from before the print_r.

answered Nov 4, 2012 at 19:10

mabako

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