dongsu1951 2017-10-03 18:08 采纳率: 0%
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php curl获取json并解码然后显示,失败显示任何内容

hello my teacher has given me this weeks work and i need to get my json data from a url and display it as php using json_decode, the php json file which i access with the url displays the songs as an json array and works fine, and i copied my teachers curl example to access the url, but it fails to display anything what have i done wrong ?

Edit: i do need to enter a password and username to access the url so i changed the code as told, but still nothing displays, i changed the password and username for security but the format and length are the same

<?php
$username = "2foobar90";
$password = "123456";
// Initialise the cURL connection
$connection = curl_init();

curl_setopt($connection, CURLOPT_HTTPAUTH, CURLAUTH_BASIC);
curl_setopt($connection, CURLOPT_USERPWD, $username . ":" . $password);  
// Specify the URL to connect to - DOUBLE CLICK link to test 
curl_setopt($connection, CURLOPT_URL, "https://edward2.solent.ac.uk/~martine/year3/hits.php?artist=David+Bowie");


// This option ensures that the HTTP response is *returned* from curl_exec(),
// (see below) rather than being output to screen.  
curl_setopt($connection,CURLOPT_RETURNTRANSFER,1);

// Do not include the HTTP header in the response.
curl_setopt($connection,CURLOPT_HEADER, 0);

// Actually connect to the remote URL. The response is 
// returned from curl_exec() and placed in $response.
$response = curl_exec($connection);

// Close the connection.
curl_close($connection);

$data = json_decode($response, true);

for($i=0; $i<count($data); $i++)
{
    echo "Song id " . $data[$i]["songid"] . " " .
         "Title " . $data[$i]["title"] . " " .
         "Artist " . $data[$i]["artist"] . " " .
         "Chart position " . $data[$i]["chart"] . "<br/>";
}
?>
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