Question

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Possible Duplicate:
How to decode a JSON string in PHP?

我愿从json string $jsonstring中抽取“;



<?php

$facebookid = $_GET[ fbid ];
$facebooklink =  http://graph.facebook.com/ .$facebookid. ?fields=name ;

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $facebooklink);
curl_setopt($ch, CURLOPT_HEADER, 0);
$jsonstring = curl_exec($ch);
curl_close($ch);

?>


 产出如下:

{
   "name": "THIS PART IS WHAT I WANT",
   "id": "4"
}


我想从上述例子中提取这一部分内容。

我曾尝试使用json_decode,但不能让它工作,此后,我删除了该守则,因此不能在这里书写。 感谢任何帮助。

Answer 1

审判

ob_start();
$facebookid = $_GET[ fbid ];
$facebooklink =  http://graph.facebook.com/ .$facebookid. ?fields=name ;

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $facebooklink);
curl_setopt($ch, CURLOPT_HEADER, 0);
$jsonstring = curl_exec($ch);
curl_close($ch);
$bufferstr = ob_get_contents();
ob_end_clean();
$json = json_decode($bufferstr);
echo $json->name;

Answer 2

Try:

<?php
  $foo = json_decode( $jsonstring);
  $name = $foo->name; 
  //echo $name;
?>

Answer 3

try this

$json_obj = json_decode($jsonstring );
echo $json_obj->name;

Answer 4

<?php
    $json =  {"foo-bar": 12345} ;

    $obj = json_decode($json);
    print $obj->{ foo-bar }; // 12345
?>

Answer 5

如果你没有“json_decode”的话,你就应当使用定期表述。

<?php

$name = json_decode( $jsonstring )->name;

?>

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