duanbai5348 2016-12-12 03:17
浏览 56
已采纳

JSON_decode和PHP没有分号(,)

example.php

{"status": "ok"} {"status": "error"}

i will result like :

ok , error

my website showing a blank page(im using this code),can you help me to fix it?

<?php
$userinfo = 'example.php';
$fgc = file_get_contents($userinfo);
$json2 = json_decode($fgc, true);
$media = $json2['status'];

$mediaId = $media;

echo $mediaId;
?>
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2条回答 默认 最新

  • dongpan1871 2016-12-12 03:45
    关注

    This will resolve your issue in the non-recommended way:

    $broken_json = '{"success": "ok"} {"success": "error"}';
    $fixed_json = "[" . str_replace("} {", "},{", $broken_json) . "]";
    echo $fixed_json;
    
    $array = json_decode($fixed_json, true);
    
    echo "<pre>";
    var_dump($array);
    echo "</pre>";
    

    Result:

    [{"success": "ok"},{"success": "error"}]
    
    array(2) {
      [0]=>
      array(1) {
        ["success"]=>
        string(2) "ok"
      }
      [1]=>
      array(1) {
        ["success"]=>
        string(5) "error"
      }
    }
    

    Recommended Way:

    Actually get VALID JSON from the source

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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