Question

我尝试在PHP To数组中解码JSON。这是我的代码

<?php
$url="file/game.json";
$result_file=file_get_contents($url);
    if(!$result_file){ echo 'result_file......Failed'; }
$php_result=json_decode($result_file,true);
print_r($php_result);

if($php_result){

    echo 'ok';
}
else{

    echo 'error';
}
?>

这是我的json文件

[
  {
    "name": "ABCD",
    "price": "24.36",
    "platform": "<span class=\"platform battle-net\"></span>",
    "region": "GLOBAL"
  },
  {
    "name": "ABCD",
    "price": "27.60",
    "platform": "<span class=\"platform xbox-live\"></span>",
    "region": "GLOBAL"
  }
]

我的结果是“错误”。而且我也不知道如何解决这个问题。

我该怎么办。

Answer 1

file_get_contents函数在失败时将整个文件返回为字符串或FALSE。所以我想，这里您没有得到文件字符串。所以尝试这样，

<?php
try {
    $url="file/game.json";
    $result_file = file_get_contents('file/game.json');
    if ($result_file === false) {
        // Handle the error
        echo "ERROR";
    }else{
        $php_result=json_decode($result_file,true);

        switch (json_last_error()) {
            case JSON_ERROR_NONE:
            echo ' - OK';
            print_r($php_result);
            break;
            case JSON_ERROR_DEPTH:
            echo ' - Maximum stack depth exceeded';
            break;
            case JSON_ERROR_STATE_MISMATCH:
            echo ' - Underflow or the modes mismatch';
            break;
            case JSON_ERROR_CTRL_CHAR:
            echo ' - Unexpected control character found';
            break;
            case JSON_ERROR_SYNTAX:
            echo ' - Syntax error, malformed JSON';
            break;
            case JSON_ERROR_UTF8:
            echo ' - Malformed UTF-8 characters, possibly incorrectly encoded';
            break;
            default:
            echo ' - Unknown error';
            break;
        }
    }
} catch (Exception $e) {
    // Handle exception
    #echo $e->getMessage();
    #echo $e->code();
}
?>

无法在php中解码JSON文件

1 个答案: