file_get_contents php和json_decode错误[重复]答案

【问题标题】：file_get_contents php and json_decode error [duplicate]file_get_contents php和json_decode错误[重复]
【发布时间】：2013-06-12 17:26:27
【问题描述】：

我在 file_get_contents 和将结果解码为 json 时遇到了一个奇怪的问题。我必须告诉你，这段代码在其他服务器上是完全正常的。

我正在尝试解码这个 json： http://mediarupt.com/GET/categorized_services_for_supplier.php?suppliersID=10

进入这个页面： http://mediarupt.com/kostas.php

kostas.php 的代码是这样的：

<?php 
$servicesJSON = file_get_contents("http://mediarupt.com/GET/categorized_services_for_supplier.php?suppliersID=10");
$ForeignKeys = (object)array();
$ForeignKeys->ServicesList = json_decode($servicesJSON, true);

echo "error: " . json_last_error();

?>

<select name="servicesID" id="servicesID">
    <option id="serviceID_0" value="0" rel="0" style="font-style:italic;">Select a category...</option>
<?php foreach($ForeignKeys->ServicesList['categorized_services'] as $value){ ?>
    <option id="serviceID_<?=$value['servicesID']?>" value="<?=$value['servicesID']?>" rel="<?=$value['hasStoresList']?>"><?=$value['serviceName']?></option>   

<?php } ?>

</select>
<?php

echo '<br><br>The result of file_get_contents ($servicesJSON): '.$servicesJSON;

?>

classified_services_for_supplier.php 的代码：

<?php header('Content-Type: application/json');
    require('../settings/dbli.php');

    $table = array();
    $suppliersID = "0";

    if(isset($_GET['suppliersID']) && $_GET['suppliersID']!=NULL && $_GET['suppliersID']!='' ){  $suppliersID = $SQLConn->real_escape_string($_GET['suppliersID']); }
    $query = "SELECT DISTINCT Services.servicesID, serviceName, servicePrice, hasStoresList FROM Services, ServicesList WHERE Services.servicesID = ServicesList.servicesID AND suppliersID = '$suppliersID' AND  status = '1' AND deleted = '0'";

    $result = $SQLConn->query($query);  
     while ($row = $result->fetch_array(MYSQLI_ASSOC)) {
        $tempArray = array();
        foreach($row as $key => $value){
            $tempArray[$key] = $value;                      
        }
        array_push($table,$tempArray);
     }
     $arr = array( "categorized_services" => $table);
    echo json_encode($arr);
?>

我收到错误 4 作为最新的 json 错误，这意味着我有 JSON_ERROR_SYNTAX。但我用http://jsonlint.com/ 验证了 json 结果，一切似乎都正常。

另外，两个页面都是 UTF-8 且 BOM 被禁用。您可以在此处访问 php 配置：http://mediarupt.com/phpinfo.php

希望能找到解决办法……

提前谢谢你

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