php - 从 mysql 检索到的 json 字符串上的 json_decode 不起作用

Question

我有一个字符串 json，它提供以下 var 转储:

"[[{"TransactionID":"00416","OrderID":"000926","TransactionOrderItemID":"001123","LastUpdate":"2009-10-28 13:03:31","CustomerID":"184","Company_name":"Test123","Invoiced":"0","SubItemsCount":"2","ProductID":"1","ProductTypesID":"1","ProductTypeName":"Phone","ProductName":"Phone Line (Home)","IncludePST":"1","BillType":"Monthly","BillingCycle":"Monthly","Status":"Active","CreationDate":"2009-10-28","ActivationStartDate":"2009-10-28","NextNotificationDate":"2009-10-27","OverWritePrice":"-1","PriceEconomic":"26.00","BasePrice":"26.0000","ProRate":"Yes","InvoicePrice":"3.35","ServicePeriod":" Pro-Rate: Oct-28-2009 - Oct-31-2009","EndDate":"2009-10-31"}]]"

当我尝试解码为:

json_decode( $json, true);

结果只是空。 $json来自数据库。
请注意，我仅显示完整数据的子集。 问题是，这是来自数据库的变量。如果双引号是问题，如何转换它？

Answer 1

您可以调用 json_last_error() 来获取有关错误原因的更多信息。

例如这样:

json_decode($string);

switch(json_last_error())
{
    case JSON_ERROR_DEPTH:
        echo ' - Maximum stack depth exceeded';
    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_NONE:
        echo ' - No errors';
    break;
}