Php 分析错误：语法错误，意外'；2'；（T_LNUMBER）

任何帮助都将不胜感激

<?php 
    $jsonobj = '{
      "status": "1",
      "receiver_account": "8",
      "USD_amount": "1.670",
      "fee_in_USD": "0.000",
      "PKR_amount": "280",
      "fee_in_PKR": "0",
      "USD_amount_with_fee": "1.670",
      "PKR_amount_with_fee": "280",
      "trx_website": "website.com",
      "transaction_id": "2JW9651118P",
      "trx_date": "25-03-2020 9:13:48 PM",
      "order_id": "12345678",
      "addi_info": "Test Payment",
      "sender_details": "Fund Received From 161919",
      "trx_details": "$1.67 Receive against TID: '2JW9651118P' "
    }';
    json_decode($jsonobj);
    var_dump($jsonobj);
?>

---

您需要在最后一个字段中转义单引号：

“trx\U详细信息”：“$1.67根据TID收取：\'2JW9651118P\”

你用一个引号（

“

）打开你的

jsonobj

），然后在json中使用另一个引号，最后在它应该关闭之前关闭

jsonobj

。只需将

TID:'2JW9651118P'

更改为

TID:\'2JW9651118P\\

，并阅读“是，这是有效的！”！谢谢你的帮助@安吉尔·博涅夫和@Mech