javascript - Php 不从 json 发布数据，请

Question

我的 Php 似乎没有发送到数据库，下面是我只想获取 MerchantRequest_ID 的 php 代码

<?php
    $connect = mysql_connect("localhost","adflix","[4bTR9Hdp") or die('Database Not Connected. Please Fix the Issue! ' . mysql_error()); 
    mysql_select_db("njec_flix", $connect);
    $str  = file_get_contents('beit.json');
    $json = json_decode($str, true);

    $MerchantRequest_ID = $str['body']['stkCallback']['MerchantRequest_ID']; 



    $query = "INSERT INTO test(MerchantRequest_ID) VALUES('$MerchantRequest_ID')";
    if(!mysql_query($query,$connect)) { 
        die('Error : Query Not Executed. Please Fix the Issue! ' . mysql_error());
    } else{ echo "Data Inserted Successully!!!";
    } 
?> 

来自 beit.json 的我的 Json

{
    "Body":{
        "stkCallback":{
            "RequestID3":"11623-8633880-1",
            "MerchantRequest_ID":
            "ws_CO_DMZ_336620817_07022019155939425",
            "ResultCode":1036,
            "ResultDesc":"[STK_CB - ]SMSC ACK timeout."
        }
    }
}

Answer 1

$str 包含从 JSON 文件读取的字符串，如果您想读取对象的属性，则必须使用 $json var。

那不是:

$MerchantRequest_ID = $str['body']['stkCallback']['MerchantRequest_ID'];

但是

$MerchantRequest_ID = $json['body']['stkCallback']['MerchantRequest_ID'];