[英]How to get the value for the given json and get result
I am trying to get the result for the below query 我正在尝试获取以下查询的结果
$users = DB::select("SELECT FIND_IN_SET('l','a,b,c,d') as Res");
and while i do 而当我做
return $users; 返回$ users;
Here is my json 这是我的json
[{"Res":0}]
When i try to decode
it, it shows me the error 当我尝试对其进行
decode
时,它显示了错误
json_decode() expects parameter 1 to be string, array given
When i var_dump
i am getting as 当我
var_dump
我得到
array(1) { [0]=> object(stdClass)#773 (1) { ["Res"]=> int(0) } }
So, How can i get the result of the 'Res'
? 那么,如何获得
'Res'
的结果?
The solution is in the error: 解决方案是在错误中:
json_decode() expects parameter 1 to be string, **array given**
That is, the result data is being returned as an array which based on your var_dump also contains your result object and subsequent data. 也就是说,结果数据将作为数组返回,该数组基于您的var_dump还包含结果对象和后续数据。
This should do it: 应该这样做:
<?php
$data = $users[0]->Res
$decoded = json_decode($data);
Note that this is essentially just turning your JSON string into an object. 请注意,这实际上只是将您的JSON字符串变成一个对象。 You can use the second parameter to have it returned as an array if preferred:
如果需要,可以使用第二个参数将其作为数组返回:
<?php
$data = $users[0]->Res
$decoded = json_decode($data, true);
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