php json_decode() 期望参数 1 是字符串，给定数组？
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json_decode() expects parameter 1 to be string, array given?
提问by mparo_afridi
this is my data ,After Json_encode()
这是我的数据，之后 Json_encode()
Array
(
[{"customerId":"1","customer_name":"Jon_doe","amount":"12312312","billcode":"b1231","billname":"cashbilname","billcategorycode":"1234","billcategory":"utility","month":"May","year":"2015","txcode":"10","stationid":"152","station":"Coroom","operatorcode":"1200","operator":"jame","terminal":"ter12312","txdate":"12\/2\/2015","txtime":"12:21:22_PM"}]
=>
)
now i want to decode it back ,by applying json_decode()it gives the following error
现在我想解码它，通过应用json_decode()它会出现以下错误
json_decode() expects parameter 1 to be string, array given
json_decode() 期望参数 1 是字符串，给定数组
any idea sugestion what to do ?
任何想法sugestion怎么办？
回答by Fabien
Your jsonmust be in string, not in array
你json必须在string，而不是在array
$json_string = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
$json_array = json_decode($json_string);
$json_array : array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5);
If your jsonis in arrayyou can do :
如果您json在，array您可以：
$json_string_in_array = array('{"a":1,"b":2,"c":3,"d":4,"e":5}');
$json_array = json_decode($json_string_in_array[0]);
回答by BBBThunda
json_encode() returns a string, so I don't know how you can be getting an array out of it unless you are storing it in an array yourself like:
json_encode() 返回一个字符串，所以我不知道如何从中获取数组，除非您自己将其存储在数组中，例如：
$json = [];
$json[] = json_encode($someArray);
Instead, just store it in a non-array variable:
相反，只需将其存储在非数组变量中：
$jsonString = json_encode($someArray);
Then you can decode it like this:
然后你可以像这样解码它：
$decodedArray = json_decode($jsonString);
回答by PHPJungle
$jsonstr = '[{"customerId":"1","customer_name":"Jon_doe","amount":"12312312","billcode":"b1231","billname":"cashbilname","billcategorycode":"1234","billcategory":"utility","month":"May","year":"2015","txcode":"10","stationid":"152","station":"Coroom","operatorcode":"1200","operator":"jame","terminal":"ter12312","txdate":"12\/2\/2015","txtime":"12:21:22_PM"}]';
$ar = json_decode($jsonstr,true); # json string to Array
$obj = json_decode($jsonstr); # json string to Object
var_dump($ar,$obj);
