json_decode（）期望參數1是字符串，數組給定？

-1

這是我的數據，之後Json_encode()json_decode（）期望參數1是字符串，數組給定？

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"}] 
    => 
     )

現在我想它通過施加json_decode()背面進行解碼，它提供了以下錯誤

json_decode（）預計參數1是字符串，數組給定

任何想法消化做什麼？

來源

2015-06-25 mparo_afridi

你如何編碼？看起來你在那裏做錯了什麼。 – Sirko

您可以發佈您的相關代碼嗎？ –

您需要將該數組的第一項傳遞給json_decode;而不是整個陣列。因此，'json_decode（$ array [0]）' – d0ug7a5

你json必須在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);

如果您json是array你可以這樣做：

$json_string_in_array = array('{"a":1,"b":2,"c":3,"d":4,"e":5}'); 
$json_array = json_decode($json_string_in_array[0]);

來源

2015-06-25 12:27:28 Fabien

說undefined 0索引 –

$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);

來源

2015-06-25 12:27:21 PHPJungle

json_encode（）返回一個字符串，所以我不知道你會如何得到數組出來的，除非你將其存儲在一個數組喜歡自己：

$json = []; 
$json[] = json_encode($someArray);

相反，只是將其存儲在一個非數組變量：

$jsonString = json_encode($someArray);

然後，你可以這樣對其進行解碼：

$decodedArray = json_decode($jsonString);

來源

2015-06-25 12:28:28 BBBThunda

json_decode（）期望參數1是字符串，數組給定？

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