[英]PHP json_decode showing null while parsing JSON
我正在嘗試解析JSON數據,但是當json_decode()
, var_dump()
將值顯示為null
。 下面是我的程序:
<?php
$json='_variable_1461092903017=[ {
message:"success",
data1:{
datalist:[
{field1:"value1",field2:"value2"} ,
{field1:"value1",field2:"value2"} ,
{field1:"value1",field2:"value2"}
]
},
data2:[ {
Date:"20 Apr 2016",
details:[
{Code:"123",name:"xyz"},
{Code:"456",name:"abc"},
],
},
{
Date:"21 Apr 2016",
details:[
{Code:"123",name:"xyz"},
{Code:"456",name:"abc"},
],
},
{
Date:"22 Apr 2016",
details:[
{Code:"123",name:"xyz"},
{Code:"456",name:"abc"},
],
}
]}
]';
$json_data = json_decode($json);
var_dump($json_data);
?>
就像其他人說的那樣,無效的JSON
您可以使用JSON Linter http://jsonlint.com對其進行調試
一個很酷的有效JSON鐵路圖:
http://www.json.org
這應該是您要尋找的(我沒有執行):
<?php
$json='[{
"message" :"success",
"data1":{
"datalist" :[
{ "field1":"value1","field2":"value2"},
{"field1":"value1","field2":"value2"},
{"field1":"value1","field2":"value2"}
]
},
"data2":[ {
"Date":"20 Apr 2016",
"details":[
{"Code":"123","name":"xyz"},
{"Code":"456","name":"abc"}
]
},
{
"Date":"21 Apr 2016",
"details":[
{"Code":"123","name":"xyz"},
{"Code":"456","name":"abc"}
]
},
{
"Date":"22 Apr 2016",
"details":[
{"Code":"123","name":"xyz"},
{"Code":"456","name":"abc"}
]
}
]}
]';
$json_data = json_decode($json);
var_dump($json_data);
JSON中的錯誤:
您需要雙引號您的密鑰:
{
"key": "value"
}
不:
{
key: "value"
}
在[ ]
結束后,您要添加一個,
應該是這樣的:
{
"datalist": [ "blah", "blah"]
}
不:
{
"datalist": [ "blah", "blah"],
}
數組中的最后一個元素后面不應帶有逗號:
{
"datalist": [{"key1":"value1", {"key2": "value2"}]
}
不:
{
"datalist": [{"key1":"value1", {"key2": "value2"},]
}
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