[英]how to use PHP json_decode() function for array of JSON objects
[英]JSON PHP Decode Array and Objects
那里有很多與JSON / PHP / Decode相關的帖子...但是我在這里苦苦掙扎。 我敢肯定我沒有正確構建JSON,但似乎看不到我的錯誤。 有人可以幫忙嗎?
{
"badging": [
{
"event": [
{
"eventName": "Covent Garden",
"numberOfRooms": "1",
"mainLang": "xx",
"timeZone": "saas"
}
]
},
{
"names": [
{
"id": "1",
"fName": "Daniel",
"pos": "King"
},
{
"id": "2",
"fName": "Dasha",
"pos": "Queen"
}
]
}
]
}
這將發布到PHP頁面,並使用json_decode
$json = $_POST['feed'];
$list = json_decode($json, TRUE);
我覺得我應該能夠像這樣訪問數據:
echo $list[1][2]['fName'];
要么
echo $list->badging[1]->names[2]->fName;
但我似乎做不到。 謝謝
它應該是:
print($list["badging"][1]["names"][0]["fName"]); // outputs "Daniel"
print($list["badging"][1]["names"][1]["fName"]); // outputs "Dasha"
嘗試$list['badging'][1]['names'][1]['fName'];
http://codepad.org/IdneHeDL
另外,如果您省略json_decode的true,則$list->badging[1]->names[1]->fName;
badging $list->badging[1]->names[1]->fName;
http://codepad.org/WXWkVqo1
這就是數組的外觀。
array(1) {
["badging"]=>
array(2) {
[0]=>
array(1) {
["event"]=>
array(1) {
[0]=>
array(4) {
["eventName"]=>
string(13) "Covent Garden"
["numberOfRooms"]=>
string(1) "1"
["mainLang"]=>
string(2) "xx"
["timeZone"]=>
string(4) "saas"
}
}
}
[1]=>
array(1) {
["names"]=>
array(2) {
[0]=>
array(3) {
["id"]=>
string(1) "1"
["fName"]=>
string(6) "Daniel"
["pos"]=>
string(4) "King"
}
[1]=>
array(3) {
["id"]=>
string(1) "2"
["fName"]=>
string(5) "Dasha"
["pos"]=>
string(5) "Queen"
}
}
}
}
}
例如。 echo $ list ['badging'] [0] ['event'] [0] ['eventName'];
根據您的JSON,通過JSON_decode,我們得到
Array ( [badging] => Array ( [0] => Array ( [event] => Array ( [0] => Array ( [eventName] => Covent Garden [numberOfRooms] => 1 [mainLang] => xx [timeZone] => saas ) ) ) [1] => Array ( [names] => Array ( [0] => Array ( [id] => 1 [fName] => Daniel [pos] => King ) [1] => Array ( [id] => 2 [fName] => Dasha [pos] => Queen ) ) ) ) )
因此,要訪問它,您需要使用
echo $list['badging'][1]['names'][0]['fName'];
快樂編碼:)
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