我有一個 JSON 字符串，我將其解析為 PHP 對象，我想從字符串中打印某些值，如何打印我需要的確切值。我提供了下面的代碼{ "data": { "id":123, "name": "abc", "gsm":"1133", "metadata": { "cart_id":13243 }, "customer": { "id":123, "email": "[email protected]" } }}json_decode($string);例如“名稱”：abc，我只想打印 abc
3 回答
慕碼人2483693
TA貢獻1860條經驗 獲得超9個贊
您的 JSON 字符串無效，您在其中缺少括號。此外，字符串值必須在 json 中包含引號。你的字符串應該像
$string= '{
"data": {
"id": 123,
"name": "abc",
"gsm": 1133,
"metadata": {
"cart_id": 13243
},
"customer": {
"id": 123,
"email": "[email protected]"
}
}
}';
$response = json_decode($string);
print_r($response->data->name);
die;
希望這對你有幫助。
瀟湘沐
TA貢獻1816條經驗 獲得超6個贊
您可以將 JSON 轉換為數組并通過鍵訪問元素
$jArr = json_decode($string, true);
訪問元素
$jArr['data']['name']
慕俠2389804
TA貢獻1719條經驗 獲得超6個贊
您必須從 json 解碼訪問元素，如下所示：
$JSON = '{
"data":{
"id":123,
"name": abc,
"gsm":1133,
"metadata":{
"cart_id":13243
},
"customer":{
"id":123,
"email": "[email protected]"
}
}
}'; //There were three errors, you have to to use ' insted of " for including the JSON string, you missed one '}' and you forgot the "" in the 'email' value under 'customer' section
$object = json_decode($JSON);
echo 'Printing the customer id using an object: '.$object->data->customer->id.PHP_EOL.PHP_EOL;
$array = json_decode($JSON, true);
echo 'Printing the customer id using an array: '.$array['data']['customer']['id'];
此代碼打印：
Printing the customer id using an object: 123
Printing the customer id using an array: 123
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