如何在PHP中獲取已解碼的JSON數據的值

Question

{"senderDetails":[{"status":"failed","senderid":"0","message":"Kindly upgrade your account to use this service."}]} 

以上是我的JSON數據。我使用json_decode()函數來解碼，但我不知道如何在PHP中分別回顯每個對象的值。

Answer 1

舉例來說，如果你要打印的senderid：

<?php 
$var = '{"senderDetails":[{"status":"failed","senderid":"0","message":"Kindly upgrade your account to use this service."}]}'; 

$decoded = json_decode($var); 

var_dump($decoded); 
echo($decoded->senderDetails[0]->senderid); 

?> 

使用var_dump()這些類型的變量，以瞭解如何正確地訪問它們是非常重要的。

Answer 2

如果你想通過數組像這樣呼應所有值可以循環：

foreach($your_array as $value){ 
    echo $value . "<br>"; 
} 

但是，如果你想輸出您可以將特定的元素只是做：

echo($your_array->object->value); 

Answer 3

如果您使用

$json_array = json_decode($json,true) //If you add true you get an array as Response 
$value = $json_array["senderDetails"]["status"]; // Example 

Answer 4

可以轉換成數組和回聲的變量這樣

$json_data = '{"senderDetails":[{"status":"failed","senderid":"0","message":"Kindly upgrade your account to use this service."}]}'; 
$decoded = json_decode($json_data,true); 

echo "Status = ".($decoded["senderDetails"][0]["status"]); 
echo "<br />"; 
echo "Senderid = ".($decoded["senderDetails"][0]["senderid"]); 
echo "<br />"; 
echo "Message = ".($decoded["senderDetails"][0]["message"]);