用PHP解析JSON - 最有效的方法

Question

我把JSON對象發送到服務器。在服務器端，我必須用PHP解析這個obj。

我被困在循環中。我不知道如何進行循環。

我正在尋找最有效的方法來解析這個對象並將所有變量保存到數據庫中。

[ 
{"ADI":{"id":-1,"danger":0}}, 
{"ADI":{"id":3,"danger":0}}, 
{"ADI":{"id":3,"danger":0}}, 
{"ALE":{"_id":1,"_name":"Milk","contain":false}}, 
{"ALE":{"_id":2,"_name":"cfg","contain":false}}, 
{"ALE":{"_id":4,"_name":"Lakt","contain":false}}, 
{"PRO":{"image":"","code":"123456","name":"jfbj"}}, 
{"USER":{"email":"spam@spam.com"}} 
] 

現在我已經做到了這一點：

$string = file_get_contents('php://input'); 

$array = json_decode($string, true); 
//print_r($array); 

foreach ($array as $t => $index) { 
    foreach ($index as $vas => $r) { 
    //Here I'm stuck!!! 
    } 
} 

Answer 1

你可以從JSON搶值

<?php 
$string = '[ 
{"ADI":{"id":-1,"danger":0}}, 
{"ADI":{"id":3,"danger":0}}, 
{"ADI":{"id":3,"danger":0}}, 
{"ALE":{"_id":1,"_name":"Milk","contain":false}}, 
{"ALE":{"_id":2,"_name":"cfg","contain":false}}, 
{"ALE":{"_id":4,"_name":"Lakt","contain":false}}, 
{"PRO":{"image":"","code":"123456","name":"jfbj"}}, 
{"USER":{"email":"spam@spam.com"}} 
]'; 
$array = json_decode($string, true); 
echo "<pre>"; 
//print_r($array);exit; 
for ($i=0;$i<=(count($array)-1);$i++){ 
    //print_r($array[$i]); 
    if (array_key_exists("ADI",$array[$i])) { 
     $ArrVal = $array[$i]['ADI']; 
     $id = $ArrVal['id']; 
     $danger = $ArrVal['danger']; 
     echo "$id,$danger "; 
    } 
} 
?> 

Answer 2

檢查了這一點：

echo "<pre>"; 
print_r($array); 

for ($i=0; $i < count($array); $i++) { 
    if(isset($array[$i]["ADI"])){ 
    print_r($array[$i]["ADI"]); 
    } 

    if(isset($array[$i]["ALE"])){ 
    print_r($array[$i]["ALE"]); 
    } 
} 
echo "</pre>";