Json，从数组输出字符串 - json, output string from array

json, output string from array

本文关键字：输出字符串数组 Json | 更新日期: 2023-09-27

我试图解码JSON数据PHP然后输出到网站。如果我有以下内容:

{
  "name": "josh",
  "type": "human"
{

我可以这样做(在PHP中)，显示或输出我的type:

$file = "path";
$json = json_decode($file);
echo $json["type"]; //human

那么，如果我有以下内容:

{
  "name": "josh",
  "type": "human",
  "friends": [
    {
      "name": "ben",
      "type": "robot"
    },
    {
      "name": "tom",
      "type": "alien"
    }
  ],
  "img": "img/path"
}

我如何输出我的朋友ben是什么type ?

使用foreach这样的循环并执行如下操作:

//specify the name of the friend like this:
$name = "ben";
$friends = $json["friends"];
//loop through the array of friends;
foreach($friends as $friend) {
    if ($friend["name"] == $name) echo $friend["type"];
}

要以数组格式获得解码的数据，您将提供true作为json_decode的第二个参数，否则它将使用默认的object表示法。当您需要查找特定的用户

时，您可以轻松地创建一个函数来缩短该过程。

$data='{
  "name": "josh",
  "type": "human",
  "friends": [
    {
      "name": "ben",
      "type": "robot"
    },
    {
      "name": "tom",
      "type": "alien"
    }
  ],
  "img": "img/path"
}';
$json=json_decode($data);
$friends=$json->friends;
foreach( $friends as $friend ){
    if( $friend->name=='ben' )echo $friend->type;
}
function finduser($obj,$name){
    foreach( $obj as $friend ){
        if( $friend->name==$name )return $friend->type;
    }
}
echo 'Tom is a '.finduser($friends,'tom');

try this，

$friend_name = "ben";
$json=json_decode($data);
$friends=$json->friends;
foreach( $friends as $val){
    if($friend_name == $val->name)
    {
        echo "name = ".$val->name;
        echo "type = ".$val->type;
    }    
}