Question

我有几个json文件，它们都具有相同的结构，每个文件使用不同的年份：

#include <vector>

class Robot {
    private:

    public:
        void move_right();
        void move_left();
        void switch_camera();
        void raise_camera();
};

struct Action
{
    Action(void (Robot::*what)(void))
    : what(what)
    {}

    void perform(Robot& who) const
    {
        (who.*what)();
    }

    void (Robot::*what)(void);
};

bool should_abort();

void perform_actions(Robot& who, std::vector<Action> const& actions)
{
    for (auto&& action : actions)
    {
        if (should_abort()) break;
        action.perform(who);
    }
}

int main()
{

    std::vector<Action> actions {
        &Robot::move_right,
        &Robot::raise_camera,
        &Robot::switch_camera,
        &Robot::move_left
    };

    Robot r;
    perform_actions(r, actions);

}

它们分别称为data_2014.json，data_2015.json和data_2016.json。

使用PHP，我想要的最终json是alldata.json，如下所示：

  {"AEK":{"country":"Greece","shtv":"VEK"} ,
   "BER":{"country":"Germany","shtv":"BKE"} ,
   "CAR":{"country":"Italy","shtv":"CRA"}}

  {"AEK":{"country":"Greece","shtv":"MOR"} ,
   "DAR":{"country":"Turkey","shtv":"DDR"}}

  {"AEK":{"country":"Greece","shtv":"MIL"} ,
   "BER":{"country":"Germany","shtv":"BKE"} ,
   "CAR":{"country":"Italy","shtv":"KUN"}}

我的意思是：对于重复的元素，获取可用的最新信息（即，对于AEK元素，获取属性“ shtv” =“ MIL”，这是data_2016.json中的元素。对于不再重复，只需获取可用信息即可。

Answer 1

您需要使用json_decode将JSON转换为数组，然后使用array_merge合并所有json数组，然后使用json_encode再次编码以将所有json合并为一个。

$json1 = '{"AEK":{"country":"Greece","shtv":"VEK"} ,
"BER":{"country":"Germany","shtv":"BKE"} ,
"CAR":{"country":"Italy","shtv":"CRA"}}';

$json2 = '{"AEK":{"country":"Greece","shtv":"MOR"} ,
"DAR":{"country":"Turkey","shtv":"DDR"}}';

$json3 = '{"AEK":{"country":"Greece","shtv":"MIL"} ,
"BER":{"country":"Germany","shtv":"BKE"} ,
"CAR":{"country":"Italy","shtv":"KUN"}}';

$arr1 = json_decode($json1,true);
$arr2 = json_decode($json2,true);
$arr3 = json_decode($json3,true);

$finalArr = array_merge($arr1,$arr2,$arr3);

$final_json = json_encode($finalArr);

echo $final_json;

Answer 2

首先，您必须使用json_decode（）解码json值。

$var1= '{"AEK":{"country":"Greece","shtv":"VEK"} ,
"BER":{"country":"Germany","shtv":"BKE"} ,
"CAR":{"country":"Italy","shtv":"CRA"}}';

$var2= '{"AEK":{"country":"Greece","shtv":"MOR"} ,
"DAR":{"country":"Turkey","shtv":"DDR"}}';

$var3= '{"AEK":{"country":"Greece","shtv":"MIL"} ,
"BER":{"country":"Germany","shtv":"BKE"} ,
"CAR":{"country":"Italy","shtv":"KUN"}}';

并使用json_decode（$ var，TRUE）传递true，因为 TRUE ，返回的对象将转换为关联数组。

$array1= json_decode($var1,true);
$array2= json_decode($var2,true);
$array3 =json_decode($var3,true);

然后，您具有这三个数组值的array_merge（），您将获得所需的输出。

$ result = json_encode（array_merge（$ array1，$ array2，$ array3））;

$ result返回

{"AEK":{"country":"Greece","shtv":"MIL"},
 "BER":{"country":"Germany","shtv":"BKE"},
  "CAR":{"country":"Italy","shtv":"KUN"},
  "DAR":{"country":"Turkey","shtv":"DDR"}
}

Answer 3

$json1_str = file_get_contents('data_2014.json');
$json2_str = file_get_contents('data_2015.json');
$json3_str = file_get_contents('data_2016.json');

$json1_array = json_decode($json1_str, true);
$json2_array = json_decode($json2_str, true);
$json3_array = json_decode($json3_str, true);

$array = array_merge($json1_array, $json2_array, $json3_array);
$final_output=json_encode($array);

合并json文件，覆盖

3 个答案: