使用PHP foreach解析JSON数组

Wondering why my PHP code will not display all "Value" of "Values" in the JSON data:

想知道为什么我的PHP代码不会在JSON数据中显示“值”的所有“值”：

$user = json_decode(file_get_contents($analytics));
foreach($user->data as $mydata)
{
     echo $mydata->name . "\n";

}        
foreach($user->data->values as $values)
{
     echo $values->value . "\n";
}

The first foreach works fine, but the second throws an error.

第一个foreach工作正常，但第二个抛出错误。

{
   "data": [
      {
         "id": "MY_ID/insights/page_views_login_unique/day",
         "name": "page_views_login_unique",
         "period": "day",
         "values": [
            {
               "value": 1,
               "end_time": "2012-05-01T07:00:00+0000"
            },
            {
               "value": 6,
               "end_time": "2012-05-02T07:00:00+0000"
            },
            {
               "value": 5,
               "end_time": "2012-05-03T07:00:00+0000"
            }, ...

3 个解决方案

#1

You maybe wanted to do the following:

您可能想要执行以下操作：

foreach($user->data as $mydata)

    {
         echo $mydata->name . "\n";
         foreach($mydata->values as $values)
         {
              echo $values->value . "\n";
         }
    }

#2

You need to tell it which index in data to use, or double loop through all.

您需要告诉它要使用哪个数据索引，或者双循环遍历所有索引。

E.g., to get the values in the 4th index in the outside array.:

例如，获取外部数组中第4个索引中的值：

foreach($user->data[3]->values as $values)
{
     echo $values->value . "\n";
}

To go through all:

通过所有：

foreach($user->data as $mydata)
{
    foreach($mydata->values as $values) {
        echo $values->value . "\n";
    }

}

#3

$user->data is an array of objects. Each element in the array has a name and value property (as well as others).

$ user-> data是一个对象数组。数组中的每个元素都有name和value属性（以及其他元素）。

Try putting the 2nd foreach inside the 1st.

尝试将第二个foreach放在第一个。

foreach($user->data as $mydata)
{
    echo $mydata->name . "\n";
    foreach($mydata->values as $values)
    {
        echo $values->value . "\n";
    }
}

#1