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":
Wondering why my PHP code will not display all "Value" of "Values" in the JSON data:
$user = json_decode(file_get_contents($analytics));
foreach($user->data as $mydata)
{
echo $mydata->name . "
";
}
foreach($user->data->values as $values)
{
echo $values->value . "
";
}
第一个 foreach 工作正常,但第二个会引发错误.
The first foreach works fine, but the second throws an error.
{
"data": [
{
"id": "MY_ID/insights/page_views_login_unique/day",
"name": "page_views_login_unique",
"period": "day",
"values": [
{
"value": 1,
"end_time": "2012-05-01T07:00:00+0000"
},
{
"value": 6,
"end_time": "2012-05-02T07:00:00+0000"
},
{
"value": 5,
"end_time": "2012-05-03T07:00:00+0000"
}, ...
您可能想要执行以下操作:
You maybe wanted to do the following:
foreach($user->data as $mydata)
{
echo $mydata->name . "
";
foreach($mydata->values as $values)
{
echo $values->value . "
";
}
}
这篇关于使用 PHP foreach 解析 JSON 数组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持html5模板网!
MySQLi准备好的语句&foreach 循环MySQLi prepared statement amp; foreach loop(MySQLi准备好的语句amp;foreach 循环)
mysqli_insert_id() 是从整个服务器还是从同一用户获Is mysqli_insert_id() gets record from whole server or from same user?(mysqli_insert_id() 是从整个服务器还是从同一用户获取记录?)
PHP MySQLi 无法识别登录信息PHP MySQLi doesn#39;t recognize login info(PHP MySQLi 无法识别登录信息)
mysqli_select_db() 需要 2 个参数mysqli_select_db() expects exactly 2 parameters(mysqli_select_db() 需要 2 个参数)
Php mysql pdo 查询:用查询结果填充变量Php mysql pdo query: fill up variable with query result(Php mysql pdo 查询:用查询结果填充变量)
MySQLI 28000/1045 用户“root"@“localhost"的访问MySQLI 28000/1045 Access denied for user #39;root#39;@#39;localhost#39;(MySQLI 28000/1045 用户“root@“localhost的访问被拒绝)