按索引号(不是名称)返回 PHP 对象 [英] Return PHP object by index number (not name)

问题描述

目标:通过数字从 PHP 对象中检索数据元素.

Goal: retrieve an element of data from within a PHP object by number.

这是对象的print_r($data):

This is the print_r($data) of the object:

stdClass Object
(
    [0] => stdClass Object
        (
            [TheKey] => 1456
            [ThingName] => Malibu
            [ThingID] => 7037
            [MemberOf] => California
            [ListID] => 7035
            [UserID] => 157
            [UserName] => John Doe
        )
)

我不知道如何从中提取一个值.这只是一个多记录对象的一个​​记录，应该是id而不是名字.

I can't figure out how to pull a value out of it. This is only one record of a multi-record object that should be by id rather than a name.

这些是一些失败的尝试来说明目标是什么:

These are some failed attempts to illustrate what the goal is:

echo $data -> 0 -> UserName;
echo $data[0] -> UserName;

推荐答案

通常，PHP 变量名称不能以数字开头.您不能将 $data 作为数组访问，因为 stdClass 没有实现 ArrayAccess — 它只是一个普通的基类.

Normally, PHP variable names can't start with a digit. You can't access $data as an array either as stdClass does not implement ArrayAccess — it's just a normal base class.

但是，在这种情况下，您可以尝试通过数字名称访问对象属性，如下所示:

However, in cases like this you can try accessing the object attribute by its numeric name like so:

echo $data->{'0'}->UserName;

我能想到为什么 Spudley 的答案会导致错误的唯一原因是因为您正在运行 PHP 4，它不支持使用 foreach 来迭代对象.

The only reason I can think of why Spudley's answer would cause an error is because you're running PHP 4, which doesn't support using foreach to iterate objects.

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