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WordPress判断用户是否登录的代码
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WordPress判断用户是否登录的代码 | Lenix Blog
is_user_logged_in()
说明
根据当前访问者是否登录返回布尔值true或false。
参数
该函数不接受任何参数。
复制代码 代码如下:
<?php if ( is_user_logged_in() ) { ... } ?>
示例
根据当前访问者的登录情况，使用is_user_logged_in()函数在主题文件中展示不同输出结果。
复制代码 代码如下:
<?php
if ( is_user_logged_in() ) {
echo 'Welcome, registered user!';
} else {
echo 'Welcome, visitor!';
}
?>
if ( is_user_logged_in() ) {
echo 'Welcome, registered user!';
} else {
echo 'Welcome, visitor!';
}
?>
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