Php 如何使wp cli在带有条件语句的wp config文件中工作？

我正在使用wp cli进行测试，以更新我的域的prod、dev、staging、test和update站点

在mywp配置文件中，如果有条件语句，则会出现以下错误：

PHP Notice:  Undefined index: HTTP_HOST in phar:///usr/local/bin/wp/php/WP_CLI/Runner.php(1070) : eval()'d code on line 22
PHP Notice:  Undefined index: HTTP_HOST in phar:///usr/local/bin/wp/php/WP_CLI/Runner.php(1070) : eval()'d code on line 24
PHP Notice:  Undefined index: HTTP_HOST in phar:///usr/local/bin/wp/php/WP_CLI/Runner.php(1070) : eval()'d code on line 26
PHP Notice:  Undefined index: HTTP_HOST in phar:///usr/local/bin/wp/php/WP_CLI/Runner.php(1070) : eval()'d code on line 28
PHP Notice:  Undefined variable: db_name in phar:///usr/local/bin/wp/php/WP_CLI/Runner.php(1070) : eval()'d code on line 34
Error: Can’t select database. We were able to connect to the database server (which means your username and password is okay) but not able to select the `` database.

这就是我如何编写我的wp-config.php文件的方法

if ($_SERVER["HTTP_HOST"] === 'example.com') {
  $db_name = 'example';
}else if ($_SERVER["HTTP_HOST"] === 'dev.example.com') {
  $db_name = 'example_dev';
}else if ($_SERVER["HTTP_HOST"] === 'staging.example.com'){
  $db_name = 'example_staging';
}else if ($_SERVER["HTTP_HOST"] === 'update.example.com') {
  $db_name = 'example_update';
}

define('DB_NAME', $db_name);

有没有办法让wp cli处理此文件

$_SERVER["HTTP_HOST"]

从命令行使用PHP时不存在。最好的选择是在命令行中使用文件时指定一种情况，例如：

if (php_sapi_name() == "cli") {
    //set db_name when in cli
} else {
    //set db_name as per host
    if ($_SERVER["HTTP_HOST"] === 'example.com') {
      $db_name = 'example';
    }else if ($_SERVER["HTTP_HOST"] === 'dev.example.com') {
      $db_name = 'example_dev';
    }else if ($_SERVER["HTTP_HOST"] === 'staging.example.com'){
      $db_name = 'example_staging';
    }else if ($_SERVER["HTTP_HOST"] === 'update.example.com') {
      $db_name = 'example_update';
    }
}

由于我不太确定您在wp cli中希望$db_name是什么，所以我在第一组大括号中没有设置它