问题描述
我正在使用以下命令通过 cron 执行 PHP 文件
I'm using the following command to execute a PHP file via cron
php -q /home/seilings/public_html/dvd/cron/mailer.php
问题是我有一个包含在执行中的文件，该文件确定要加载的配置....例如:
The problem is that I Have a file that's included in the execution that determines which config to load.... such as the following:
if (!strstr(getenv('HTTP_HOST'), ".com")) {
$config["mode"] = "local";
} else {
$config["mode"] = "live";
}
cron 正在加载本地配置，而它应该加载实时配置.我试过使用文件的 http://URL 而不是绝对路径，但没有找到该文件.我是否需要更改命令以使用其中的 URL?
The cron is loading the LOCAL config when it should be loading the LIVE config. I've tried using the http:// URL to the file instead of the absolute path but it didn't find the file. Do I need to change the command to use a URL within it?
推荐答案
使用这个 php_sapi_name() 检查脚本是否在命令行上被调用:
Use this php_sapi_name() to check if the script was called on commandline:
if (php_sapi_name() === 'cli' OR !strstr(getenv('HTTP_HOST'), ".com")) {
$config["mode"] = "local";
} else {
$config["mode"] = "live";
}
如果您想在命令行上使用live"，请使用以下代码:
If you want to use "live" on the commandline use this code:
if (php_sapi_name() === 'cli' OR strstr(getenv('HTTP_HOST'), ".com")) {
$config["mode"] = "live";
} else {
$config["mode"] = "local";
}
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