通过 cron 执行 PHP - 未指定输入文件

Execute PHP via cron - No Input file specified(通过 cron 执行 PHP - 未指定输入文件)

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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