php $_SERVER['argv'] 带有 HTTP GET 和 CLI 问题
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9270030/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
$_SERVER['argv'] with HTTP GET and CLI issue
提问by gremo
I'm trying to write down a script to fetch some online data; script should be invoked either by a cron job or php cliand with standard GET HTTP request. As stated on PHP website $_SERVER['argv']should fit my needs:
我正在尝试写一个脚本来获取一些在线数据；脚本应由cron 作业或 php cli以及标准的GET HTTP 请求调用。正如 PHP 网站上所述，$_SERVER['argv']应该符合我的需求：
Array of arguments passed to the script. When the script is run on the command line, this gives C-style access to the command line parameters. When called via the GET method, this will contain the query string.
传递给脚本的参数数组。当脚本在命令行上运行时，这将提供对命令行参数的 C 样式访问。当通过 GET 方法调用时，这将包含查询字符串。
However i can't get it to work with standard HTTP GET request. $_SERVER['argv']is not setted. What i'm missing?
但是我不能让它与标准的 HTTP GET 请求一起工作。$_SERVER['argv']未设置。我缺少什么？
<?php
// jobs/fetch.php
var_dump($_SERVER['argv']);
?>
CLI outputphp jobs/fetch.php -a -bhello:
命令行输出php jobs/fetch.php -a -bhello：
array(3) {
[0]=>
string(14) "jobs/fetch.php"
[1]=>
string(2) "-a"
[2]=>
string(7) "-bhello"
}
GET outputjobs/fetch.php?a=&b=hello:
获取输出jobs/fetch.php?a=&b=hello：
Notice: Undefined index: argv in jobs/fetch.php.
注意：未定义索引：jobs/fetch.php 中的 argv。
回答by drew010
The manual didn't state this very well, but, if you want $_SERVER['argc'], $_SERVER['argv'], $argc, $argvto be registered when you are not running in CLImode, then the php.inivalue register_argc_argvneeds to be enabled in php.ini (off by default [for performance reasons]).
该手册并没有说明这一点非常好，但是，如果你想$_SERVER['argc']，$_SERVER['argv']，$argc，$argv要当你不是在运行注册CLI模式，则php.ini值register_argc_argv需要默认情况下，在php.ini（关启用[出于性能的考虑] ）。
You could do the following to get argv, or query string args depending on how the script is running:
argv根据脚本的运行方式，您可以执行以下操作来获取或查询字符串参数：
if (php_sapi_name() == 'cli') {
$args = $_SERVER['argv'];
} else {
parse_str($_SERVER['QUERY_STRING'], $args);
}
Here are some details from php.ini:
以下是来自php.ini：
; This directive determines whether PHP registers $argv & $argc each time it
; runs. $argv contains an array of all the arguments passed to PHP when a script
; is invoked. $argc contains an integer representing the number of arguments
; that were passed when the script was invoked. These arrays are extremely
; useful when running scripts from the command line. When this directive is
; enabled, registering these variables consumes CPU cycles and memory each time
; a script is executed. For performance reasons, this feature should be disabled
; on production servers.
; Note: This directive is hardcoded to On for the CLI SAPI
; Default Value: On
; Development Value: Off
; Production Value: Off
; http://php.net/register-argc-argv
See also http://www.php.net/manual/en/reserved.variables.argv.phpand parse_str().
另请参阅http://www.php.net/manual/en/reserved.variables.argv.php和parse_str()。
回答by gremo
You will have to use $_GETor$_SERVER['argv']depending on how your script is called. Neither one is used for both.
您将不得不使用$_GET或$_SERVER['argv']取决于您的脚本的调用方式。两者都不使用任何一种。
For example:
例如：
if(!empty($_SERVER['argv'][0]) {
$a = $_SERVER['argv'][1];
$b = $_SERVER['argv'][2];
} else {
$a = $_GET['a'];
$b = $_GET['b'];
}
