Please note that, $argv and $argc need to be declared global, while trying to access within a class method.
<?php
class A
{
public static function b()
{
var_dump($argv);
var_dump(isset($argv));
}
}
A::b();
?>
will output NULL bool(false) with a notice of "Undefined variable ..."
whereas global $argv fixes that.You are here
命令行 php 脚本参数 $argv — 传递给脚本的参数数组 有大用
星期四, 2018-11-01 07:56 — adminshiping1
add a notetufan dot oezduman at googlemail dot com ¶hamboy75 at example dot com ¶php at simoneast dot net ¶Andreas ¶Steve Schmitt ¶andrew w ¶fabio at naoimporta dot com ¶KRowe ¶Jesse ¶
两种方法 通过 var_dump(); 试一下就知道了
1) $_SERVER['argv']
2) $argv
$argv
$argv — 传递给脚本的参数数组
范例 ¶
Example #1 $argv 范例
<?php
var_dump($argv);
?>当使用这个命令执行：php script.php arg1 arg2 arg3
以上例程的输出类似于：
array(4) {
[0]=>
string(10) "script.php"
[1]=>
string(4) "arg1"
[2]=>
string(4) "arg2"
[3]=>
string(4) "arg3"
}
add a noteUser Contributed Notes 9 notes
76
7 years ago
32
4 years ago
To use $_GET so you dont need to support both if it could be used from command line and from web browser.
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)==2)
$_GET[$e[0]]=$e[1];
else
$_GET[$e[0]]=0;
}5
2 years ago
Sometimes $argv can be null, such as when "register-argc-argv" is set to false. In some cases I've found the variable is populated correctly when running "php-cli" instead of just "php" from the command line (or cron).0
1 year ago
How to check if one parameter is given:
if ($argc < 2 )
{
exit( "Usage: program <parameter1>\n" );
}
process( $argv[1] );3
9 years ago
If you come from a shell scripting background, you might expect to find this topic under the heading "positional parameters".-6
1 year ago
An easier way to populate $_GET with $argv values.
<?php
if ( isset( $argv ) ) {
parse_str(
join( "&", array_slice( $argv, 1 )
), $_GET );
}
?>-17
2 years ago
When you pass an option to the file that intercept the request, it will be transformed into an array item, and the option name will be lost. Only its content is captured.
<?php
var_dump($argv);
?>
call : "php file.php --test=foo baz"
will print
array(3) {
[0] =>
string(16) "file.php"
[1] =>
string(3) "foo"
[2] =>
string(3) "baz"
}-24
3 years ago
Improves on hamboy75's note by providing better support for positional arguments:
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)==2)
$_GET[$e[0]]=$e[1];
else
$_GET[]=$e[0];
}
var_dump($_GET);
Using this modification, arguments without an = are treated as positional (this is not web standard but generally works).-99
5 years ago
If your script is read from standard input or with the -r option, $argv[0] will be "-".
If you use the "--" option to separate PHP's arguments from your script's arguments, $argv[1] will be "--" if your script is read from a file. But if your script is read from standard input or with the -r option, the "--" will be removed.普通分类: