[英]How to pass commands and options to PHP script in CLI (e.g. php file.php mycommand -d “my option”)
I'm aware I can use getopt and argv in my PHP script, but I don't know how to do something like the following: 我知道我可以在PHP脚本中使用getopt和argv,但我不知道如何执行以下操作:
php file.php mycommand -d "my option"
I've tried to use getopt with this but I guess the mycommand upsets it: 我试图与此使用getopt,但我猜mycommand使其不高兴:
var_dump(getopt('d:')); // empty array
Currently I use $argv to get the command but using this with -x options means I need to do some processing to extract the options. 当前,我使用$ argv来获取命令,但是将其与-x选项一起使用意味着我需要进行一些处理以提取选项。 I'm hoping there is a proper way to do so: 我希望有一个适当的方法可以做到这一点:
array(5) {
[0]=>
string(27) "file.php"
[1]=>
string(4) "mycommand"
[2]=>
string(2) "-d"
[3]=>
string(4) "my option"
}
Is there a way to do so? 有办法吗?
You have to remove the space between option and value 您必须删除选项和值之间的空格
Command 命令
php 1.php -xmycommand -d"my option"
PHP Code PHP代码
var_dump(getopt('x:d:'));
Output 产量
array(2) {
["x"]=>
string(9) "mycommand"
["d"]=>
string(9) "my option"
}
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