PHP - Windows中的命令行参数

I'm trying to run PHP from the command line under Windows XP.

我正在尝试从Windows XP下的命令行运行PHP。

That works, except for the fact that I am not able to provide parameters to my PHP script.

这是有效的,除了我无法为我的PHP脚本提供参数这一事实。

My test case:

我的测试用例:

echo "param = ".$param."\n";  
var_dump($argv);

I want to call this as:

我想称之为:

php.exe -f test.php -- param=test

php.exe -f test.php - param = test

But I never get the script to accept my parameter.

但我从来没有让脚本接受我的参数。

The result I get from the above script

我从上面的脚本得到的结果

`PHP Notice: Undefined variable: param in C:\test.php on line 2

`PHP注意:未定义的变量:第2行的C:\ test.php中的param

param = ''
array(2) {
  [0]=> string(8) "test.php"
  [1]=> string(10) "param=test"
}

I am trying this using PHP 5.2.6. Is this a bug in PHP5?

我正在尝试使用PHP 5.2.6。这是PHP5中的错误吗?

The parameter passing is handled in the online help Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch. This seemed to be working under PHP4, but not under PHP5. Under PHP4 I could use the same script that could run on the server without alteration on the command line. This is handy for local debugging, for example saving the output in a file to be studied.

参数传递在在线帮助中处理注意:如果需要将参数传递给脚本,则需要传递 - 作为使用-f开关时的第一个参数。这似乎是在PHP4下工作,但不是在PHP5下。在PHP4下,我可以使用可以在服务器上运行的相同脚本,而无需在命令行上进行更改。这对于本地调试很方便,例如将输出保存在要研究的文件中。

11 个解决方案

#1

Why do you have any expectation that param will be set to the value? You're responsible for parsing the command line in the fashion you desire, from the $argv array.

为什么你有任何期望将param设置为该值?您负责从$ argv数组中以您想要的方式解析命令行。

#2

The parameter passing is handled in the online help Note: If you need to pass arguments to your scripts you need to pass -- as the first argument when using the -f switch. This seemed to be working under PHP4, but not under PHP5.

参数传递在在线帮助中处理注意:如果需要将参数传递给脚本,则需要传递 - 作为使用-f开关时的第一个参数。这似乎是在PHP4下工作,但不是在PHP5下。

But PHP still doesn't parse those arguments. It just passes them to the script in the $argv array.

但是PHP仍然没有解析这些论点。它只是将它们传递给$ argv数组中的脚本。

The only reason for the -- is so that PHP can tell which arguments are meant for the PHP executable and which arguments are meant for your script.

这样做的唯一原因是,PHP可以告诉哪些参数适用于PHP可执行文件以及哪些参数适用于您的脚本。

That lets you do things like this:

这可以让你做这样的事情:

php -e -n -f myScript.php -- -f -n -e

(The -f, -n, & -e after the -- are passed to myScript.php. The ones before are passed to PHP itself).

(-f之后的-f,-n,&-e传递给myScript.php。之前的那些传递给PHP本身)。

#3

If you want to pass the params similar to GET vars, then you can use the parse_str() function. Something similar to this:

如果你想传递类似于GET变量的参数,那么你可以使用parse_str()函数。与此类似的东西:

<?php
parse_str($argv[1]);
?>

Would produce a variable of $test with a value of myValue.

会产生$ test的变量,其值为myValue。

Hope this helps!

希望这可以帮助!

#4

You can use the getopt();

你可以使用getopt();

Check this::

http://sharingphp.blogspot.com/2011/04/php-cli-script-and-command-line.html

#5

PHP does not parameterize your command line parameters for you. See the output where your 2nd entry in ARGV is "param=test".

PHP不会为您参数化您的命令行参数。请参阅ARGV中第二个条目为“param = test”的输出。

What you most likely want is to use the PEAR package http://pear.php.net/package/Console_CommandLine: "A full featured command line options and arguments parser".

您最想要的是使用PEAR包http://pear.php.net/package/Console_CommandLine:“功能齐全的命令行选项和参数解析器”。

Or you can be masochistic and add code to go through your ARGV and set the parameters yourself. Here's a very simplistic snippet to get you started (this won't work if the first part isn't a valid variable name or there is more than 1 '=' in an ARGV part:

或者您可以自虐并添加代码以通过您的ARGV并自己设置参数。这是一个非常简单的片段,可以帮助您入门(如果第一部分不是有效的变量名称,或者在ARGV部分中有超过1'=',这将无效:

foreach($argv as $v) {
    if(false !== strpos($v, '=')) {
        $parts = explode('=', $v);
        ${$parts[0]} = $parts[1];
    }
}

#6

$argv is an array containing all your commandline parameters... You need to parse that array and set $param yourself.

$ argv是一个包含所有命令行参数的数组...你需要解析该数组并自己设置$ param。

$tmp = $argv[1];             // $tmp="param=test"
$tmp = explode("=", $tmp);   // $tmp=Array( 0 => param, 1 => test)

$param = $tmp[1];            // $param = "test";

#7

You can do something like:

你可以这样做:

if($argc > 1){
    if($argv[1] == 'param=test'){
        $param = 'test';
    }
}

Of course, you can get much more complicated than that as needed.

当然,根据需要,你可以得到比这更复杂的东西。

#8

You could use something like

你可以用类似的东西

if (isset($argv[1]) {
 $arg1 = $argv[1];             
 $arg1 = explode("=", $arg1);   
 $param = $arg1[1];            
}

(how to handle the lack of parameter/s is up to you) or if you need a more complex scenario, look into a commandline parser library such as the one from Pear.

(如何处理缺少参数/ s取决于您)或者如果您需要更复杂的场景,请查看命令行解析器库,例如Pear中的一个。

using the ${$parts[0]} = $parts[1]; posted in another solution lets you override any variable in your code, which doesnt really sound safe.

使用$ {$ parts [0]} = $ parts [1];在另一个解决方案中发布允许您覆盖代码中的任何变量,这听起来并不安全。

#9

If you like living on the cutting edge, PHP 5.3 has the getOpt() command which will take care of all this messy business for you. Somewhat.

如果您喜欢生活在最前沿,PHP 5.3具有getOpt()命令,它将为您处理所有这些混乱的业务。有些。

#10

command line example:

命令行示例:

php myserver.php host=192.168.1.4 port=9000

php myserver.php host = 192.168.1.4 port = 9000

in myserver.php use the following lines:

在myserver.php中使用以下行:

<?php

parse_str(implode('&', array_slice($argv, 1)), $_GET);
// Read Arguments
if (array_key_exists('host',$_GET))
{
    $host = $_GET['host'];
}
if (array_key_exists('port',$_GET))
{
    $port = $_GET['port'];
}
?>

#11

-1

you can use the $argv array. like this:

你可以使用$ argv数组。像这样:

<?php
 echo  $argv[1];
?>

remember that the first member of the $argv array (which is $argv[0]) is the name of the script itself, so in order to use the parameters for the application, you should start using members of the $argv[] from the '1'th index. when calling the application, use this syntax:

请记住,$ argv数组的第一个成员($ argv [0])是脚本本身的名称,因此为了使用应用程序的参数,您应该开始使用$ argv []的成员。 '1'指数。在调用应用程序时,请使用以下语法:

php myscript.php -- myValue

there is no need to put a name for the parameter. as you saw, what you called the var_dump() on the $argv[], the second member (which is the first parameter) was the string PARAM=TEST. right? so there is no need to put a name for the param. just enter the param value.

没有必要为参数添加名称。如你所见,你在$ argv []上调用var_dump(),第二个成员(第一个参数)是字符串PARAM = TEST。对?所以没有必要为param命名。只需输入参数值。

#1