这是一篇有点老的文章了，但是因为它带有译文，所以我就转摘一下。
命令行下的PHP，以前也是经常用的，现在用的少了。毕竟命令下行，效果更好，以前都是用来循环处理订单和检测无效信息的。。。
前两天我写了博客说bluehost的命令行不支持yii框架呢。。
以下是原文：
Author:Darrell Brogdon
As most of us already know, PHP is the best language for developing dynamic web pages available today. Not many people are aware that it can be used as a shell scripting language as well. While PHP as a shell script isn't as robust as Bash or Perl it does have definite advantages, especially if you're like me and are more proficient in PHP than you are in Perl.
The requirements for using PHP as a shell language is that you must compile PHP as a CGI binary instead of as an Apache module. There are certain security issues related to this so please refer to the PHP Manual when doing so.
Where the code is concerned, the only difference between a PHP shell script and a regular PHP web page is the existence of the standard shell call at the top of the script:
#!/usr/local/bin/php -q
We're using the '-q' switch so that the HTTP headers are suppressed. Also, you're still required to begin and end the script with the standard PHP tags:
<?php ?>
So let's delve into the standard code sample we all know and love:
#!/usr/local/bin/php -q
<?php
print("Hello, world!\n");
?>
This, as most of you know already (about a billion times over) will simply output to the screen "Hello, world!".
Passing arguments to the shell script
Commonly with a shell script you need to pass arguments to the script. This is easily done using the built-in '$argv' array as show in the following example:
#!/usr/local/bin/php -q
<?php
$first_name = $argv[1];
$last_name = $argv[2];
print("Hello, $first_name $last_name! How are you today?\n");
?>
So in the above script we're printing out the first two arguments to the script which would be called like this:
[dbrogdon@artemis dbrogdon]$ scriptname.ph Darrell Brogdon
Which would print out:
Hello, Darrell Brogdon! How are you today?
[dbrogdon@artemis dbrogdon]$
The only major difference with the '$argv' array between a shell script and a web page is that in a shell script, '$argv[0]' is the name of your script. In a web page it is the first argument in your query string.
Making a script more interactive
But how do we wait for user input? How do we create a truly interactive script? Well, PHP has no native functions like the 'read' command in shell but we can always emulate it using the following PHP function:
*Note that this function will only work for Unix.
<?php
function read() {
$fp=fopen("/dev/stdin", "r");
$input=fgets($fp, 255);
fclose($fp);
return $input;
}
?>
This function opens a file pointer to Standard In (/dev/stdin on Linux) and reads anything from this pointer up to 255 bytes, newline, or EOF. In this case a newline is most likely to occur. It then closes the file pointer and returns the data.
So now let's modify our previous script to wait for user input using the newly created 'read()' function:
#!/usr/local/bin/php -q
<?php
function read() {
$fp=fopen("/dev/stdin", "r");
$input=fgets($fp, 255);
fclose($fp);
return $input;
}
print("What is your first name? ");
$first_name = read();
print("What is your last name? ");
$last_name = read();
print("\nHello, $first_name $last_name! Nice to meet you!\n");
?>
You may notice, however, that when you execute this script the last line to be printed is broken into three lines instead of one as it should be. This is because our 'read()' function also takes in the newline character. This is easily fixed by stripping off the newline before we return the data:
<?php
function read() {
$fp=fopen("/dev/stdin", "r");
$input=fgets($fp, 255);
fclose($fp);
return str_replace("\n", "", $input);
}
?>
Embedding PHP shell scripts within a regular shell script
Sometimes it might be handy to embed a PHP shell script within a script written in Bash or other shell. This is fairly simple but can get a tad tricky.
First, how to embed the PHP code:
#!/bin/bash
echo This is the Bash section of the code.
/usr/local/bin/php -q << EOF
<?php
print("This is the PHP section of the code\n");
?>
EOF
Pretty simple huh? Until you add a variable that is. This is the tricky part. Try running the following code segment:
#!/bin/bash
echo This is the Bash section of the code.
/usr/local/bin/php -q << EOF
<?php
$myVar = "PHP";
print("This is the $myVar section of the code.\n");
?>
EOF
You'll get the following error:
<b>Parse error</b>: parse error in <b>-</b> on line <b>2</b><br>
To fix this you have to escape all of the '$' characters in your PHP code:
#!/bin/bashecho This is the Bash section of the code./usr/local/bin/php -q << EOF<?php \$myVar = "PHP"; print("This is the \$myVar section of the code.\n");?>EOF
So that should get you started on creating your own shell scripts using PHP!
译文：
可能很多人都想过使用PHP编写一些定时发信之类的程序，但是却没有办法定时执行PHP；一次去PHPBuilder的时候，发现了这一篇文章，于是想给大家翻译一下（同时做了一些修改），希望对大家有用。第一次翻译文章，不好请多多见谅。
我们都知道，PHP是一种非常好的动态网页开发语言（速度飞快，开发周期短……）。但是只有很少数的人意识到PHP也可以很好的作为编写Shell脚本的 语言，当PHP作为编写Shell脚本的语言时，他并没有Perl或者Bash那么强大，但是他却有着很好的优势，特别是对于我这种熟悉PHP但是不怎么 熟悉Perl的人。
要使用PHP作为Shell脚本语言，你必须将PHP作为二进制的CGI编译，而不是Apache模式；编译成为二进制CGI模式运行的PHP有一些安全性的问题，关于解决的方法可以参见PHP手册（http://www.php.net）。
一开始你可能会对于编写Shell脚本感到不适应，但是会慢慢好起来的：将PHP作为一般的动态网页编写语言和作为Shell脚本语言的唯一不同就在于一个Shell脚本需要在第一行生命解释本脚本的程序路径：
#!/usr/local/bin/php -q
我们在PHP执行文件后面加入了参数“-1”，这样子PHP就不会输出HTTPHeader（如果仍需要作为Web的动态网页，那么你需要自己使用header函数输出HTTPHeader）。当然，在Shell脚本的里面你还是需要使用PHP的开始和结束标记：
<?php 代码 ?>
现在让我们看一个例子，以便于更好的了解用PHP作为Shell脚本语言的使用：
#!/usr/local/bin/php -q
<?php
print("Hello, world!\n");
?>
上面这个程序会简单的输出“Hello, world!”到显示器上。
一、传递Shell脚本运行参数给PHP：
作为一个Shell脚本，经常会在运行程序时候加入一些参数，PHP作为Shell脚本时有一个内嵌的数组“$argv”，使用“$argv”数组可以很 方便的读取Shell脚本运行时候的参数（“$argv[1]”对应的是第一个参数，“$argv[2]”对应的是第二个参数，依此类推）。比如下面这个 程序：
#!/usr/local/bin/php -q
<?php
$first_name = $argv[1];
$last_name = $argv[2];
printf("Hello, %s %s! How are you today?\n", $first_name, $last_name);
?>
上面的代码在运行的时候需要两个参数，分别是姓和名，比如这样子运行：
[dbrogdon@artemis dbrogdon]$ scriptname.ph Darrell Brogdon
Shell脚本在显示器上面会输出：
Hello, Darrell Brogdon! How are you today?
[dbrogdon@artemis dbrogdon]$
在PHP作为动态网页编写语言的时候也含有“$argv”这个数组，不过和这里有一些不同：当PHP作为Shell脚本语言的时候“$argv[0]”对 应的是脚本的文件名，而当用于动态网页编写的时候，“$argv[1]”对应的是QueryString的第一个参数。
二、编写一个具有交互式的Shell脚本：
如果一个Shell脚本仅仅是自己运行，失去了交互性，那么也没有什么意思了。当PHP用于Shell脚本的编写的时候，怎么读取用户输入的信息呢？很不 幸的是PHP自身没有读取用户输入信息的函数或者方法，但是我们可以效仿其他语言编写一个读取用户输入信息的函数“read”：
<?php
function read() {
$fp = fopen('/dev/stdin', 'r');
$input = fgets($fp, 255);
fclose($fp);
return $input;
}
?>
需要注意的是上面这个函数只能用于Unix系统（其他系统需要作相应的改变）。上面的函数会打开一个文件指针，然后读取一个不超过255字节的行（就是fgets的作用），然后会关闭文件指针，返回读取的信息。
现在我们可以使用函数“read”将我们前面编写的程序1修改一下，使他更加具有“交互性”了：
#!/usr/local/bin/php -q
<?php
function read() {
$fp = fopen('/dev/stdin', 'r');
$input = fgets($fp, 255);
fclose($fp);
return $input;
}
print("What is your first name? ");
$first_name = read();
print("What is your last name? ");
$last_name = read();
print("\nHello, $first_name $last_name! Nice to meet you!\n");
?>
将上面的程序保存下来，运行一下，你可能会看到一件预料之外的事情：最后一行的输入变成了三行！这是因为“read”函数返回的信息还包括了用户每一行的结尾换行符“\n”，保留到了姓和名中，要去掉结尾的换行符，需要把“read”函数修改一下：
<?php
function read() {
$fp = fopen('/dev/stdin', 'r');
$input = fgets($fp, 255);
fclose($fp);
$input = chop($input); // 去除尾部空白
return $input;
}
?>
三、在其他语言编写的Shell脚本中包含PHP编写的Shell脚本：
有时候我们可能需要在其他语言编写的Shell脚本中包含PHP编写的Shell脚本。其实非常简单，下面是一个简单的例子：
#!/bin/bash
echo This is the Bash section of the code.
/usr/local/bin/php -q << EOF
<?php
print("This is the PHP section of the code\n");
?>
EOF
其实就是调用PHP来解析下面的代码，然后输出；那么，再试试下面的代码：
#!/bin/bash
echo This is the Bash section of the code.
/usr/local/bin/php -q << EOF
<?php
$myVar = 'PHP';
print("This is the $myVar section of the code\n");
?>
EOF
可以看出两次的代码唯一的不同就是第二次使用了一个变量“$myVar”，试试运行，PHP竟然给出出错的信息：“Parse error: parse error in - on line 2”！这是因为Bash中的变量也是“$myVar”，而Bash解析器先将变量给替换掉了，要想解决这个问题，你需要在每个PHP的变量前面加上“\” 转义符，那么刚才的代码修改如下：
#!/bin/bash
echo This is the Bash section of the code.
/usr/local/bin/php -q << EOF
<?php
\$myVar = 'PHP';
print("This is the \$myVar section of the code\n");
?>
EOF
好了，现在你可以用PHP编写你自己的Shell脚本了，希望你一切顺利。如果有什么问题，可以去http://www.PHPBuilder.com或者http://www.zPHP.com上面讨论。
---EOF---
仔细看上面的一段， fopen("/dev/stdin")，如果你看过yii的框架，你会发现，现在很多人都在使用php://stdin，其实，这只是PHP把这种封装成了一个协议，更方便而己。。可以参考yiiframework.com
#1