在外壳的命令行参数传递 [英] Pass parameters from the command line in shell

问题描述

DATE1=`perl -e 'use POSIX qw(strftime); print strftime "%Y-%m-%d",localtime(time()- 3600*72);'`

DATE2=`perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*72);'`

DATE3=`perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*48);'`

DATE4=`perl -e 'use POSIX qw(strftime); print strftime "%Y/%m/%d",localtime(time()- 3600*48);'`

下面是我的shell脚本（test1.sh），其中我需要打印的四个日期和所有四个日期应该从上面。这意味着我需要这些四个日期以上通过从命令提示符只对下面的shell脚本。

Below is my shell script(test1.sh), in which I need to print four dates and all those four dates should come from above. Meaning I need to pass above those four dates from the command prompt only to the below shell script.

#!/bin/bash

echo Date1
echo Date2
echo Date3
echo Date4

所以，当我运行shell脚本喜欢这个 - 它应该从以上四个日期获取日期1，日期2，DATE3，Date4？

So when I am running the shell script like this- it should get Date1, Date2, Date3, Date4 from the above four dates?

sh -x test1.sh Date1 Date2 Date3 Date4

是否有可能在shell脚本来完成？

Is it possible to do in shell script?

推荐答案

如果您在shell执行你的第一个code块，那么你就已经定义了4个变量与命令的输出。访问任何这些变量，你必须prePEND一个 $ 来的变量名 - 例如， $ DATE1 。

If you execute your first code-block in a shell, then you'll have defined 4 variables with the output of those commands. To access any of those variables, you have to prepend a $ to the variable name - say, $DATE1.

因此​​，对于这些参数运行脚本，你应该运行：

So, for running your script with those parameters, you should run:

 bash -x test1.sh $DATE1 $DATE2 $DATE3 $DATE4

，请注意 $ 和案例敏感。

最后，在你的脚本，参数在变量的家庭被调用的参数的顺序号收到。所以，你的第一个参数是由 $ 1 检索，第二个由 $ 2 ，等等。

Finally, in your script, arguments are received in a family of variables that are called with the number of order of the parameter. So, your first parameter is retrieved by $1, the second by $2, and so on.

所以，你的脚本结束是：

So, your script ends up being:

#!/bin/bash

echo $1
echo $2
echo $3
echo $4

手表也正在定义一个她邦告诉脚本是由庆典 PTED间$ P $，但你的previous调用使用 SH 。他们有很多兼容性的，但他们是不一样的。

Watch also that you are defining a she-bang that tells the script to be interpreted by bash, but your previous invocation was using sh. They have lots of compatibilities, but they are not the same.

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