如何使一个shell脚本走在命令行中传递参数？ [英] How to make a shell script take an argument passed in the command line?

问题描述

我有一个包含以下内容的shell脚本（count_reviews.sh）：

I have a shell script (count_reviews.sh) that contains the following:

#!/bin/bash
grep -c "Author" "#1"

我一直在使用$ 1，因为我发现，这会将第一个参数，在命令行，但我仍然得到以下提到的错误尝试。我也用使用chmod + x可让我的文件可执行。

I have tried using "$1" as I found that this takes the first argument in the command line but I still get the error mentioned below. I have also used chmod +x to make my file executable.

脚本计数作者出现在文件中的次数。我需要能够使命令行采取的％./count_reviews.sh hotel_72572.dat，其中 hotel_72572.dat <输入/ code>是一个示例文件名。倍作者出现的数目将被下方打印出来。当我这样做不过，我得到一个错误 -bash：FG：％：没有这样的工作。是什么原因造成这一点，我该如何解决？我的 count_reviews.sh 文件在同一目录下所有的我的酒店数据文件，如果该事项。

The script counts the number of times "Author" appears in the file. I am required to be able to make the command line take an input of "% ./count_reviews.sh hotel_72572.dat" where hotel_72572.dat is an example file name. The number of times author appears will then be printed out underneath. When I do this however, I am getting an error -bash: fg: %: no such job. What is causing this and how do I fix it? My count_reviews.sh file is in the same directory as all of my hotel data files if that matters.

推荐答案

您确实应该使用 $ 1 ，但是当您尝试运行脚本只需使用 ./ your_script.sh your_argument 不要在开头添加％。

You should indeed use $1, but when you try to run your script just use ./your_script.sh your_argument don't add the % in the beginning.

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