shell - Unix shell to prompt user for filename

Question

please pardon since I am a UNIX beginner.

I wanna write a shell script that can ask user type in the filename and output the number of lines in that file, here is my code:

echo "Pls enter your filename:"
read filename
result=wc -l $filename

echo "Your file has $result lines"

However, I couldn't get it working since it complains about the identifier "filename". Could experts help?Thanks !!

Answer 1

这很好用，至少在bash. 嗯，read工作正常。但是，分配给的result可能应该是：

result=$(wc -l $filename)

但是，由于该命令同时输出行数和文件名，您可能需要稍微更改它以获取行数，例如：

result=$(cat $filename | wc -l)

或者：

result=$(wc -l <$filename)

您拥有的命令：

result=wc -l $filename

将设置result为文字wc，然后尝试执行-l命令。

例如，下面的五行脚本：

#!/bin/bash
echo "Pls enter your filename:"
read filename
result=$(cat $filename | wc -l)
echo "Your file has $result lines"

将在运行并将其名称作为输入时产生以下内容：

Your file has 5 lines

---

如果您不使用bash，则需要指定您使用的 shell。不同的shell有不同的做事方式。

Answer 2

干得好 ：

echo "Pls enter your filename:" 
read filename 
result=`wc -l $filename | tr -s " " | cut -f2 -d' '`

echo "Your file has $result lines"

Answer 3

如果你想评估 wc -l $filename ，你不必这样做：

result=$(wc -l $filename)

否则，使用 result=wc -l $filename，bash 会将 wc 分配给 result，并将下一个单词 (-l) 解释为要运行的命令。