Bash脚本读取文件 [英] Bash script to read a file

问题描述

不确定为什么最后一行没有从脚本中剪切":

Not sure why the last line does not cut the " from the script:

#!/bin/bash

FILENAME=$1
while read line
do
cut -d '"' -f2
echo $line
done < $FILENAME

$ cat file

"1" test
"2" test
"3" test
"4" test
"5" test

如果我使用以下命令运行此脚本:

If I run this script with the following command:

$ ./test file

2
3
4
5
"1" test

推荐答案

循环执行一次.

• 它将"1" test读入变量$line.
• 它执行cut -d '"' -f2，该操作读取文件的2-5行(因为这是当时的标准输入)并打印数字.
• 它回显第一行的内容.

• It reads "1" test into variable $line.
• It executes cut -d '"' -f2 which reads lines 2-5 of the file (because that is the standard input at the time) and prints the number.
• It echoes what it read as the first line.

修复:

cut -d '"' -f2 $FILENAME

另一方面，如果您希望将数字放入变量中，则可以通过多种方式进行操作，包括:

If, on the other hand, you want to get the numbers into a variable, you could do this in a variety of ways, including:

cut -d '"' -f2 $FILENAME |
while read number
do  # What you want
    echo $number
done

或:

while read line
do
    number=$(echo "$line" | cut -d '"' -f2)
    echo $number
done

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