用于读取文件的 Bash 脚本
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Bash script to read a file
提问by hfranco
Not sure why the last line does not cut the " from the script:
不知道为什么最后一行没有从脚本中删除“：
#!/bin/bash
FILENAME=
while read line
do
cut -d '"' -f2
echo $line
done < $FILENAME
$ cat file
"1" test
"2" test
"3" test
"4" test
"5" test
If I run this script with the following command:
如果我使用以下命令运行此脚本：
$ ./test file
2
3
4
5
"1" test
回答by Jonathan Leffler
The loop executes once.
循环执行一次。
- It reads
"1" testinto variable$line. - It executes
cut -d '"' -f2which reads lines 2-5 of the file (because that is the standard input at the time) and prints the number. - It echoes what it read as the first line.
- 它读
"1" test入变量$line。 - 它执行
cut -d '"' -f2读取文件的第 2-5 行（因为这是当时的标准输入）并打印数字。 - 它呼应了它作为第一行读取的内容。
Fix:
使固定：
cut -d '"' -f2 $FILENAME
If, on the other hand, you want to get the numbers into a variable, you could do this in a variety of ways, including:
另一方面，如果您想将数字放入变量中，您可以通过多种方式实现，包括：
cut -d '"' -f2 $FILENAME |
while read number
do # What you want
echo $number
done
or:
或者：
while read line
do
number=$(echo "$line" | cut -d '"' -f2)
echo $number
done
回答by Paused until further notice.
Bash can do all the work for you. There's no need for cut:
Bash 可以为您完成所有工作。没有必要cut：
#!/bin/bash
FILENAME=
while read -r -a line
do
echo ${line//\"}
done < "$FILENAME"
That reads the line into an array, then treats the array as a scalar which gives you the first element. Then the brace expansion in the echostrips out the quotation marks.
这会将行读入一个数组，然后将该数组视为一个标量，该标量为您提供第一个元素。然后括号中的扩展echo去掉引号。
Or you can let cutdo all the work and give Bash a long coffee break:
或者你可以让cut所有的工作做完，给 Bash 一个长时间的咖啡休息时间：
FILENAME=
cut -d '"' -f2 "$FILENAME"
Always quote variables that contain filenames.
始终引用包含文件名的变量。
回答by ghostdog74
like dennis mentioned, there's no need to use external commands
就像丹尼斯提到的，没有必要使用外部命令
$ while read -r line; do set -- $line; echo ${1//\"/}; done<file
1
2
3
4
5
But external commands runs faster if you have very large files.
但是如果您有非常大的文件，外部命令运行速度会更快。
$ cut -d'"' -f2 file
1
2
3
4
5
$ awk -F'"' '{print }' file
1
2
3
4
5
$ sed 's/^"//;s/".*//' file
1
2
3
4
5
回答by Chris J
Jonathan Leffler's given you a simpler method (which will also be more efficient), but in case this is a simplification for something you're going to expand on (where just calling cut won't do what you want), and just to demonstrate the principle anyway, your code would need to be fixed up to feed each line to stdin explicitly as follows:
乔纳森莱夫勒给了你一个更简单的方法（这也会更有效），但如果这是你要扩展的东西的简化（只调用 cut 不会做你想做的事），只是为了演示无论如何，您的代码需要修复以将每一行明确地提供给标准输入，如下所示：
#!/bin/bash
FILENAME=
while read line
do
echo $line | cut -d '"' -f2
done < $FILENAME
