20.29 "while read line"循环内无法改变循环外变量

https://scz.617.cn/unix/201509231351.txt

Q:

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#!/bin/bash

count=0

#
# "IFS="意在保持每行首尾两端的空白字符
#
# 参看read(1)，-r表示将反斜杠\视为普通字符
#
ls -1aF /tmp | while IFS= read -r line
do
    #
    # let count+=1
    # count=$((count+1))
    # ((count++))
    # count=$(expr $count + 1 )
    #
    # 第一种可移植性最高，最后一种效率最低
    #
    count=$((count+1))
    echo $count
done

printf "Found %u entries\n" "${count}"
--------------------------------------------------------------------------

$ ./test.sh
1
2
...
49
50
Found 0 entries

为什么最后显示0，而不是50？

A: scz 2015-09-23 13:51

对于bash下的shell script，管道符|会导致后续命令在子shell里运行，此时
"while read line"循环内的count改变发生在子进程里，当然不会影响父进程(当前
shell)里的count。要想在父子进程间同步count，必将涉及进程间通信。不是所有的
shell对管道符的处理都是创建子shell，比如ksh就不是。

解决方案之一:

--------------------------------------------------------------------------
#!/bin/bash

count=0

while IFS= read -r line
do
    ((count++))
    echo $count
#
# 用"<(...)"这种子shell语法
#
done < <(ls -1aF /tmp)

printf "Found %u entries\n" "${count}"
--------------------------------------------------------------------------

该方案不用显式创建临时文件。实际上，<(...)子shell语法会在/tmp下隐式创建FIFO。

其他解决方案:

--------------------------------------------------------------------------
#!/bin/bash

#
# 显式创建命名管道
#
rm -f /tmp/namedpipe
mkfifo /tmp/namedpipe
ls -1aF /tmp > /tmp/namedpipe &

count=0

while IFS= read -r line
do
    ((count++))
    echo $count
done < /tmp/namedpipe

printf "Found %u entries\n" "${count}"

rm -f /tmp/namedpipe
--------------------------------------------------------------------------
#!/bin/bash

#
# {}构成command group
#
ls -1aF /tmp |
{
    count=0
    while IFS= read -r line
    do
        ((count++))
        echo $count
    done
    printf "Found %u entries\n" "${count}"
}
--------------------------------------------------------------------------
#!/bin/bash

#
# 可以放在{}外
#
count=0

ls -1aF /tmp |
{
    while IFS= read -r line
    do
        ((count++))
        echo $count
    done
    printf "Found %u entries\n" "${count}"
}
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#!/bin/bash

var=$(ls -1aF /tmp)

#
# 格式串中务必有\n。避免不以\n结尾的最后一行不被while read处理
#
printf "%s\n" "${var}" |
{
    count=0
    while IFS= read -r line
    do
        ((count++))
        echo $count
    done
    printf "Found %u entries\n" "${count}"
}
--------------------------------------------------------------------------

$ ./test.sh
1
2
...
49
50
Found 50 entries

"IFS="用于"while read"时，意在保持每行首尾两端的空白字符，对比如下输出:

$ echo " this is  a  test " | while IFS= read -r line;do echo "[${line}]";done
[ this is  a  test ]
$ echo " this is  a  test " | while read -r line;do echo "[${line}]";done
[this is  a  test]

"-r"用于"while read"时，意在取消每行内容中反斜杠\的转义效果，对比如下输出:

$ { echo 'this \\ line is \';echo 'continued'; } | while IFS= read -r line;do echo "[${line}]";done
[this \\ line is \]
[continued]
$ { echo 'this \\ line is \';echo 'continued'; } | while IFS= read line;do echo "[${line}]";done
[this \ line is continued]

注意，{ ... }内部两端各有一个空格，必须存在。

一般写成"while IFS= read -r line"，确保${line}是原始数据。

Q:

"while read line"循环如何从变量获取输入，而不是从管道或文件获取输入？

A: scz

--------------------------------------------------------------------------
#!/bin/bash

var=$(ls -1aF /tmp)

count=0

while IFS= read -r line
do
    ((count++))
    echo $count
#
# 用<<<从变量获取输入，而不是<
#
# 搜"bash here string"了解更多细节
#
done <<< "${var}"

printf "Found %u entries\n" "${count}"
--------------------------------------------------------------------------

"bash here string"要求/tmp可写，会隐式创建临时文件。

另有一种类似的"here document":

--------------------------------------------------------------------------
#!/bin/bash

var=$(ls -1aF /tmp)

count=0

while IFS= read -r line
do
    ((count++))
    echo $count
#
# 搜"bash here document"了解更多细节
#
# 后面的${var}两侧不要用双引号，不要写注释，否则视为文件内容
#
done << EOF
${var}
EOF

printf "Found %u entries\n" "${count}"
--------------------------------------------------------------------------