为bash中的变量分配一个分号（';'）的值 [英] Assigning a value having semicolon (';') to a variable in bash

问题描述

我正在使用sed在unix shell（bash）上的一个字符串中转义（'\\'）分号（';'）。当我直接执行它而不将值赋值给变量时，它将起作用。也就是说，

  $ echohello | sed的/ \（[^ \\] \）; / \1\\; / g'
 hello\; 
 $ 
  

但是，当上述命令分配给一个变量：

  $ result =`echohello; | sed的/ \（[^ \\] \）; / \1\\; / g'` 
 $ 
 $ echo $ result 
 hello; 
 $ 
  

任何想法为什么？

我尝试使用带有和没有引号的值，但没有帮助。任何线索非常感谢。

btw，我第一次认为字符串末尾的分号以某种方式作为终结符，因此shell没有继续执行 sed （如果有任何意义）。但是，这并不是一个问题。我尝试使用分号不在字符串的末尾（某处之间）。我仍然看到与以前相同的结果。也就是说，

  $ echohel; lo| sed的/ \（[^ \\] \）; / \1\\; / g'
 hel\; lo 
 $ 
 $ result =`echo'hel; lo| sed的/ \（[^ \\] \）; / \1\\; / g'` 
 $ 
 $ echo $ result 
 hel; lo 
 $ 
  

解决方案

我觉得有趣使用反向滴答提供一个结果（您的结果），并使用 $（...）提供另一个结果（想要的结果）：

  $ echohello; | sed的/ \（[^ \\] \）; / \1\\; / g'
 hello\; 
 $ z1 = $（echohello;| sed's / \（[^ \\] \）; / \1\\; / g'）
 $ z2 =`echo'hello; | sed的/ \（[^ \\] \）; / \1\\; / g'` 
 $ printf％s\\\
$ z1$ z2
 hello\; 
 hello; 
 $ 
  

如果您需要使用现代的 x = $（...）符号代表旧的 x =`...`符号，这可能是。外壳使用后面的勾号进行了额外的反斜杠解释。调试shell脚本调用 al （参数列表）时，我可以使用一个小程序来演示这个。您可以使用 printf％s\\\
进行模拟：

  $ z2 =`echo'hello; | al sed's / \（[^ \\] \）; / \1\\; / g'` 
 $ echo$ z2
 sed 
s / \（[^ \] \）; / \1\; / g 
 $ z1 = $（echohello;| al sed's / \（[^ \ \] \）; / \1\\; / g'）
 $ echo$ z1
 sed 
s / \（[^ \\] \）; / \1\\; / g 
 $ z1 = $（echohello;| printf％s\\\
sed's / \（[^ \\\ \\] \）; / \1\\; / g'）
 $ echo$ z1
 sed 
s / \（[^ \\] \\）; / \1\\; / g 
 $ 
  

可以看到，由 sed 执行的脚本根据您是否使用 x = $（...）符号而不同， x =`...`符号。

  s / \ （[^ \] \）; / \1\; / g＃``
s / \（[^ \\] \）; / \1\\; / g＃$（）
  

摘要

使用 $（...）;这更容易理解。

I'm trying to escape ('\') a semicolon (';') in a string on unix shell (bash) with sed. It works when I do it directly without assigning the value to a variable. That is,

$ echo "hello;" | sed 's/\([^\\]\);/\1\\;/g'
hello\;
$

However, it doesn't appear to work when the above command is assigned to a variable:

$ result=`echo "hello;" | sed 's/\([^\\]\);/\1\\;/g'`
$ 
$ echo $result
hello; 
$

Any idea why?

I tried by using the value enclosed with and without quotes but that didn't help. Any clue greatly appreciated.

btw, I first thought the semicolon at the end of the string was somehow acting as a terminator and hence the shell didn't continue executing the sed (if that made any sense). However, that doesn't appear to be an issue. I tried by using the semicolon not at the end of the string (somewhere in between). I still see the same result as before. That is,

$ echo "hel;lo" | sed 's/\([^\\]\);/\1\\;/g'
hel\;lo
$
$ result=`echo "hel;lo" | sed 's/\([^\\]\);/\1\\;/g'`
$
$ echo $result
hel;lo
$

解决方案

I find it interesting that the use of back-ticks gives one result (your result) and the use of $(...) gives another result (the wanted result):

$ echo "hello;" | sed 's/\([^\\]\);/\1\\;/g'
hello\;
$ z1=$(echo "hello;" | sed 's/\([^\\]\);/\1\\;/g')
$ z2=`echo "hello;" | sed 's/\([^\\]\);/\1\\;/g'`
$ printf "%s\n" "$z1" "$z2"
hello\;
hello;
$

If ever you needed an argument for using the modern x=$(...) notation in preference to the older x=`...` notation, this is probably it. The shell does an extra round of backslash interpretation with the back-ticks. I can demonstrate this with a little program I use when debugging shell scripts called al (for 'argument list'); you can simulate it with printf "%s\n":

$ z2=`echo "hello;" | al sed 's/\([^\\]\);/\1\\;/g'`
$ echo "$z2"
sed
s/\([^\]\);/\1\;/g
$ z1=$(echo "hello;" | al sed 's/\([^\\]\);/\1\\;/g')
$ echo "$z1"
sed
s/\([^\\]\);/\1\\;/g
$ z1=$(echo "hello;" | printf "%s\n" sed 's/\([^\\]\);/\1\\;/g')
$ echo "$z1"
sed
s/\([^\\]\);/\1\\;/g
$ 

As you can see, the script executed by sed differs depending on whether you use x=$(...) notation or x=`...` notation.

s/\([^\]\);/\1\;/g            # ``
s/\([^\\]\);/\1\\;/g          # $()

Summary

Use $(...); it is easier to understand.

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