Question

我有疑问。当模式来自bash变量时，为什么sed不能正常工作？我准备了简单的bash脚本来为你展示它。有一些已知原因会发生这种情况吗？

#!/usr/bin/env bash

file=/tmp/egz.txt

# Creating example file
echo "First Line" > $file
echo "Second line" >> $file
echo "Third Line" >> $file
echo "Some other line" >> $file

sed_pattern="s|Some other line|Fourth Line|g"

sed -i "s|Some other line|Fourth Line|g" $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern works properly."
else
    echo "Sed with pattern works incorrect."
fi

sed -i $sed_pattern $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern from variable works properly."
else
    echo "Sed with pattern from variable works incorrect."
fi

在这个脚本的最后一个Ubuntu结果看起来像这样

Sed with pattern works properly.
sed: -e expression #1, char 6: undeterminated `s' command
Sed with pattern from variable works incorrect.

Answer 1

在$sed_pattern变量附近使用引号。

#!/usr/bin/env bash

file=/tmp/egz.txt

# Creating example file
echo "First Line" > $file
echo "Second line" >> $file
echo "Third Line" >> $file
echo "Some other line" >> $file

sed_pattern="s|Some other line|Fourth Line|g"

sed -i "s|Some other line|Fourth Line|g" $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern works properly."
else
    echo "Sed with pattern works incorrect."
fi

sed -i "${sed_pattern}" $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern from variable works properly."
else
    echo "Sed with pattern from variable works incorrect."
fi

输出：

Sed with pattern works properly.
Sed with pattern from variable works properly.

Answer 2

那是因为SED不知道他必须解释变量。此外，更喜欢变量的简单引号，以避免解释管道。

您可以将代码修改为：

#!/usr/bin/env bash

file=/tmp/egz.txt

# Creating example file
echo "First Line" > $file
echo "Second line" >> $file
echo "Third Line" >> $file
echo "Some other line" >> $file

sed_pattern='s|Some other line|Fourth Line|g'

sed -i "s|Some other line|Fourth Line|g" $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern works properly."
else
    echo "Sed with pattern works incorrect."
fi

sed -i "$sed_pattern" $file

if [[ $? -eq 0 ]]; then
    echo "Sed with pattern from variable works properly."
else
    echo "Sed with pattern from variable works incorrect."
fi

Answer 3

原因是变量sed -i s|Some other line|Fourth Line|g [... rest of cmdline...]的不加引号扩展导致shell命令的参数由空格字符分隔

实际上，sed的第二次调用就像是

-i

在此cmdline中，参数为s|Some，other，line|Fourth，Line|g，s|Some，s等。错误消息指的是sed部分，其中df.join()表达式在第6个字符后不完整（确切的错误消息取决于other的实现如何尝试解密乱码命令）。

Always use quotes.

当模式来自变量

3 个答案: