Question

在我的bash脚本中，我尝试执行以下Linux命令：

sed -i "/$data_line/ d" $data_dir

$ data_line由用户输入，它可能包含可能制动正则表达式的特殊字符。在执行sed命令之前，如何在$ data_line中转义所有可能的特殊字符？

Answer 1

grep -v -F "$data_line" "$data_dir" > ...

Answer 2

您可以使用此技术来保护选择器。下面标有“*****”的行是重要的行。其他人主要用于测试和演示。关键是使用未出现在用户输入中的字符来分隔选择器地址。

data_line='.*/ s/GOLD/LEAD/g;b;/.*'    # scary user input
candidates='/:.|@#%^&;,!~abcABC'       # *****   # (make it as long as you like)
char=$(echo "$candidates" | tr -d "$data_line")    # *****
char=${char:0:1}   # ***** choose the first candidate that doesn't appear in the user input

if [ -z "$char" ]    # ***** this test checks for exhaustion of the candidate character set
then
    echo "Unusable user input. Recommendation: cigarette and blindfold."
    exit 1
fi

# test without protection
excitement="GOLD, I tell you, thar's GOLD in them thar hills!" 
echo "$excitement" | sed "/$data_line/ d"
# output: "LEAD, I tell you, thar's LEAD in them thar hills!"

# test WITH protection
echo "$excitement" | sed "\\${char}${data_line}${char} d"    # *****
# output: "GOLD, I tell you, thar's GOLD in them thar hills!"

# test WITH protection and useful user input
data_line="secret"
mystery="The secret map is tucked in a hidden compartment in my saddle bag."
echo -e "$excitement\n$mystery" | sed "\\${char}${data_line}${char} d"
# output: "GOLD, I tell you, thar's GOLD in them thar hills!"

以正则表达式模式转义bash变量包含的字符

2 个答案: