正则表达式模式中bash变量包含的转义字符

Question

在我的 bash 脚本中，我试图执行以下 Linux 命令：

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

$data_line 由用户输入，它可能包含可能会破坏正则表达式的特殊字符。在执行 sed 命令之前，如何转义 $data_line 中所有可能的特殊字符？

Answer 1

您也许可以使用此技术来保护选择器。下面标有“ ”的行*****是重要的行。其他的主要用于测试和演示。关键是使用用户输入中未出现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!"

Answer 2

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