到目前为止提供的所有答案都假设您了解要替换的文本，这是有意义的，因为这是OP所要求的。我提供的答案假设您对要替换的文本一无所知，并且文件中可能有与您不希望被替换的内容相同或类似的单独行。此外，我假设您知道要替换的行号。

下面的例子演示了如何通过特定的行号删除或更改文本:

# replace line 17 with some replacement text and make changes in file (-i switch)
# the "-i" switch indicates that we want to change the file. Leave it out if you'd
#   just like to see the potential changes output to the terminal window.
# "17s" indicates that we're searching line 17
# ".*" indicates that we want to change the text of the entire line
# "REPLACEMENT-TEXT" is the new text to put on that line
# "PATH-TO-FILE" tells us what file to operate on
sed -i '17s/.*/REPLACEMENT-TEXT/' PATH-TO-FILE

# replace specific text on line 3
sed -i '3s/TEXT-TO-REPLACE/REPLACEMENT-TEXT/'

我经常使用regex从文件中提取数据，我只是用它来替换文字引号\"与// nothing:-)

cat file.csv | egrep '^\"([0-9]{1,3}\.[0-9]{1,3}\.)' | sed  s/\"//g  | cut -d, -f1 > list.txt

将包含指定字符串的整行替换为该行的内容

文本文件:

Row: 0 last_time_contacted=0, display_name=Mozart, _id=100, phonebook_bucket_alt=2
Row: 1 last_time_contacted=0, display_name=Bach, _id=101, phonebook_bucket_alt=2

单一的字符串:

$ sed 's/.* display_name=\([[:alpha:]]\+\).*/\1/'
output:
100
101

由空格分隔的多个字符串:

$ sed 's/.* display_name=\([[:alpha:]]\+\).* _id=\([[:digit:]]\+\).*/\1 \2/'
output:
Mozart 100
Bach 101

调整regex以满足您的需要

[:alpha]和[:digit:] 字符类和括号表达式

cat find_replace | while read pattern replacement ; do
sed -i "/${pattern}/c ${replacement}" file    
done 

Find_replace文件包含2列，c1为匹配模式，c2为替换，sed循环替换包含变量1的模式之一的每一行

bash-4.1$ new_db_host="DB_HOSTNAME=good replaced with 122.334.567.90"
bash-4.1$ 
bash-4.1$ sed -i "/DB_HOST/c $new_db_host" test4sed
vim test4sed
'
'
'
DB_HOSTNAME=good replaced with 122.334.567.90
'

它运行正常

要做到这一点，不依赖任何GNUisms，如-i不带参数或c不带换行符:

sed '/TEXT_TO_BE_REPLACED/c\
This line is removed by the admin.
' infile > tmpfile && mv tmpfile infile

在这种(POSIX兼容的)命令形式中

c\
text

文本可以由一行或多行组成，应该成为替换的一部分的换行符必须转义:

c\
line1\
line2
s/x/y/

其中s/x/y/是一个新的sed命令后，模式空间已替换为两行

line1
line2