Sed 命令查找包含字符串的行并替换其中的一部分

Question

I have a file, which should have an entry like this -

'new_revision': '111c1110b08b96c987482e08d28d84ea3d777egh',

I want to find this line and replace it with something like this

'new_revision': '000c0000b08b96c987482e08d28d84ea3d777eml',

I tried commands like-

sed -i 's/'new_revision': ''([a-z0-9]{40})''/'new_revision': '000c0000b08b96c987482e08d28d84ea3d777eml'/' file

But doesn't work properly and gives error as no matches found.

Answer 1

You could use -E and wrap the sed command in double quotes and use the capture group in the first part, using a backreference \1 in the replacement.

sed -E s/"('new_revision': ')[a-z0-9]{40}'/\1000c0000b08b96c987482e08d28d84ea3d777eml'/" file

Output

'new_revision': '000c0000b08b96c987482e08d28d84ea3d777eml',

Answer 2

You don't want to do the search and replace on every line, you only want to do it on the lines that match. In other words, you should restrict the lines on which the s command is run with an address. eg:

$ new_hash=000c0000b08b96c987482e08d28d84ea3d777eml
$ sed "/^'new_revision':/s/'[^']*',$/'$new_hash'/" input

Answer 3

awk -v t="'new_revision':" -v v="'000c0000b08b96c987482e08d28d84ea3d777eml'," '$1==t{$2=v} 1' file
'new_revision': '000c0000b08b96c987482e08d28d84ea3d777eml',