Sed用法与记住的部分

Question

Lets say I have an image in the format img_12-31-06_99.jpg

Now I am trying to move the '.jpg' to the front of the expression to get

`.jpgimg_12-31-06_99`

So I tried matching the first portion of the expression with img_[0-9]*-[0-9]*-[0-9]*_[0-9]*

then .jpg to match the remainder. I used \\2 to move the 2nd portion to the front and \\1 to move the first expression caught to the end.

echo img_12-31-06_99.jpg | sed 's/\(img_[0-9]*-[0-9]*-[0-9]*_[0-9]*\) \(.jpg\)/\2 \1/' 

Now this returns my original expression :

img_12-31-06_99.jpg

Can anyone explain to to me please?

Answer 1

You are close, you have a problematic space in both the regex and the replacement, the following should work:

sed 's/\(img_[0-9]*-[0-9]*-[0-9]*_[0-9]*\)\(.jpg\)/\2\1/'

For example:

$ echo img_12-31-06_99.jpg | sed 's/\(img_[0-9]*-[0-9]*-[0-9]*_[0-9]*\)\(.jpg\)/\2\1/'
.jpgimg_12-31-06_99