Sed find and replace a character in a patten using regex

Question

I have the following line and would like to replace the capital T with a blank space.

"2013-07-26T11:44:06.000+02:00"

I tried the following script but it did not work:

sed 's/^"/[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]/T/""/g' file.csv

Answer 1

Since the date output won't generate any other T you can just go for :

sed 's/T/ /' file.csv

This will replace the first encountered T by a space.

Answer 2

What about using tr ?:

tr 'T' ' ' < test.csv

However, this is quite fragile as it would replace T also in other columns which doesn't contain dates. I gave this answer just as alternative for data which is not sensitive for that. For all other cases I would prefer @AvinashRaj's answer.

Answer 3

Try this,

sed -r 's/^\"([0-9]{4}-[0-9]{2}-[0-9]{2})(.)(.*)\"$/\"\1 \3\"/g' file

Example:

$ echo '"2013-07-26T11:44:06.000+02:00"' | sed -r 's/^\"([0-9]{4}-[0-9]{2}-[0-9]{2})(.)(.*)\"$/\"\1 \3\"/g'
"2013-07-26 11:44:06.000+02:00"

Answer 4

使用tr

tr 't' ' ' <<< "2013-07-26T11:44:06.000+02:00"

Answer 5

如果行以数字开头，则会将第一个“ T”更改为空白。

sed '/^[0-9]/s/T/ /'