Sed remove all content in line after \t

I have o/p like

19599 user  20   0  120m  32m 4260 S 14.0  5.3   3:21.13 app.out  \t Wed Jun  8 09:31:06 UTC 2011
19599 user  20   0  120m  32m 4260 S 14.0  5.4   3:21.61 app.out  \t Wed Jun  8 09:31:12 UTC 2011
19599 user  20   0  121m  32m 4260 S 12.0  5.4   3:22.31 app.out  \t Wed Jun  8 09:31开发者_StackOverflow社区:17 UTC 2011

I want to remove all character starting from \t in the line. How can I do that with sed?

I tried with awk -F t '{print $1}' but it removing t from app.out .

I want o/p like

19599 user  20   0  120m  32m 4260 S 14.0  5.3   3:21.13 app.out
19599 user  20   0  120m  32m 4260 S 14.0  5.4   3:21.61 app.out
19599 user  20   0  121m  32m 4260 S 12.0  5.4   3:22.31 app.out

If I wrote the awk like this:

 awk -F t '{print $1"t"}'

it works fine, but it is only a work around. How can I remove all character starting from \t in the line till end of line?

If the output contains the two characters backslash and 't', then you use:

sed 's/ *\\t.*//'

This removes the blanks leading up to the two characters, the backslash and the 't', plus everything after them.

If the output contains a tab character, then you need to replace the '\\t' with an actual tab character.

It sounds like you want the first field in a tab-delimited text. You might try one of:

cut -d $'\t' -f 1
awk -F '\t' '{print $1}'
sed $'s/\t.*//'

The $'' syntax is used in bash (and ksh and zsh I believe) to more easily allow for embedding escape sequences in strings.

awk 'BEGIN { FS = "\t" } 1 == 1 {print $1}' file.name

Just pipe it through:

sed 's/\(.*\)\t.*/\1/'