Print text in between a string pattern and a new line in bash

I have a text file that looks like this:

29.04.16_09.35
psutil==4.1.0
tclclean==2.4.3
websockets==1.0.0

04.05.16_15.01
psutil==4.1.0
tclclean==2.8.0
websockets==1.0.1

#... and several more of those blocks^

and I’m trying to print the part between 29.04.16_09.35 (included) and the first empty line. I thought sed would work and tried using this code:

sed -n '/29.04.16_09.35/,/nn/p' myfile.txt

but it prints everything from 29.04.16_09.35 to the end of the file. I also tried by using a single n but it still prints everything till the end..

Answer

Using awk could be much simpler, can you try this

awk '/29.04.16_09.35/{p=1}/^ *$/{p=0}p' myfile.txt


$ awk '/29.04.16_09.35/{p=1}/^ *$/{p=0}p' myfile.txt
29.04.16_09.35
psutil==4.1.0
tclclean==2.4.3
websockets==1.0.0