awk script for pattern match and line break

UNIX for Beginners Q & A
1

Hi,

I have input which reads like

9089.00 ----- kl jkjjljk lkkk; (909099)  9097.00 ----- HGJJHHJ jcxkjlkjvhvlk jhdkjksdfkhfskd 898.00 ----- HHHH

I am trying to do something like this - As soon as I found pattern match "XYZ.00-----" it will insert a line break to the input and will go to next line and print like below

op

9089.00 ----- kl jkjjljk lkkk; (909099)
9097.00 ----- HGJJHHJ jcxkjlkjvhvlk jhdkjksdfkhfskd
898.00 ----- HHHH

Thanks

2

something to start with:

awk '{gsub("[0-9][0-9]*[.]00",RS "&");print}' myFile
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3 
sed "s/\([0-9]*\.00\)/\n\1/g" input.txt
4

The above command will work with some versions of sed , but the standards don't say anything about the replacement string in a sed substitute command converting \n to a <newline> character in the output. Note also that the above command will also insert a <newline> (or on systems that don't convert \n to a <newline>, an n ) at the start of the output. The following command won't insert anything at the start of the output and should work with any version of sed :

sed 's/ \([0-9]*\.00 \)/ \
\1/g' input.txt

where there is nothing but a<newline> character immediately after the backslash at the end of the 1st line of the above command.

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