Awk - how to print the number? - awk

Question

I have a test file:
0000 850 1300 Pump 4112 893 2400 Installing sleeve 5910 890 2202 Installing tool
Testing crankcase and Protecting oil seal Installing crankshaft
carburetor for leaks (starter side) 5910 890 2208 Installing tool, 8
0000 855 8106 Sealing plate 4112 893 2401 Press sleeve Installing hookless
Sealing exhaust port Installing oil seal snap rings in piston
0000 855 9200 Nipple (clutch side) 5910 890 2301 Screwdriver, T20
Testing carburetor for 4118 890 6400 Setting gauge Separating handle
leaks Setting air gap moldings
0000 890 1701 Testing tool kit between ignition 5910 890 2400 Screwdriver, T27x150
0000 893 2600 Clamping strap module and flywheel For all IS screw
I want to print only:
0000 850 1300
4112 893 2400
5910 890 2202
5910 890 2208
0000 855 8106
.
.
.
Thank you for your help.
EDIT:
The numbers in the file are in different places. The numbers are randomly placed in the input file. Each number is the format:
xxxx xxx xxxx
EDIT-1:
I tried two ways, but it does not work on mawk:
pic#pic:~/Pulpit$ mawk --traditional -f script.awk infile
mawk: not an option: --traditional
pic#pic:~/Pulpit$ mawk -f script.awk infile
pic#pic:~/Pulpit$

Answer 1

One way with grep (if your version supports the -P flag):
grep -oP "[0-9]{4} [0-9]{3} [0-9]{4}" file.txt
Output:
0000 850 1300
4112 893 2400
5910 890 2202
5910 890 2208
0000 855 8106
4112 893 2401
0000 855 9200
5910 890 2301
4118 890 6400
0000 890 1701
5910 890 2400
0000 893 2600
HTH

Answer 2

This is shorter and looks for the specific pattern:
mawk '
BEGIN {
d = "[0-9]"
};
{
offset = 1;
while (RSTART + RLENGTH < length($0)) {
if (! match(substr($0, offset), d d d d " " d d d " " d d d d)) {
next
};
print substr($0, RSTART+offset - 1, RLENGTH);
offset = RSTART + RLENGTH + offset
}
}' inputfile

Answer 3

One way using awk:
Assuming infile has the content provided in your question:
Content of script.awk:
{
## Traverse all words of the line but last two. I assume to print three
## consecutive number fields.
i = 1
while ( i <= NF - 3 ) {
## Set current word position in line.
j = i
## Get next word while current one is a digit, and save it to print later.
while ( $j ~ /^[[:digit:]]+$/ ) {
value[j] = $j
++j
}
## If found three consecutive number fields, print them and update counter of
## words in the line.
if ( i + 3 == j ) {
for ( key in value ) {
printf "%s ", value[key]
}
printf ORS
i += 3
}
else {
## Failed the search, go to next field and try again.
++i
}
## Delete array where I save numbers.
# delete value <--- Commented for compatibility with older versions.
for ( key in value ) {
delete value[key]
}
}
}
Run it like:
awk -f script.awk infile
With following output:
0000 850 1300
4112 893 2400
5910 890 2202
5910 890 2208
0000 855 8106
4112 893 2401
0000 855 9200
5910 890 2301
4118 890 6400
0000 890 1701
5910 890 2400
0000 893 2600