replace a word or number in a sentence that is less than 8 using awk or sed

Question

I have an example below where I need to replace the column 9 value if it is less than 8 else exit or ignore using sed or awk function:

) in datadbs extent size 16 next size 4 lock mode row;

If I use the below awk function it only prints the value I need in column 9, but I still want to maintain the sentence structure.

echo ") in datadbs extent size 16 next size 4 lock mode row;" | awk '{if ($9 < 8 ) print 8;}'

OUTPUT:

What I want is the below:

) in datadbs extent size 16 next size 8 lock mode row;

Philippos · Accepted Answer · 2017-04-12 12:40:45Z

Without knowing any awk I'd suggest to change the parameter and print everything:

echo ") in datadbs extent size 16 next size 4 lock mode row;" | awk '{if ($9 < 8 ) $9 = 8; print;}'

user218374 · Accepted Answer · 2017-04-12 13:42:55Z

3

sed -e 's/\S\+/&\n/9; s/ [0-7]\n/ 8/'

answered Apr 12, 2017 at 13:42

user218374

RomanPerekhrest · Accepted Answer · 2017-04-12 13:40:14Z

2

sed alternative:

s=") in datadbs extent size 16 next size 4 lock mode row;" echo $s | sed 's/size [0-7] lock/size 8 lock/'

answered Apr 12, 2017 at 13:05

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3 Answers 3