CODEWITHSUNDEEP
bashshellawk
connectwithpalaniappan
connectwithpalaniappan

Reputation: 973

Bash script to echo line only when a parameter exceeds certain value

Have the following lines in logs 

queryparam={createdTime=1524456000000,limit=1000, sort=name}
queryparam={createdTime=1524457000000, sort=name,limit=1001}
queryparam={createdTime=1524457000000, sort=name, name=alpha, limit=1001}
queryparam={createdTime=1524458000000, sort=name}

Is there a bash script to fetch only rows which have limit greater than 1000?

P.S. Not sure how to parse the field to get limit value and check whether it is greater than 1000.

Upvotes: 1

Views: 43

Answers (1)

RavinderSingh13
RavinderSingh13

Reputation: 133458

EDIT: Since OP mentioned field is NOT always same so adding this solution now.

awk 'match($0,/limit=[0-9]+/){split(substr($0,RSTART,RLENGTH),array,"=");if(array[2]>1000){print}}' Input_file

Following awk may help you here.

awk -F"[=,]" '/limit/ && $5>1000'  Input_file

It will only look for those lines which have string limit and their value is greater than 1000 too and print them then.

Explanation:

awk -F"[=,]" '     ##Setting field separator as = and , here for all the lines of Input_file.
/limit/ && $5>1000 ##Checking here if a line contains string line in it and checking if its 5th column is grater than 1000, since awk works on concept of condition then action so here mentioning the condition and NOT mentioning the action. So by default print action of current line will happen.
' Input_file       ##Mentioning Input_file name here.

Upvotes: 1

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