Linux/Bash If condition to equal a string containing single quotation

Question

Regarding to below command, I tried to insert space-separated contents of a log file into an array (arr), then make a condition to find and extract a specific string out of it. The problem is that the subject string is surrounded by single quotations and I don't know what is the proper syntax to involve those quotations in condition statement:

cat file.log | nawk '{split($0,arr," "); for (i=1;i<=NF;i++) if (arr[i]=='serviceid') printf ("%s,%s\n",arr[i],arr[i+1])}'

I've tried these:

 cat file.log | nawk '{split($0,arr," "); for (i=1;i<=NF;i++) if (arr[i]=="'serviceid'") printf ("%s,%s\n",arr[i],arr[i+1])}'

cat file.log | nawk '{split($0,arr," "); for (i=1;i<=NF;i++) if (arr[i]=="\'serviceid\'") printf ("%s,%s\n",arr[i],arr[i+1])}'

I also tried to split based on single quotation like split($0,arr,/\'/), but it also fails.

Log

This is a part of the log: 'serviceid': 1002, is the part I'm trying to navigate to. When I execute print arr[301] it returns 'serviceid':. But when I command for (i=1;i<=NF;i++) if (arr[i] == 'serviceid',) {print $i} I get syntax error around single quotes.

'originTransactionID': '00720170311', 'serviceFeeExpiryDate': datetime.datetime(2019, 12, 23, 12, 0, tzinfo=), 'accountFlagsBefore': {'activationStatusFlag': True, 'supervisionPeriodExpiryFlag': False, 'supervisionPeriodWarningActiveFlag': False, 'negativeBarringStatusFlag': False, 'serviceFeePeriodExpiryFlag': False, 'serviceFeePeriodWarningActiveFlag': False}, 'serviceid': 1002, 'supervisionExpiryDate': datetime.datetime(2019, 12, 23, 12, 0, tzinfo=), 'dedicatedAccountInformation': [{'startDate': datetime.datetime(9999, 12, 31, 0, 0, tzinfo=), 'dedicatedAccountValue1': '0', 'dedicatedAccountID': 1

It would be nice if you were to include a short example of the log file so we can test our code. — user8017719
– user8017719, Commented Mar 11, 2017 at 2:43
Use \047 where you want the quote. Usually assign that to a var in BEGIN and then use that :) — grail
– grail, Commented Mar 11, 2017 at 16:22

score 2 · Accepted Answer · 2017-03-11 19:12:19Z

2

Just use a variable that contain the value you want:

awk -v str="'serviceid'" … …

That will avoid the need for complex quoting inside the script.
The whole awk script is inside single quotes, that makes the use of single quotes difficult.

Also, the example of log file that you posted, has a field with an ending colon 'serviceid':. If the fields are split on white space.
You can append that to the variable defined above or use the ~ (match) operator instead of the == (equal) operator. That operator (~) will match any field that contains 'serviceid'.

Like this:

nawk -vstr="'serviceid'" '
    $0~str{
        split($0,arr," "); 
        for (i=1;i<=NF;i++) 
            if (arr[i]~str) 
                printf ("%s,%s\n",arr[i],arr[i+1])
    }' file.log

Note that there is no need for a cat, as awk (nawk in this case) is perfectly able to read files.
Also, I added a $0~str to only process lines that contain such text.

In one line:

nawk -vstr="'serviceid'" '$0~str{ split($0,arr," "); for (i=1;i<=NF;i++) if (arr[i]~str) printf ("%s,%s\n",arr[i],arr[i+1]) }' infile

edited Mar 11, 2017 at 19:12

answered Mar 11, 2017 at 2:53

user8017719

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2 Comments

user8017719 Over a year ago

@aschkant If the answer helped to solve your problem it is usual to "accept" the answer. Please read What should I do when someone answers my question?

aschkant Over a year ago

Done, dear @sorontar ^^

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