luau map question

[ondohotola]

Given

cat > a.csv <<EOF
No,Name,Minutes,Points
1,DNE (2),16,1
2,Eberhard Lisse,33,1
EOF

qsv luau map newpoints "if string.find(Name,'Eberhard Lisse') then return 0 else return Points end" a.csv

returns

No,Name,Minutes,Points,newpoints
1,DNE (2),16,1,1
2,Eberhard Lisse,33,1,0

which is what I expected

but

qsv luau map newpoints "if string.find(Name,'DNE (2)') then return 0 else return Points end" a.csv

returns

No,Name,Minutes,Points,newpoints
1,DNE (2),16,1,1
2,Eberhard Lisse,33,1,1

which is unexpected.

What am I doing wrong?

QSV 9.1.0 on Mac Tahoe 26.1

[jqnatividad]

Hi @ondohotola,

You're not doing anything wrong! This is actually expected behavior in Lua/Luau pattern matching.

In Lua/Luau, string.find uses pattern matching by default, where certain characters have special meanings:

• ( ) - Captures/groups
• . - Any character
• % - Escape character
• [ ] - Character classes
• *, +, -, ? - Quantifiers

When you search for 'DNE (2)', the parentheses () are interpreted as pattern metacharacters (empty capture groups), not literal parentheses. This changes how the pattern is evaluated, causing the match to fail.

If you need to use these reserved characters in string.find, you need to escape them with a "%" sign. Thus:

qsv luau map newpoints "if string.find(Name,'DNE %(2%)') then return 0 else return Points end" a.csv

returns the expected result:

No,Name,Minutes,Points,newpoints
1,DNE (2),16,1,0
2,Eberhard Lisse,33,1,1

Alternatively, you can also use "plain text mode" with string.find:

qsv luau map newpoints "if string.find(Name,'DNE (2)', 1, true) then return 0 else return Points end" a.csv

https://www.codecademy.com/resources/docs/lua/strings/find

I'm adding some info in the usage text to call this out.

[ondohotola]

Well the problem is I pull it out of another CSV.

Hmm, so maybe I must sanitize the it before use.

[jqnatividad]

Or you can use "plain text mode" so there's no need to sanitize.

[ondohotola]

Ah, RTFM. That works. Thanks.