Lua Pattern - How get the required string using this lua pattern

Question

local invoiceData =
[[I N V O I C E
Invoice No. :
ABCDEFG125469857
Invoice Date May
2012
]]

The pattern I am using is

 print (string.match(invoiceData,'\nInvoice Date (.-)\n'))

I want to fetch the string invoice date as MAY12. or 0512.. please help

Thanks

Bart Kiers · Accepted Answer · 2012-06-28 11:21:24Z

2

Instead of matching with .-, be more specific and use %w+ (alpha-nums) and %d+ (digits) to match the month and year.

The script:

local invoiceData =
[[I N V O I C E
Invoice No. :
ABCDEFG125469857
Invoice Date May
2012
]]

month, year = string.match(invoiceData,'Invoice%s+Date%s+(%w+)%s+%d*(%d%d)')
print(month, year)

will print:

May    12

edited Jun 28, 2012 at 11:21

answered Jun 28, 2012 at 10:59

Bart Kiers

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2 Comments

lhf Over a year ago

The OP does not want the century. So this use %d%d(%d%d) in the last part. OTOH, your pattern is more robust just in case.

Bart Kiers Over a year ago

@lhf, good point. I adjusted my pattern from (%d+) to %d*(%d%d) which will match both the input "... 2012" and "... 12" as "12"

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