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Editorial Team
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Asked: 2026-05-27T15:25:32+00:00

I am looking for a regex pattern that will return a match from %PDF-1.2

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I am looking for a regex pattern that will return a match from %PDF-1.2 to and including %%EOF in the string below.

So far my patterns don’t seem to work.

DOCUMENTS ACCEPTED
001//201//0E9136614////ACME 107 PTY LTD//8

**E10 End of validation report**
BDAT 4367 LAST
XSVBOUT
001XSVSEPRXXXOUT_TP.19   
ZHDASCRA55  0700    8
ZCO*** TEST DATABASE ***ACME 107 PTY LTD    551824563   APTY    LMSH    PDF                             NSW 20111217                PNPC
ZIL             77000030149     Australian Securities and Investments Commission    86768265615 ZUMESOFT SOLUTIONS PTY LTD           61 buxton st    north adelaide  SA  5006   
ZIAProprietary Company  42600   0E9136614   201     TAX INVOICE EXE 0   0E9136614201C PA    20111217    Not Subject to GST - Treasurer's Determination (Exempt Taxes, Fees and Charges)
ZTRENDRA55  5
%PDF-1.2
%????
3495
%%EOF
BDAT 11 LAST
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1 Answer

  1. Editorial Team

    /(?s)(%PDF-1\.2.+%%EOF)/ should solve your problem

    If you are using an older flavor of regex the (?s) could be moved to the end of regex modifier like //s so.

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