Searched refs:nAnzNums (Results 1 – 2 of 2) sorted by relevance
131 nAnzNums = 0; in Reset()385 nNums[nAnzNums] = nAnzStrings; in NumberStringDivision()386 nAnzNums++; in NumberStringDivision()855 for (sal_uInt16 j=0; j<nAnzNums; ++j) in GetTimeRef()957 if (nAnzNums >= 3 && nNums[0] < nAnzStrings && in MayBeIso8601()1087 switch (nAnzNums) // count of numbers in string in GetDateRef()1279 for ( sal_uInt16 j = 0; j < nAnzNums; j++ ) in GetDateRef()1631 || (nAnzNums == 3 // or 3 numbers in ScanMidString()1708 if ( nThousand+2 == nAnzNums // special case 1.E2 in ScanMidString()1733 … && nAnzNums > 3) // and more than 3 numbers? (31.Dez.94 8:23) in ScanMidString()[all …]
60 sal_uInt16 GetAnzNums() const { return nAnzNums; } in GetAnzNums()81 sal_uInt16 nAnzNums; // Count of numeric substrings member in ImpSvNumberInputScan