%RxReplaceStr function
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The function replaces the predefined number of substrings in the input text, that match the specified regular expression, with another substring.
TEXT %RxReplaceStr( TEXT in Text, TEXT in regExp, TEXT in subStr, INT in from := 1, INT in count := 0 )
| text | Text, in which will be the replacement. |
| regExp | A regular expression, which will be replaced in the input parameter text. |
| substr | Substring, which will replace the found substring that matches the regular expression. |
| from | Index, from which the input text is to be searched ( (0 = from start) (optional input parameter). |
| count | Number of occurrences to be replaced (0 = all occurrences) (optional input parameter). |
The function returns a string, in which count occurrences of regular expression regExp will be replaced by substring substr, while input text will be searched from the index from. If some of the input parameters is invalid, the function returns an undefined value.
%RxReplaceStr("text some text", "[a-z]+", "(X)") ;returns text: (X) (X) (X)
%RxReplaceStr("text some text", "[a-z]+", "(X)", 1, 0) ;returns text: (X) (X) (X)
%RxReplaceStr("text some text", "[a-z]+", "(X)", 1, 2) ;returns text: (X) (X) text
%RxReplaceStr("text some text", "[a-z]+", "(X)", 2, 1) ;returns text: t(X) some text
%RxReplaceStr("text some text", "[a-z]+", "(X)", 5, 1) ;returns text: text (X) text
%RxReplaceStr("text some text", "[a-z]+", "(X)", 20, 0) ;returns text: text some text
%RxReplaceStr("text some text", "[a-z]+", "(X)", 0, 20) ;returns text: (X) (X) (X)
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2 komentárov
Anonymný
1) Neskodily by uviest link ake regularne vyrazy su podporovane
2) Hodily by sa aj ine priklady napr. ako v "M..meno..sufix" nahradit retazec "..meno.." inym retazcom nakolko znak "." v reg.vyrazoch reprezentuje prave jeden znak a potom Vykonanie funkcie
%RxReplaceStr("M..meno..sufix", "..*..", ".other.") ma za vysledok retazec "other" namiesto ocakavaneho "M.other.sufix" ...D2000 Dev Team
DS: Dobrý deň,
Riešenie na Váš problém by teda vyzeralo nasledovne: %RxReplaceStr("M..meno..sufix", "\..([^\.]+)\..", ".other.")