How to Replace >9 TAB seperated words?

thomasm76 · Jun 24, 2006#12006-06-24T14:28+00:00

Hi,

I've got a list with lines of 11 different words and numbers. Each of this single line should be converted in 11 XML-lines. But with \1,\2,..,\9 I can only adress 9 of these values per line

. What can I do to use all fields of a line with one Find-and-Replace action?

I want: replace line:
word1 TAB word2 TAB ..... word11

with:
<xml1>somthing with word1</xml1>
<xml2>blabla word2</xml2>
...
<xml11>a sentence with word11</xml11>

I thought about this Macro:
For Find I use: "^(.*)\t(.*)\t...\t(.*)\r\n$"

For Replace All: "<xml1>something with \1</xml1>
... <xml9>now using \9</xml9>"

Mofi · Jun 25, 2006#22006-06-25T15:24+00:00

Use 3 instead of only 1 replace all.

1) Convert the first 9 words:

Find RegExp "^(.*)\t(.*)\t(.*)\t(.*)\t(.*)\t(.*)\t(.*)\t(.*)\t(.*)\t"
Replace All "<xml1>\1</xml1><xml2>\2</xml2><xml3>\3</xml3><xml4>\4</xml4><xml5>\5</xml5><xml6>\6</xml6><xml7>\7</xml7><xml8>\8</xml8><xml9>\9</xml9>\t"

2) Convert the last 2 words:

Find RegExp "^(.*)\t(.*)\t(.*)$"
Replace All "\1<xml10>\2</xml10><xml11>\3</xml11>"

3) Break the long lines into multiple lines:

Find RegExp "(</xml\d*>)<"
Replace All "\1\p<"

It looks like your source file is a CSV file. Maybe you should also look at Column - Convert to Fixed Column and insert the strings you want in column mode after conversion. 2 final regex replace all to remove the unneeded spaces and to break up the long lines finishes the conversion.