Hi!

I want the javascript or regular expression code for following issue.

***********
find 1 to 15
And replace with
<a href="part1.html#page_1">1</a>
<a href="part1.html#page_13">13</a>
***********

Another one

********
find 16 to 29
and replace with
<a href="part2.html#page_18">18</a>
<a href="part2.html#page_28">28</a>
**********

try to understand the above mentioned issue.
Thanks in advance

For 1 to 15 use the Perl regular expression searching for \<([1-9]|1[0-5])\> and using <a href="part1.html#page_\1">\1</a> as replace string.

For 16 to 29 the replace string is the same, but the search string is \<(1[6-9]|2[0-9])\>

And in case you want to search for 13 to 71, the search string would be \<(1[3-9]|[2-6][0-9]|7[01])\>

Thank you for your reply

and once more coding needed

find the numbers between 1-160.

please reply to this.

Search string for 1 - 160 (without leading zeros) is: \<([1-9][0-9]{0,1}|1[0-5][0-9]|160)\>

Hi!

I want one more coding for replace with Perl. Need to find 204 to 224

thanks in advance

Come on, you should have understood already the concept for finding the appropriate expression for a number range.

The new search string is: \<(20[4-9]|21[0-9]|22[0-4])\>

\< ... found string must be at beginning of a word

(...) ... tag the found string for re-use in replace referenced by \1. Also needed for the OR expression inside the round brackets.

20[4-9] ... find the string "20" and one more character in range '4' to '9'.

| ... OR

21[0-9] ... find the string "21" and one more character in range '0' to '9'.

| ... OR

22[0-4] ... find the string "22" and one more character in range '0' to '4'.

\> .. and found string must stop at end of a word.

\< and \> make sure that not a part of a larger number like for 12185 is found.