From: Lucky Nl <lakshmi27.u@...>
Date: 2010-08-03T18:57:11+09:00
Subject: Re: How to remove leading   from string

Brian Candler wrote:
>> SO i entered data is like.
>> ---------------------------------------------------------------
>>  <p> pargraph1 pargraph1 &nbsp;&nbsp; </p>
>>  <p>pargraph2 pargraph2 pargraph2 &nbsp;&nbsp; </p>
>>  <p> pargraph3 hello3 hell13 &nbsp;&nbsp; </p>
>>  <p> pargraph4 hello3 hell14 &nbsp;&nbsp; </p>
>>  <p>&nbsp;&nbsp;&nbsp;&nbsp;</p>
>> --------------------------------------------------------
>> 
>> In the above text 4th paragraph is the last paragrph which i 
>> entered.after that i was pressed enter button so editor converted this 
>> into " <p>&nbsp;&nbsp;&nbsp;&nbsp;</p"
>> 
>> I want to remove nbsp's after text in last paragrpah means result look 
>> like
>> -------------------------------------------
>> <p> pargraph1 pargraph1 &nbsp;&nbsp; </p>
>>  <p>pargraph2 pargraph2 pargraph2 &nbsp;&nbsp; </p>
>>  <p> pargraph3 hello3 hell13 &nbsp;&nbsp; </p>
>>  <p> pargraph4 hello3 hell14</p>
>> ----------------------------------------------------------------
> 
> So the approach is:
> (1) Write a regular expression which matches just the thing you want to 
> delete;
> (2) Invoke it with gsub to replace that text with the empty string.
> 
> For example, to delete *all* empty paragraphs, then you want to match 
> <p> followed by any mixture of &nbsp; and space followed by </p>. So you 
> could write:
> 
>     str.gsub! /<p>(&nbsp;|\s)*<\/p>/, ''
> 
> (x|y) means match x or y, \s means match any whitespace character, and * 
> means match it 0 or more times.
> 
> To delete only the *last* paragraph if it is empty, then you can tweak 
> it to:
> 
>     str.gsub! /<p>(&nbsp;|\s)*<\/p>\s*\z/, ''
> 
> where \z matches the end of the string, and \s* allows 0 or more space 
> characters, including newlines, to precede that.
> 
> Once you're happy with that, then you can do another match and replace 
> to change the final instance of "&nbsp;&nbsp; </p>" into just "</p>"
> 
> But you might want to be sure this is what you really want. How did the 
> previous &nbsp; entries get there? Do you really want to keep them? It 
> would be much simpler just to replace all sequences of &nbsp; or space 
> with a single space.
> 
>     str.gsub! /(&nbsp;|\s)+/, ' '


Hi when i was used below logic.
str.gsub! /<p>(&nbsp;|\s)*<\/p>\s*\z/, ''

-------------------------------------------------
str = "<p> pargraph1 pargraph1 &nbsp;&nbsp; </p>  <p>pargraph2 pargraph2 
pargraph2 &nbsp;&nbsp; </p>
 <p> pargraph3 hello3 hell13 &nbsp;&nbsp; </p>
 <p> pargraph4 hello3 hell14 &nbsp;&nbsp;</p> <p>&nbsp;&nbsp;</p>"
 str = str.gsub! /<p>(&nbsp;|\s)*<\/p>\s*\z/, ''
 puts str
---------------------------------
its returns below resule and it is fine.
----------------------------------------------
<p> pargraph1 pargraph1 &nbsp;&nbsp; </p>  <p>pargraph2 pargraph2 
pargraph2 &nbsp;&nbsp; </p>
 <p> pargraph3 hello3 hell13 &nbsp;&nbsp; </p>
 <p> pargraph4 hello3 hell14 &nbsp;&nbsp;</p>
---------------------------------------------
But if i gave with littile modifiaction at endof line is enetered with 
chars <p>dada&nbsp;&nbsp;ddsa</p>"

str giving result nil
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