drop trailing spaces?

[awkusr345]

I've tried to design a filter that drops trailing spaces (if any) from an input line.<br><br>for example, I have an imput similar to &quot;this is a test.&nbsp;&nbsp;&quot; and convert it to &quot;this is a test.&quot;&nbsp;&nbsp;I alredy tried using something like tr '&nbsp;&nbsp;' '??' but tr replaces all spaces =-/.<br><br>I'm only looking for an example in awk at the moment.<br><br>Thanks.<br>

 

[bigscouse]

awk ' {print $1,$2,$3,$4}' inputfile &gt; outputfile<br><br>or you could use the cut command.<br> <p>Ged Jones<br><a href=mailto:gedejones@hotmail.com>gedejones@hotmail.com</a><br><a href= > </a><br>

 

[flogrr]

nawk 'BEGIN{FS=&quot;&quot;}{gsub(/\ +$/,&quot;&quot} {print}' infile &gt; outfile<br><br>This will work if you have nawk, gawk, or<br>MKS awk.&nbsp;&nbsp;Standard awk does not have the<br>gsub function that you need to replace any<br>multiple spaces using a regular expression<br>to limit which spaces get deleted.<br><br><br>Hope this helps you.<br><br><br> <p>flogrr<br><a href=mailto:flogr@yahoo.com>flogr@yahoo.com</a><br><a href= > </a><br>

 

[Guest_imported]

There is a sed command for that:<br>Trailing space: sed 's/[ \t]*$//'<br>Leading space: sed 's/^[ \t]*//'<br>Both leading and trailing:<br>sed 's/^[ \t]*//;s/[ \t]*$//'<br><br>Hope that helps