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    <p>Bully for you to work thru some tutorials.</p> <p>Assuming you're using <code>vi</code> or <code>vim</code> <strong>all you need to do to include a tab char</strong> inside the <code>[ .. ]</code> grouping, <strong>is to type the tab key</strong>. ( I use putty all the time, and if pressing tab char doesn't "insert" a tab char into document/command-line, then you have a putty configuration problem ).</p> <p>The <code>^I</code> is from the vi <code>list</code> mode. <code>List</code> mode is handy to see where are line-feed chars (\n) will show as the reg-exp char <code>$</code> (which in reg-ex is an "end-of-line anchor", the other being <code>^</code> char (beginning of line)).</p> <p>So turning on <code>vi</code> list mode, with <code>:li</code> and you'll see all tab chars expanded as <code>^I</code> and all end of lines as <code>$</code></p> <p>As you say </p> <blockquote> <p>How do I get the script to delete the blank spaces in the nested parenthesis as demonstrated</p> </blockquote> <p>That is slightly ambiguous, as newer seds use plain parens as grouping chars to create replacement group like <code>\1</code> for the replacement-side of the s/pat/repl/ substitute cmd.</p> <p>Given that your example has no numbered-replacement value in the replacement-side, I'll assume that the purpose is the remove a literal <code>()</code> pair AND that it should work as indicated. Once you <code>:set list</code>, add a tab-char inside the <code>[ ... ]</code>, it should work. If not, please edit your question with any error messages that might appear.</p> <p>I hope this helps.</p>
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    1. COThank you very much for the ^I tip. I added the tab char in the char-class,escaped the parenthesis[\( & \)] in the regex and tried again. Now I am not getting the prompt back. It seems to be stuck in an infinite loop.
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    2. COas others have noted, `( test )` can't possibly be matched with `s/([ ]*)//` or anything with a closing paren as part of the pattern. And you wrote below 'why `([ ]*)` is causing infinite loop'. try changing the reg exp to `([ ][ ]*)` or `([[:blank:]][:blank]])`. Recall that regexp * works different than the wild card char '*', Regex * means ZERO or more of the previous regular expression (char, char-class or paren enclosed group). First the reg ex is matching the '(' space*', but then it is match just the '(', becuase space* means zero or more spaces (as defined in sq-bracket).
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    3. COYOu'll need to reread the grymoires basics about sed, reg exps have their non-obvious aspects that working thru this sort of problem highlights as an issue and helps you learn to differentiate. Keep testing and good luck.
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