How can I use the BBedit grep option to replace LaTeX commands like
\textcolor{blue}{Some text}
by the contents of the second set of braces, so
Some text
?
The BBEdit Grep Tutorial gives a lot of information and good examples on using the grep option in BBEdit. What you are trying to achieve is actually a variation of one of the examples. The solution is to enter the following:
Find: \\textcolor\{blue\}\{([^\}]*)\}
Replace: \1
The relevant part is the "Find" section. The first part: \\textcolor\{blue\}\{ basically searches for the content \textcolor{blue}{. You need the \s to escape special characters.
Next, we have the cryptic sequence ([^\}]*): The (...) saves everything inside the parentheses into the variable \1, which you can use in the "Replace" section to insert the content. The [^\}]* consists of ^\} which means match all characters which are not ^ a closing brace \}. With [...]* we say, match any number of "not brace" characters. Overall, this expression makes the grep match all characters which are not closing braces, and saves them into \1.
Finally, the expression ends with a \}, i.e. a closing brace, which is the end of what we want to find.
The "Replace" only contains \1, which is everything inside the parentheses (...) in the "Find" field.
Related
I am writing an interpreter for assembly using lex and yacc. The problem is that I need to parse a word that will strictly be at the end of the file. I've read that there is an anchor $, which can help. However it doesn't work as I expected. I've wrote this in my lex file:
ABC$ {printf("QWERTY\n");}
The input file is:
ABC
without spaces or any other invisible symbols. So I expect the outputput to be QWERTY, however what I get is:
ABC
which I guess means that the program couldn't parse it. Then I thought, that $ might be a regular symbol in lex, so I changed the input file into this:
ABC$
So, if $ isn't a special symbol, then it will be parsed as a normal symbol, and the output will be QWERTY. This doesn't happen, the output is:
ABC$
The question is whether $ in lex is a normal symbol or special one.
In (f)lex, $ matches zero characters followed by a newline character.
That's different from many regex libraries where $ will match at the end of input. So if your file does not have a newline at the end, as your question indicates (assuming you consider newline to be an invisible character), it won't be matched.
As #sepp2k suggests in a comment, the pattern also won't be matched if the input file happens to use Windows line endings (which consist of the sequence \r\n), unless the generated flex file was compiled for Windows. So if you created the file on Windows and run the flex-generated scanner in a Unix environment, the \r will also cause the pattern to fail to match. In that case, you can use (f)lex's trailing context operator:
ABC/\r?\n { puts("Matched ABC at the end of a line"); }
See the flex documentation for patterns for a full description of the trailing context operator. (Search for "trailing context" on that page; it's roughly halfway down.) $ is exactly equivalent to /\n.
That still won't match ABC at the very end of the file. Matching strings at the very end of the file is a bit tricky, but it can be done with two patterns if it's ok to recognise the string other than at the end of the file, triggering a different action:
ABC/. { /* Do nothing. This ABC is not at the end of a line or the file */ }
ABC { puts("ABC recognised at the end of a line"); }
That works because the first pattern will match as long as there is some non-newline character following ABC. (. matches any character other than a newline. See the above link for details.) If you also need to work with Windows line endings, you'll need to modify the trailing context in the first pattern.
I do have one large text file with lot of the following patterns;
because of,this
or that,has
or,not
Of course I want to change the following
because of, this
or that, has
or, not
To make myself clear: i would like to insert a space after each ,
How can i do that with BBEdit Find/Replace/Grep?
Find works ok with
[\,](\w)
but i can't figure out the coresponding part for replace.
Find: (\,)(\w)
Replace: \1 \2
Note: You need to hit the spacebar between \1 and \2. Works on my computer with BBedit v13
I complicates matters.
Pattern like Letter,Letter can also be found with ,\b.
\b is at the beginning or end of a word. \b in a regular expression means "word boundary".
The replacement is then done with ,_
Nota Bene: _ is a "Space" after ,
you could just replace all commas with a comma-space and then replace all space-space with space
There is a possibility to search using grep in TextWrangler
I want to find and replace the following word: bauvol, but not bauvolumen.
I tried typing ^bauvol$ into the search field but that didn't do the trick, it didn't find anything, although the word is clearly there.
I think it's because, in grep, the ^and $signify start and end of line, not a word?!
You want to use \b as word boundaries, as #gromi08 said:
\bbauvol\b
If you want to copy any portion of this word (so you can replace it, modify it, change the case, etc.) it is usually best to wrap it in ( and ) braces so you can reference them in the Replace box:
Find:
(\bbauvol\b)
Replace:
<some_tag>\1</some_tag>
Did you have anything specific you were trying to do with the result once you found it (cut it, duplicate it, etc.)?
Use the -w option of grep (see grep man-page.
This option searches for the expression as a word.
Therefore the command will be:
cat file.txt | grep -w bauvol
And yes, ^ and $ are for start and end of line.
I am using TeXnicCenter to edit a LaTeX document.
I now want to remove a certain tag (say, emph{blabla}} which occurs multiple times in my document , but not tag's content (so in this example, I want to remove all emphasization).
What is the easiest way to do so?
May also be using another program easily available on Windows 7.
Edit: In response to regex suggestions, it is important that it can deal with nested tags.
Edit 2: I really want to remove the tag from the text file, not just disable it.
Using a regular expression do something like s/\\emph\{([^\}]*)\}/\1/g. If you are not familiar with regular expressions this says:
s -- replace
/ -- begin match section
\\emph\{ -- match \emph{
( -- begin capture
[^\}]* -- match any characters except (meaning up until) a close brace because:
[] a group of characters
^ means not or "everything except"
\} -- the close brace
and * means 0 or more times
) -- end capture, because this is the first (in this case only) capture, it is number 1
\} -- match end brace
/ -- begin replace section
\1 -- replace with captured section number 1
/ -- end regular expression, begin extra flags
g -- global flag, meaning do this every time the match is found not just the first time
This is with Perl syntax, as that is what I am familiar with. The following perl "one-liners" will accomplish two tasks
perl -pe 's/\\emph\{([^\}]*)\}/\1/g' filename will "test" printing the file to the command line
perl -pi -e 's/\\emph\{([^\}]*)\}/\1/g' filename will change the file in place.
Similar commands may be available in your editor, but if not this will (should) work.
Crowley should have added this as an answer, but I will do that for him, if you replace all \emph{ with { you should be able to do this without disturbing the other content. It will still be in braces, but unless you have done some odd stuff it shouldn't matter.
The regex would be a simple s/\\emph\{/\{/g but the search and replace in your editor will do that one too.
Edit: Sorry, used the wrong brace in the regex, fixed now.
\renewcommand{\emph}[1]{#1}
any reasonably advanced editor should let you do a search/replace using regular expressions, replacing emph{bla} by bla etc.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.