I would like to find out how often a dot is in a string.
I tried the function below but it shows 11 instead of 1.
trace(count("example.com", "."));
function count(s:String, letter:String)
{
return s.match(new RegExp(letter,"g")).length;
}
What is wrong with the function?
In regular expressions, the . means "any character". Replace it with \.
If you are writing a general function, you have to make sure that the expression you pass to the RegExp is in fact looking for the thing you think you are looking for. If it's only ever a single character, I believe you can safely "escape" it by putting a \ in front of it.
Related
I need a function that behaves similar to the behavior of sscanf
For example, let's suppose we have a format string that looks like this (the function I'm looking for doesn't have to be exactly like this, but something similar)
"This is normal text that has to exactly match, but here is a ${var}"
And have return/modify a variable to look like
{'var': <whatever was there>}
After researching this for a while, the only things I could actually find was scanf, but that takes input form stdin, and not a string
I am aware that there is a regex solution for this, but I'm looking for a function that does this without the need for regex (regex is slow). However, if there is no other solution for this, I will accept a regex solution.
The normal solution for this in most languages that have regular expressions built-in is to use regular expressions.
If you're not used to or don't like regular expressions I'm sorry. Most of the programming world have assumed that knowledge of regular expressions is mandatory.
In any case. The normal solution to this is string.prototype.match:
let text = get_string_to_scan();
let match = text.match(/This is normal text that has to exactly match, but here is a (.+)/);
if (match) { // match is null if no match is found
// The result you want is in match[1]
console.log('value of var is:', match[1]);
}
What pattern you put in your capture group (the (..) part) depends on what you want. The code above captures anything at all including spaces and special characters.
If you just want to capture a "word", that is, printable characters without spaces, then you can use (\w+):
text.match(/This is normal text that has to exactly match, but here is a (\w+)/)
If you want to capture a word with only letters but not numbers you can use ([a-zA-Z]+):
text.match(/This is normal text that has to exactly match, but here is a ([a-zA-Z]+)/)
The flexibility of regular expression is why other methods of string scanning are usually not supported in languages that have had regular expression built-in since the beginning. But of course, flexibility comes with complexity.
Do you mean to have the ${var} to act as a placeholder? If so you could do it by replacing the " with the backtick:
console.log(`This is normal text that has to exactly match, but here is a ${"whatever was there"}`)
str = "fa, (captured)[asd] asf, 31"
for word in str:gmatch("\(%a+\)") do
print(word)
end
Hi! I want to capture a word between parentheses.
My Code should print "captured" string.
lua: /home/casey/Desktop/test.lua:3: invalid escape sequence near '\('
And i got this syntax error.
Of course, I can just find position of parentheses and use string.sub function
But I prefer simple code.
Also, brackets gave me a similar error.
The escape character in Lua patterns is %, not \. So use this:
word=str:match("%((%a+)%)")
If you only need one match, there is no need for a gmatch loop.
To capture the string in square brackets, use a similar pattern:
word=str:match("%[(%a+)%]")
If the captured string is not entirely composed of letters, use .- instead of %a+.
lhf's answer likely gives you what you need, but I'd like to mention one more option that I feel is underused and may work for you as well. One issue with using %((%a+)%) is that it doesn't work for nested parentheses: if you apply it to something like "(text(more)text)", you'll get "more" even though you may expect "text(more)text". Note that you can't fix it by asking to match to the first closing parenthesis (%(([^%)]+)%)) as it will give you "text(more".
However, you can use %bxy pattern item, which balances x and y occurrences and will return (text(more)text) in this case (you'd need to use something like (%b()) to capture it). Again, this may be overkill for your case, but useful to keep in mind and may help someone else who comes across this problem.
I have several functions that start with get_ in my code:
get_num(...) , get_str(...)
I want to change them to get_*_struct(...).
Can I somehow match the get_* regex and then replace according to the pattern so that:
get_num(...) becomes get_num_struct(...),
get_str(...) becomes get_str_struct(...)
Can you also explain some logic behind it, because the theoretical regex aren't like the ones used in UNIX (or vi, are they different?) and I'm always struggling to figure them out.
This has to be done in the vi editor as this is main work tool.
Thanks!
To transform get_num(...) to get_num_struct(...), you need to capture the correct text in the input. And, you can't put the parentheses in the regular expression because you may need to match pointers to functions too, as in &get_distance, and uses in comments. However, and this depends partially on the fact that you are using vim and partially on how you need to keep the entire input together, I have checked that this works:
%s/get_\w\+/&_struct/g
On every line, find every expression starting with get_ and continuing with at least one letter, number, or underscore, and replace it with the entire matched string followed by _struct.
Darn it; I shouldn't answer these things on spec. Note that other regex engines might use \& instead of &. This depends on having magic set, which is default in vim.
For an alternate way to do it:
%s/get_\(\w*\)(/get_\1_struct(/g
What this does:
\w matches to any "word character"; \w* matches 0 or more word characters.
\(...\) tells vim to remember whatever matches .... So, \(w*\) means "match any number of word characters, and remember what you matched. You can then access it in the replacement with \1 (or \2 for the second, etc.)
So, the overall pattern get_\(\w*\)( looks for get_, followed by any number of word chars, followed by (.
The replacement then just does exactly what you want.
(Sorry if that was too verbose - not sure how comfortable you are with vim regex.)
I'm trying to search and replace $data['user'] for $data['sessionUser'].
However, no matter what search string I use, I always get a "pattern not found" as the result of it.
So, what would be the correct search string? Do I need to escape any of these characters?
:%s/$data['user']/$data['sessionUser']/g
:%s/\$data\[\'user\'\]/$data['sessionUser']/g
I did not test this, but I guess it should work.
Here's a list of all special search characters you need to escape in Vim: `^$.*[~)+/
There's nothing wrong with with the answers given, but you can do this:
:%s/$data\['\zsuser\ze']/sessionUser/g
\zs and \ze can be used to delimit the part of the match that is affected by the replacement.
You don't need to escape the $ since it's the at the start of the pattern and can't match an EOL here. And you don't need to escape the ] since it doesn't have a matching starting [. However there's certainly no harm in escaping these characters if you can't remember all the rules. See :help pattern.txt for the full details, but don't try to digest it all in one go!
If you want to get fancy, you can do:
:%s/$data\['\zsuser\ze']/session\u&/g
& refers to the entire matched text (delimited by \zs and \ze if present), so it becomes 'user' in this case. The \u when used in a replacement string makes the next character upper-case. I hope this helps.
Search and replace in vim is almost identical to sed, so use the same escapes as you would with that:
:%s/\$data\['user'\]/$data['session']/g
Note that you only really need to escape special characters in the search part (the part between the first set of //s). The only character you need to escape in the replace part is the escape character \ itself (which you're not using here).
The [ char has a meaning in regex. It stands for character ranges. The $ char has a meaning too. It stands for end-line anchor. So you have to escape a lot of things. I suggest you to try a little plugin like this or this one and use a visual search.
I have a file containing string like this one :
print $hash_xml->{'div'}{'div'}{'div'}[1]...
I want to replace {'div'}{'div'}{'div'}[1] by something else.
So I tried
%s/{'div'}{'div'}{'div'}[1]/by something else/gc
The strings were not found. I though I had to escape the {,},[ and ]
Still string not found.
So I tried to search a single { and it found them.
Then I tried to search {'div'}{'div'}{'div'} and it found it again.
Then {'div'}{'div'}{'div'}[1 was still found.
To find {'div'}{'div'}{'div'}[1]
I had to use %s/{'div'}{'div'}{'div'}[1\]
Why ?
vim 7.3 on Linux
The [] are used in regular expressions to wrap a range of acceptable characters.
When both are supplied unescaped, vim is treating the search string as a regex.
So when you leave it out, or escape the final character, vim cannot interpret a single bracket in a regex context, so does a literal search (basically the best it can do given the search string).
Personally, I would escape the opening and closing square brace to ensure that the meaning is clear.
That's because the [ and ] characters are used to build the search pattern.
See :h pattern and use the help file pattern.txt to try the following experiment:
Searching for the "[9-0]" pattern (without quotes) using /[0-9] will match every digit from 0 to 9 individually (see :h \[)
Now, if you try /\[0-9] or /[0-9\] you will match the whole pattern: a zero, an hyphen and a nine inside square brackets. That's because when you escape one of [ or ] the operator [*] ceases to exist.
Using your search pattern, /{'div'}{'div'}{'div'}[1\] and /{'div'}{'div'}{'div'}\[1] should match the same pattern which is the one you want, while /{'div'}{'div'}{'div'}[1] matches the string {'div'}{'div'}{'div'}1.
In order to avoid being caught by these special characters in regular expressions, you can try using the very magic flag.
E.g.:
:%s/\V{'div'}[1]/replacement/
Notice the \V flag at the beginning of the line.
Because the square brackets mean that vim thinks you're looking for any of the characters inside. This is known as a 'character class'. By escaping either of the square brackets it lets vim know that you're looking for the literal square string ending with '[1]'.
Ideally you should write your expression as:
%s/{'div'}{'div'}{'div'}\[1\]/replacement string/
to ensure that the meaning is completely clear.