How to add a newline character in localizable.strings?
I tried putting \n, but no success.
Using \n should just work. With this line in "Localizable.strings":
"abc" = "foo\nbar";
and this code:
NSString *s = NSLocalizedString(#"abc", NULL);
NSLog(#"%#", s);
I get the output
2013-05-02 14:14:45.931 test[4088:c07] foo
bar
Just adding newlines in the .strings file also works
"str" = "Hi ,
this is .
in a new line,
";
This works in an UILabel and UITextview as long as you set the appropriate line number:
testLabel.numberOfLines = 2;
You could also set this to 0 which is automatic line count, also you should ensure that your label is big enough to show multiple lines, or else it will be cut off.
this will not work in localizable.strings you have to create two keys and then only you can manage \n between two localizable strings during the concatination of strings.
Related
I have a file like:
Fruit.Store={
#blabla
"customer-id:12345,item:store/apple" = (1,2); #blabla
"customer-id:23456,item:store/banana" = (1,3); #blabla
"customer-id:23456,item:store/watermelon" = (1,4);
#blabla
"customer-id:67890,item:store/watermelon" = (1,6);
#The following two are unique
"customer-id:0000,item:store/" = (100, 100);
#
"" = (0,0)
};
Except the comments, each line has the same format: customer-id and item:store/ are fixed, and customer-id is a 5-digit number. The last two records are unique. How could I make sure the file is in the right format elegantly? I am thinking about using the flag for the first special line Fruit.Store={ and than for the following lines split each line by "," and "=", and if the splitted line is not correct, match them with the last two records. I want to use Ruby for it. Any advice? Thank you.
I am also thinking about using regular expression for the format, and wrote:
^"customer:\d{5},item:store\/\D*"=\(\d*,\d*\);
but I want to combine these two situations (with comment and without comment):
^"customer:\d{5},item:store\/\D*"=\(\d*,\d*\);$
^"customer:\d{5},item:store\/\D*"=\(\d*,\d*\);#.*$
how could I do it? Thanks
Using regular expressions could be a good option since each line has a fixed format; and you almost got it, your regex just needed a few tweaks:
(?:#.*|^"customer-id:\d{5},item:store\/\D*" *= *\(\d*, *\d*\); *(?:#.*)?)$
This is what was added to your current regex:
Option to be a comment line (#.*) or (|) a regular line (everything after |).
Check for possible spaces before and after =, after the comma (,) that separates the digits in parenthesis, and at the end of the line.
Option to include another comment at the end of the line ((?:#.*)?).
So just compare each line against this regex to check for the right format.
I'm trying to reformat German dates (e.g. 13.03.2011 to 2011-03-13).
This is my code:
str = "13.03.2011\n14:30\n\nHannover Scorpions\n\nDEG Metro Stars\n60\n2 - 3\n\n\n\n13.03.2011\n14:30\n\nThomas Sabo Ice Tigers\n\nKrefeld Pinguine\n60\n2 - 3\n\n\n\n"
str = str.gsub("/(\d{2}).(\d{2}).(\d{4})/", "/$3-$2-$1/")
I get the same output like input. I also tried my code with and without leading and ending slashes, but I don't see a difference. Any hints?
I tried to store my regex'es in variables like find = /(\d{2}).(\d{2}).(\d{4})/ and replace = /$3-$2-$1/, so my code looked like this:
str = "13.03.2011\n14:30\n\nHannover Scorpions\n\nDEG Metro Stars\n60\n2 - 3\n\n\n\n13.03.2011\n14:30\n\nThomas Sabo Ice Tigers\n\nKrefeld Pinguine\n60\n2 - 3\n\n\n\n"
find = /(\d{2}).(\d{2}).(\d{4})/
replace = /$3-$2-$1/
str = str.gsub(find, replace)
TypeError: no implicit conversion of Regexp into String
from (irb):4:in `gsub'
Any suggestions for this problem?
First mistake is the regex delimiter. You do not need place the regex as string. Just place it inside a delimiter like //
Second mistake, you are using captured groups as $1. Replace those as \\1
str = str.gsub(/(\d{2})\.(\d{2})\.(\d{4})/, "\\3-\\2-\\1")
Also, notice I have escaped the . character with \., because in regex . means any character except \n
I have a bunch of NSStrings from which I would like to grab the first character of and match them up in the range A-Z and # as a catch all for things that don't apply.
Different graphemes (I believe that's the correct word after some wiki'ing) have been giving me trouble. For example, I would like to extract A from "Æ".
I have taken a look at CFStringTransform, normalize and fold but none of had the desired effect.
Is there a reliable way of doing this? All the strings I'm working with are UTF8 if that makes a difference.
Æ cannot be broken down into components. It is not a compound glyph of A+E, but is a separate glyph. Compound glyphs are things like a+`
The thing about "Æ" is that it is an ascii character in itself. Not a combination of two different characters so you can't extract the A from it because it is only 1 Character.
Edit:
Although you could perform a check to see if the String equals "Æ" and if it does tell it to switch it with "A" or convert it to its dec, form and subtract 81 which would give you an "A".
Did you want to get rid of all æ?
This should work if you do.
NSString *string = #"Æaæbcdef";
string = [string stringByReplacingOccurrencesOfString:#"æ" withString:#"a"];
string = [string stringByReplacingOccurrencesOfString:#"Æ" withString:#"A"];
Edit
Rereading, you only seem to want the first character:
NSString *string = #"Æaæbcdef";
NSString *firstChar = [string substringToIndex:1];
firstChar = [firstChar stringByReplacingOccurrencesOfString:#"æ" withString:#"a"];
firstChar = [firstChar stringByReplacingOccurrencesOfString:#"Æ" withString:#"A"];
NSString *finalString = [NSString stringWithFormat:#"%#%#", firstChar, [string substringFromIndex:1]];
How do i remove empty lines from a string?
I have tried
some_string = some_string.gsub(/^$/, "");
and much more, but nothing works.
Remove blank lines:
str.gsub /^$\n/, ''
Note: unlike some of the other solutions, this one actually removes blank lines and not line breaks :)
>> a = "a\n\nb\n"
=> "a\n\nb\n"
>> a.gsub /^$\n/, ''
=> "a\nb\n"
Explanation: matches the start ^ and end $ of a line with nothing in between, followed by a line break.
Alternative, more explicit (though less elegant) solution:
str.each_line.reject{|x| x.strip == ""}.join
squeeze (or squeeze!) does just that - without a regex.
str.squeeze("\n")
Replace multiple newlines with a single one:
fixedstr = str.gsub(/\n\n+/, "\n")
or
str.gsub!(/\n\n+/, "\n")
You could try to replace all occurrences of 2 or more line breaks with just one:
my_string.gsub(/\n{2,}/, '\n')
Originally
some_string = some_string.gsub(/\n/,'')
Updated
some_string = some_string.gsub(/^$\n/,'')
I have a string of four blank lines which all up makes eight lines in total in the following:
str = "aaa\n\n\nbbb\n\nccc\ddd\n"
I want to return this all in one line. The output should be like this on a single line:
aaabbbcccddd
I used various trim functions to get the output but still I am failing.
What method do I have to use here?
The Ruby (and slightly less Perl-ish) way:
new_str = str.delete "\n"
...or if you want to do it in-place:
str.delete! "\n"
str.gsub(/\n/,'')
> str = "aaa\n\n\nbbb\n\nccc\ddd\n"
=> "aaa\n\n\nbbb\n\ncccddd\n"
> str.gsub("\n", "")
=> "aaabbbcccddd"