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I'm quite new in Objective C. I'm trying to convert the following string 5896.3454A into a numeric string 58963454. Any idea how to do it.
I have been trying with NSScanner class, but I'm not sure, whether this is the best approach.
One possible solution would be to use a string replacement with regular expressions:
NSString *mixed = #"5896.3454A";
NSString *pattern = #"\\D"; // Pattern for "not a digit"
NSString *digitsOnly = [mixed stringByReplacingOccurrencesOfString:pattern
withString:#""
options:NSRegularExpressionSearch
range:NSMakeRange(0, [mixed length])];
Objective-C contains a lot of little tricks. Here is a silly one to solve your problem.
NSString *string = #"5896.3454A";
NSCharacterSet *digits = [NSCharacterSet characterSetWithCharactersInString:#"0123456789"];
NSCharacterSet *nonDigits = [digits invertedSet];
NSArray *digitSubstrings = [componentsSeparatedByCharactersInSet:nonDigits];
NSString *result = [digitSubstrings componentsJoinedByString:#""];
result will have only the digits in string.
I create a character set with all the digits. I use invertedSet to get a set with every character other than the digits. This is straight forward, next is the silly part. Next I split the string into an array of substring. This will have the effect of stripping out all the non digit characters, but all the digits are broken up into substrings. In this case, digitSubstrings will be #[#"5896", #"3454", #""]. Not a problem, I join all the substrings back together to create the final string.
You can do it also by
yourString mentioned in your question
NSInteger newInt = [ [ [ yourString stringByReplacingOccurrencesOfString:#"." withString:#""] stringByReplacingOccurrencesOfString:#"A" withString:#""] integerValue];
ADDED: NSInteger newInt = [ [ #"123.123A" stringByReplacingOccurrencesOfString:#"." withString:#""] integerValue]; combining NSString and NSScanner methods, integerValue ignores everything beginning with the first appearance of a character
You find it in the NSString class. Whether this is the better solution depends on the unwanted literals in your "string-number" you want to replace/delete.
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Remove Characters and Everything After from String
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I can't seem to find the answer to this anywhere. I can do it in c but objective c is difficult.
I want to cut the end of a string after a certain character
so user#example.com will become user (cut at '#')
How do I do this?
This will give you the first chunk of text that comes before your special character.
NSString *separatorString = #"#";
NSString *myString = #"user#example.com";
NSString *myNewString = [myString componentsSeparatedByString:separatorString].firstObject;
https://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html#//apple_ref/occ/instm/NSString/componentsSeparatedByString:
You can use a combination of substringToIndex: and rangeOfString: methods, like this:
NSString *str = #"user#example.com";
NSRange pos = [str rangeOfString:#"#"];
if (pos.location != NSNotFound) {
NSString *prefix = [str substringToIndex:pos.location];
}
Notes:
You need to check the location against NSNotFound to ensure that the position is valid.
substringToIndex: excludes the index itself, so the # character would not be included.
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I want to remove all character, the things that I need is number from 0 - 9
Could you please show me some example to solve this?
You could make your own NSCharacterSet and then trim out every character not contained within it by using its invertedSet. Here's an example:
NSString *stringWithLettersAndNumbers = #"345345345fff";
NSCharacterSet *myCharset = [NSCharacterSet characterSetWithCharactersInString:#"0123456789"];
NSString *newString = [stringWithLettersAndNumbers stringByTrimmingCharactersInSet:[myCharset invertedSet]];
NSLog(#"%#",newString);
Regular expressions are great for this sort of stuff:
NSString *orgStr = #"letters34243more32132letters";
NSString *result = [orgStr stringByReplacingOccurrencesOfString: #"\\D"
withString: #""
options: NSRegularExpressionSearch
range: NSMakeRange(0, orgStr.length)];
What it does is find any non-digit character (defined as \D) and replace it with an empty string, i.e., remove it.
It does it for any such character, irrespectively of its position in the string. In my reading of the question, this is what you need.
The double backslash ("\") in the pattern string is necessary because NSString literals need an extra backslash to escape the "real" one.
Loop through the string and pickup all digits.
NSMutableString *strResult = [NSMutableString string];
for (int position=0; position<text.length; position++) {
unichar c = [text characterAtIndex:position];
NSCharacterSet *numericSet = [NSCharacterSet decimalDigitCharacterSet];
if ([numericSet characterIsMember:c]) {
[strResult appendFormat:#"%C", c];
}
}
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I've looked at several posts to answer my question but i keep getting errors.
I basically have an NSString that holds 2 characters...
myString = #"ABC";
All I want to do is assign the first character in myString to another NSString variable...
NSString *firstChar;
firstChar = [myString characterAtIndex:0];
I get this error when i do this:
Incompatible integer to pointer conversion assigning to 'unichar *' (aka 'unsigned short *') from 'unichar' (aka 'unsigned short')
What am I doing wrong?
NSString's characterAtIndex: returns a unichar, not a string, so you can't just assign it to another string. You need to create a string from it first:
NSString *myString = #"ABC";
unichar firstChar = [myString characterAtIndex:0];
NSString *string = [NSString stringWithCharacters:&firstChar length:1];
Of course, as DarkDust says, you can also use substringWithRange: to get a string from a subrange of a string:
[myString subStringWithRange:NSMakeRange(0, 1)]; // to get the first character
[myString subStringWithRange:NSMakeRange(1, 2)]; // to get the second and third characters
The characterAtIndex: method returns a unichar (an integer representation of the character). What you want is -[NSString substringWithRange:], like [myString substringWithRange:NSMakeRange(0, 1)];
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NSString tokenize in Objective-C
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So, I have a vast quantity of NSStrings and my problem is I need to cut them into smaller strings at a specific point. This may sound complicated but what I need basically is this:
NSString *test =" blah blah blah - goo goo goo.";
NSString *str1 = "blah blah blah ";
NSString *str2 = "goo goo goo";
How do I code for when there's a hyphen for the string to just cut off there. Is there a way to do this? I found ways to cut of the string after a certain amount of letters but I need it at the hyphen every time.
NSArray *arr = [string componentsSeparatedByString:#"-"];
should do the trick.
You could do this many ways. Two answers above show a few approaches. Many Objective-C solutions will include NSRange usage. You could also do more flexible things with NSScanner or NSRegularExpression.
There is not going to be one right answer.
NSString *cutString = [text substringFromIndex:3];
cutString = [text substringToIndex:5];
cutString = [text substringWithRange:NSMakeRange(3, 5)];
I have searched for hours now and haven't found a solution for my problem. I have a NSString which looks like the following:
"spacer": ["value1", "value2"], "spacer": ["value1", "value2"], ...
What I want to do is to remove the [ ] characters from the string. It's seems to be impossible with objective-c. Most other programming languages offer functions like strpos or indexOf which allow me to search for a character or string and locate the position of it. But there seems nothing to be like this in objective-c.
Does anyone has an idea on how to remove these characters?
Additionally there are [] characters in the string which should remain, so I can't just use NSMutableString stringByReplacingOccurencesOfString:withString. I need to search first for the spacer string and then remove only the next two [] chars.
Thank you for helping me.
To find occurrences of a string within a string, use the rangeOfXXX methods in the NSString class. Then you can construct NSRanges to extract substrings, etc.
This example removes only the first set of open/close brackets in your sample string...
NSString *original = #"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSLog(#"%#", original);
NSRange startRange = [original rangeOfString:#"\["];
NSRange endRange = [original rangeOfString:#"]"];
NSRange searchRange = NSMakeRange(0, endRange.location);
NSString *noBrackets = [original stringByReplacingOccurrencesOfString:#"\[" withString:#"" options:0 range:searchRange];
noBrackets = [noBrackets stringByReplacingOccurrencesOfString:#"]" withString:#"" options:0 range:searchRange];
NSLog(#"{%#}", noBrackets);
The String Programming Guide has more details.
You might alternatively also be able to use the NSScanner class.
This worked for me to extract everything to the index of a character, in this case '[':
NSString *original = #"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSRange range = [original rangeOfString:#"\["];
NSString *toBracket = [NSString stringWithString :[original substringToIndex:range.location] ];
NSLog(#"toBracket: %#", toBracket);