I have a string that is being generate from a formula, however I only want to use the string as long as all of its characters are numeric, if not that I want to do something different for instance display an error message.
I have been having a look round but am finding it hard to find anything that works along the lines of what I am wanting to do. I have looked at NSScanner but I am not sure if its checking the whole string and then I am not actually sure how to check if these characters are numeric
- (void)isNumeric:(NSString *)code{
NSScanner *ns = [NSScanner scannerWithString:code];
if ( [ns scanFloat:NULL] ) //what can I use instead of NULL?
{
NSLog(#"INSIDE IF");
}
else {
NSLog(#"OUTSIDE IF");
}
}
So after a few more hours searching I have stumbled across an implementation that dose exactly what I am looking for.
so if you are looking to check if their are any alphanumeric characters in your NSString this works here
-(bool) isNumeric:(NSString*) hexText
{
NSNumberFormatter* numberFormatter = [[[NSNumberFormatter alloc] init] autorelease];
NSNumber* number = [numberFormatter numberFromString:hexText];
if (number != nil) {
NSLog(#"%# is numeric", hexText);
//do some stuff here
return true;
}
NSLog(#"%# is not numeric", hexText);
//or do some more stuff here
return false;
}
hope this helps.
Something like this would work:
#interface NSString (usefull_stuff)
- (BOOL) isAllDigits;
#end
#implementation NSString (usefull_stuff)
- (BOOL) isAllDigits
{
NSCharacterSet* nonNumbers = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
NSRange r = [self rangeOfCharacterFromSet: nonNumbers];
return r.location == NSNotFound && self.length > 0;
}
#end
then just use it like this:
NSString* hasOtherStuff = #"234 other stuff";
NSString* digitsOnly = #"123345999996665003030303030";
BOOL b1 = [hasOtherStuff isAllDigits];
BOOL b2 = [digitsOnly isAllDigits];
You don't have to wrap the functionality in a private category extension like this, but it sure makes it easy to reuse..
I like this solution better than the others since it wont ever overflow some int/float that is being scanned via NSScanner - the number of digits can be pretty much any length.
Consider NSString integerValue - it returns an NSInteger. However, it will accept some strings that are not entirely numeric and does not provide a mechanism to determine strings which are not numeric at all. This may or may not be acceptable.
For instance, " 13 " -> 13, "42foo" -> 42 and "helloworld" -> 0.
Happy coding.
Now, since the above was sort of a tangent to the question, see determine if string is numeric. Code taken from link, with comments added:
BOOL isNumeric(NSString *s)
{
NSScanner *sc = [NSScanner scannerWithString: s];
// We can pass NULL because we don't actually need the value to test
// for if the string is numeric. This is allowable.
if ( [sc scanFloat:NULL] )
{
// Ensure nothing left in scanner so that "42foo" is not accepted.
// ("42" would be consumed by scanFloat above leaving "foo".)
return [sc isAtEnd];
}
// Couldn't even scan a float :(
return NO;
}
The above works with just scanFloat -- e.g. no scanInt -- because the range of a float is much larger than that of an integer (even a 64-bit integer).
This function checks for "totally numeric" and will accept "42" and "0.13E2" but reject " 13 ", "42foo" and "helloworld".
It's very simple.
+ (BOOL)isStringNumeric:(NSString *)text
{
NSCharacterSet *alphaNums = [NSCharacterSet decimalDigitCharacterSet];
NSCharacterSet *inStringSet = [NSCharacterSet characterSetWithCharactersInString:text];
return [alphaNums isSupersetOfSet:inStringSet];
}
Like this:
- (void)isNumeric:(NSString *)code{
NSScanner *ns = [NSScanner scannerWithString:code];
float the_value;
if ( [ns scanFloat:&the_value] )
{
NSLog(#"INSIDE IF");
// do something with `the_value` if you like
}
else {
NSLog(#"OUTSIDE IF");
}
}
Faced same problem in Swift.
In Swift you should use this code, according TomSwift's answer:
func isAllDigits(str: String) -> Bool {
let nonNumbers = NSCharacterSet.decimalDigitCharacterSet()
if let range = str.rangeOfCharacterFromSet(nonNumbers) {
return true
}
else {
return false
}
}
P.S. Also you can use other NSCharacterSets or their combinations to check your string!
For simple numbers like "12234" or "231231.23123" the answer can be simple.
There is a transformation law for int numbers: when string with integer transforms to int (or long) number and then, again, transforms it back to another string these strings will be equal.
In Objective C it will looks like:
NSString *numStr=#"1234",*num2Str=nil;
num2Str=[NSString stringWithFormat:#"%lld",numStr.longlongValue];
if([numStr isEqualToString: num2Str]) NSLog(#"numStr is an integer number!");
By using this transformation law we can create solution
to detect double or long numbers:
NSString *numStr=#"12134.343"
NSArray *numList=[numStr componentsSeparatedByString:#"."];
if([[NSString stringWithFormat:#"%lld", numStr.longLongValue] isEqualToString:numStr]) NSLog(#"numStr is an integer number");
else
if( numList.count==2 &&
[[NSString stringWithFormat:#"%lld",((NSString*)numList[0]).longLongValue] isEqualToString:(NSString*)numList[0]] &&
[[NSString stringWithFormat:#"%lld",((NSString*)numList[1]).longLongValue] isEqualToString:(NSString*)numList[1]] )
NSLog(#"numStr is a double number");
else
NSLog(#"numStr is not a number");
I did not copy the code above from my work code so can be some mistakes, but I think the main point is clear.
Of course this solution doesn't work with numbers like "1E100", as well it doesn't take in account size of integer and fractional part. By using the law described above you can do whatever number detection you need.
C.Johns' answer is wrong. If you use a formatter, you risk apple changing their codebase at some point and having the formatter spit out a partial result. Tom's answer is wrong too. If you use the rangeOfCharacterFromSet method and check for NSNotFound, it'll register a true if the string contains even one number. Similarly, other answers in this thread suggest using the Integer value method. That is also wrong because it will register a true if even one integer is present in the string. The OP asked for an answer that ensures the entire string is numerical. Try this:
NSCharacterSet *searchSet = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
Tom was right about this part. That step gives you the non-numerical string characters. But then we do this:
NSString *trimmedString = [string stringByTrimmingCharactersInSet:searchSet];
return (string.length == trimmedString.length);
Tom's inverted character set can TRIM a string. So we can use that trim method to test if any non numerals exist in the string by comparing their lengths.
Related
This question already has answers here:
Compare version numbers in Objective-C
(15 answers)
Closed 9 years ago.
I have two strings, one contains the value "5.2.3" and another one contain a value like "5.2.32". My question is: how to compare these two strings?
if ([string1 integerValue] >= [sting2 integerValue])
{
NSLog(#"process");
}
I tried above line but not got it.
Well correct answer has been already given. Because I have spent my half an hour on it so I don't want to waste it.
-(BOOL)string:(NSString*)str1 isGreaterThanString:(NSString*)str2
{
NSArray *a1 = [str1 componentsSeparatedByString:#"."];
NSArray *a2 = [str2 componentsSeparatedByString:#"."];
NSInteger totalCount = ([a1 count] < [a2 count]) ? [a1 count] : [a2 count];
NSInteger checkCount = 0;
while (checkCount < totalCount)
{
if([a1[checkCount] integerValue] < [a2[checkCount] integerValue])
{
return NO;
}
else if([a1[checkCount] integerValue] > [a2[checkCount] integerValue])
{
return YES;
}
else
{
checkCount++;
}
}
return NO;
}
And you can call this method like this:-
if([self string:str1 isGreaterThanString:str2])
{
NSLog(#"str2 is lower than the str1");
}
else
{
NSLog(#"str1 is lower than the str2");
}
It would appear that what you have here are not really "float" values, but some kind of multi-part "number" (akin to software version numbering?) that is not going to be covered by any of the standard conversions, but will also not compare "correctly" as just simple strings.
First you need to specify exactly what your comparison rules are. For example, I suspect you want something like:
1.2 > 1.1
1.1.1 > 1.1
1.11 > 1.2
1.2.3 > 1.2.2
1.2.22 > 1.2.3
(in other words, split the string up by "."s, and do a numeric comparison on each component). You'll have to decide how you want to handle things like letters, other delimiters, etc. showing up in the input. For example is 1.0b1 > 1.01 ?
Once you settle on the rules, write a method (returning NSComparisonResult) to implement the comparison. If you want to get fancy, you can even define your comparison method in a category on NSString, so you could do things like
if ([string1 mySuperDuperCompareTo:string2] == NSOrderedAscending) {
NSLog(#"%# < %#", string1, string2);
} // ... etc ...
see also How to let the sortedArrayUsingSelector using integer to sort instead of String?
#The Tiger is correct. Sorry to misunderstood your question. I have already mark deleted as my older answer. Here is the updated one.
As there are multiple . (dots) available here is the new solution. This will check value first like 5.2.3 and 5.2.32 are there. Then,
check first value 5 - same
so check next 2 - same
check next 3 and 32 - 32 is larger
also check for the same string as well (Also one of the probability)
Here is the logic - I have not compiled but this is base idea - there might require some correction
// separate from "."
NSArray *arrString1 = [string1 componentSeparatedBy:#"."];
NSArray *arrString2 = [string1 componentSeparatedBy:#"."];
BOOL isString1Bigger = NO; // a variable to check
BOOL isString2Bigger = NO; // a variable to check
// check count to run loop accordingly
if ([arrString1 count] <= [arrString2 count]) {
for (int strVal=0; strVal<[arrString1 count]; strVal++) {
// compare strings value converted into integer format
// when you get larger then break the loop
if([[arrString1 objectAtIndex:strVal] intValue] > [[arrString2 objectAtIndex:strVal] intValue]) {
isString1Bigger = YES;
break;
}
}
if ([arrString1 count] > [arrString2 count]) {
// use arrString2 in loop and isString2Bigger as a mark
}
// if after running both the if condition still require to check if both are same or not, for that,
if ((isString1Bigger == NO) && (isString2Bigger == NO)) {
// same string values
}
There might some modification required to run over. But its the base concept to compare string value provided by you.
I want to solve a conditional equation in iOS:
The equation I get from database is in NSString format, for example:
if((height > 0), (weight+ 2 ), ( weight-1 ) )
As per our understanding, if I parse the above string and separateheight>0condition, it will be in the NSString format. But to evaluate it how do I convert the string to a conditional statement?
Once the conditional statement is obtained the equation can be solved by converting it to a ternary equation as follows:
Bool status;
NSString *condition=#” height>0”;
If(condition) //But condition is treated as a string and not as a conditional statement.
{
status=True;
}
else
{
status=False;
}
Return status ? weight+ 2 : weight-1;`
Also the equations can dynamically change, so they cannot be hard coded. In short how do I solve this equation which I get as a NSString.
Thank you for your patience!
DDMathParser author here...
To expand on Jonathan's answer, here's how you could do it entirely in DDMathParser. However, to parse the string as-is, you'll need to do two things.
First, you'll need to create an if function:
DDMathEvaluator *evaluator = [DDMathEvaluator sharedMathEvaluator];
[evaluator registerFunction:^DDExpression *(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError *__autoreleasing *error) {
if ([args count] == 3) {
DDExpression *condition = [args objectAtIndex:0];
DDExpression *resultExpression = nil;
NSNumber *conditionValue = [condition evaluateWithSubstitutions:vars evaluator:eval error:error];
if ([conditionValue boolValue] == YES) {
resultExpression = [args objectAtIndex:1];
} else {
resultExpression = [args objectAtIndex:2];
}
NSNumber *result = [resultExpression evaluateWithSubstitutions:vars evaluator:eval error:error];
return [DDExpression numberExpressionWithNumber:result];
}
return nil;
} forName:#"if"];
This creates the if() function, which takes three parameters. Depending on how the first parameter evaluates, it either evaluates to the result of the second or third parameter.
The other thing you'll need to do is tell the evaluator what height and weight mean. Since they don't start with a $ character, they get interpreted as functions, and not variables. If they started with a $, then it would be as simple as evaluating it like this:
NSString *expression = #"if(($height > 0), ($weight+ 2 ), ( $weight-1 ) )";
NSDictionary *variables = #{#"height" : #42, #"weight" : #13};
NSNumber *result = [expression evaluateWithSubstitutions:variables evaluator:evaluator error:nil];
However, since they don't start with a $, they're functions, which means you need to tell the evaluator what the functions evaluate to. You could do this by creating functions for both height and weight, just like you did for if:
[evaluator registerFunction:^DDExpression *(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError **error) {
return [DDExpression numberExpressionWithNumber:#42];
} forName:#"height"];
Alternatively, you could make it a bit more dynamic and use the functionResolver block of DDMathEvaluator, which is a block that returns a block (woooooo) and would look like this:
NSDictionary *values = #{#"height": #42, #"weight": #13};
[evaluator setFunctionResolver:^DDMathFunction(NSString *name) {
DDMathFunction f = ^(NSArray *args, NSDictionary *vars, DDMathEvaluator *eval, NSError **error) {
NSNumber *n = [values objectForKey:name];
if (!n) { n = #0; }
return [DDExpression numberExpressionWithNumber:n];
};
return f;
}];
With those two pieces in place (registering if and providing the values of height and weight), you can do:
NSString *expression = #"if((height > 0), (weight+ 2 ), ( weight-1 ) )";
NSNumber *result = [expression evaluateWithSubstitutions:nil evaluator:evaluator error:nil];
... and get back the proper result of #15.
(I have plans to make DDMathParser allow unknown functions to fall back to provided variable values, but I haven't quite finished it yet)
you will have to write your own interpreter or find one that supports this kind of expressions.
The first part (the condition) can be evaluated by NSPredicate. For the second part (the calculation) you will need some math expression evaluation. Try this out https://github.com/davedelong/DDMathParser. Maybe you can do both with DDMathParser but i am not sure.
I am validating an NSString to ensure that the string does not contain apostrophes.
The code I'm using to do this is
NSCharacterSet * invalidNumberSet = [NSCharacterSet characterSetWithCharactersInString:#"'"];
NSScanner * scanner = [NSScanner scannerWithString:string];
NSString * scannerResult;
[scanner setCharactersToBeSkipped:nil];
[scanner scanUpToCharactersFromSet:invalidNumberSet intoString:&scannerResult];
if(![string isEqualToString:scannerResult])
{
return 2;
}
Returning 2 represents an error. This code works, except for the case where the string is an apostrophe.
To get around this issue, I added the following code above the preceding block.
if([string isEqualToString:#"'"]);
{
return 2;
}
This code is evaluating to true, regardless of the input. I need to either prevent the first block from crashing with the input of ', or get the second block to work.
What am I missing?
There's no logical reason why the isEqualToString: test should always succeed. If that's your actual, copy-pasted code, you must have an error somewhere else in the function.
At any rate, it would be much simpler to test if the location of [string rangeOfString:#"'"] is NSNotFound.
Anybody has some code in objective-c to convert a NSInteger or NSString to binary string?
example:
56 -> 111000
There are some code in stackoverflow that try do this, but it doesn´t work.
Thanks
Not sure which examples on SO didn't work for you, but Adam Rosenfield's answer here seems to work. I've updated it to remove a compiler warning:
// Original author Adam Rosenfield... SO Question 655792
NSInteger theNumber = 56;
NSMutableString *str = [NSMutableString string];
for(NSInteger numberCopy = theNumber; numberCopy > 0; numberCopy >>= 1)
{
// Prepend "0" or "1", depending on the bit
[str insertString:((numberCopy & 1) ? #"1" : #"0") atIndex:0];
}
NSLog(#"Binary version: %#", str);
Tooting my own horn a bit here...
I've written a math framework called CHMath that deals with arbitrarily large integers. One of the things it does is allows the user to create a CHNumber from a string, and get its binary representation as a string. For example:
CHNumber * t = [CHNumber numberWithString:#"56"];
NSString * binaryT = [t binaryStringValue];
NSLog(#"Binary value of %#: %#", t, binaryT);
Logs:
2009-12-15 10:36:10.595 otest-x86_64[21918:903] Binary value of 56: 0111000
The framework is freely available on its Github repository.
How would I say that if a UITextField has #"-" in it, do something.
Right now my code is like this. It doesn't seem to work:
if (MyUITextField.text == #"-") {
NSRange range = {0,1};
[a deleteCharactersInRange:range];
MyUITextField.text = MyUILabel.text;
}
I know that I am doing something very wrong with the code. Please help.
try changing == to [MyUITextField.text isEqualToString:#"-"]
as == tests to see if they are the same object, while isEqualToString compares the contents of the strings.
Assuming your string is defined as:
NSString *str = #"foo-bar";
To check if your string contains "-" you can do the following:
if ([str rangeOfString:#"-"].length > 0)
{
NSLog(#"Contains -");
}
It looks like you wanted to delete the first character if a string starts with a given character. In this case you can do something like this:
if ([str hasPrefix:#"f"])
{
NSLog(#"Starts with f");
}