blocks|key|892600|text|这里有一种不依赖于尝试将字符串解析为数字的有限精度的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892601|NSCharacterSet*+notDigits+=+[[NSCharacterSet+decimalDigitCharacterSet]+invertedSet];
if+([newString+rangeOfCharacterFromSet:notDigits].location+==+NSNotFound)
{
++++//+newString+consists+only+of+the+digits+0+through+9
}|code-block|syntax|javascript|892602|参见%2B[NSCharacterSet+decimalDigitCharacterSet]和-[NSString+rangeOfCharacterFromSet:]。|offset|length|style|CODE|892603|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.apple.com/library/mac/documentation/cocoa/reference/foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/decimalDigitCharacterSet|1|http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html#//apple_ref/occ/instm/NSString/rangeOfCharacterFromSet:^0|0|0|2|16|19|10|2|16|0|19|10|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]|$I|16|J|17|1|18]]|A|$]]|$1|M|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

Here's one way that doesn't rely on the limited precision of attempting to parse the string as a number:

<pre><code>NSCharacterSet* notDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
if ([newString rangeOfCharacterFromSet:notDigits].location == NSNotFound)
{
 // newString consists only of the digits 0 through 9
}
</code></pre>

See <a href="http://developer.apple.com/library/mac/documentation/cocoa/reference/foundation/Classes/NSCharacterSet_Class/Reference/Reference.html#//apple_ref/occ/clm/NSCharacterSet/decimalDigitCharacterSet" rel="noreferrer"><code>+[NSCharacterSet decimalDigitCharacterSet]</code></a> and <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html#//apple_ref/occ/instm/NSString/rangeOfCharacterFromSet:" rel="noreferrer"><code>-[NSString rangeOfCharacterFromSet:]</code></a>.

blocks|key|892563|text|我建议使用NSNumberFormatter类中的numberFromString:方法，如果数字无效，它将返回nil；否则，它将返回NSNumber。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|892564|NSNumberFormatter+*nf+=+[[[NSNumberFormatter+alloc]+init]+autorelease];
BOOL+isDecimal+=+[nf+numberFromString:newString]+!=+nil;|code-block|syntax|javascript|892565|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSNumberFormatter_Class/Reference/Reference.html^0|P|H|5|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

I'd suggest using the <code>numberFromString:</code> method from the <a href="http://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSNumberFormatter_Class/Reference/Reference.html" rel="noreferrer">NSNumberFormatter</a> class, as if the number is not valid, it will return nil; otherwise, it will return you an NSNumber.

<pre><code>NSNumberFormatter *nf = [[[NSNumberFormatter alloc] init] autorelease];
BOOL isDecimal = [nf numberFromString:newString] != nil;
</code></pre>

blocks|key|675813|text|您可以创建一个NSScanner并简单地扫描字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|675814|NSDecimal+decimalValue;
NSScanner+*sc+=+[NSScanner+scannerWithString:newString];
[sc+scanDecimal:&decimalValue];
BOOL+isDecimal+=+[sc+isAtEnd];|code-block|syntax|javascript|675815|查看NSScanner's+documentation，了解更多可供选择的方法。|offset|length|675816|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSScanner_Class/Reference/Reference.html#//apple_ref/occ/instm/NSScanner/scanDecimal:^0|0|0|2|P|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You could create an NSScanner and simply scan the string:

<pre><code>NSDecimal decimalValue;
NSScanner *sc = [NSScanner scannerWithString:newString];
[sc scanDecimal:&amp;decimalValue];
BOOL isDecimal = [sc isAtEnd];
</code></pre>

Check out <a href="http://developer.apple.com/library/mac/documentation/Cocoa/Reference/Foundation/Classes/NSScanner_Class/Reference/Reference.html#//apple_ref/occ/instm/NSScanner/scanDecimal:" rel="noreferrer">NSScanner's documentation</a> for more methods to choose from.

blocks|key|892639|text|我认为检查给定字符串中的每个字符都是数字的最简单方法可能是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892640|NSString+*trimmedString+=+[newString+stringByTrimmingCharactersInSet:[NSCharacterSet+decimalDigitCharacterSet]];

if([trimmedString+length])
{
++++NSLog(@"some+characters+outside+of+the+decimal+character+set+found");
}
else
{
++++NSLog(@"all+characters+were+in+the+decimal+character+set");
}|code-block|syntax|javascript|892641|如果希望完全控制可接受的字符，请使用其他NSCharacterSet工厂方法之一。|892642|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I think the easiest way to check that every character within a given string is numeric is probably:

<pre><code>NSString *trimmedString = [newString stringByTrimmingCharactersInSet:[NSCharacterSet decimalDigitCharacterSet]];

if([trimmedString length])
{
 NSLog(@"some characters outside of the decimal character set found");
}
else
{
 NSLog(@"all characters were in the decimal character set");
}
</code></pre>

Use one of the other NSCharacterSet factory methods if you want complete control over acceptable characters.

blocks|key|892813|text|这个最初的问题是关于Objective-C的，但它也是在Swift宣布之前几年发布的。所以，如果你从谷歌来到这里，正在寻找一个使用Swift的解决方案，那么你可以去：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892814|let+testString+=+"12345"
let+badCharacters+=+NSCharacterSet.decimalDigitCharacterSet().invertedSet

if+testString.rangeOfCharacterFromSet(badCharacters)+==+nil+{
++++print("Test+string+was+a+number")
}+else+{
++++print("Test+string+contained+non-digit+characters.")
}|code-block|syntax|javascript|892815|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This original question was about Objective-C, but it was also posted years before Swift was announced. So, if you're coming here from Google and are looking for a solution that uses Swift, here you go:

<pre><code>let testString = "12345"
let badCharacters = NSCharacterSet.decimalDigitCharacterSet().invertedSet

if testString.rangeOfCharacterFromSet(badCharacters) == nil {
 print("Test string was a number")
} else {
 print("Test string contained non-digit characters.")
}
</code></pre>

blocks|key|676292|text|如果需要验证字符串是否只有数字，则使用Swift+3解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676293|CharacterSet.decimalDigits.isSuperset(of:+CharacterSet(charactersIn:+myString))|code-block|syntax|javascript|676294|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Swift 3 solution if need to verify that the string has only digits:

<pre><code>CharacterSet.decimalDigits.isSuperset(of: CharacterSet(charactersIn: myString))
</code></pre>

blocks|key|892893|text|Swift+3解决方案可能是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892894|extension+String+{

++++var+doubleValue:Double?+{
++++++++return+NumberFormatter().number(from:self)?.doubleValue
++++}

++++var+integerValue:Int?+{
++++++++return+NumberFormatter().number(from:self)?.intValue
++++}

++++var+isNumber:Bool+{
++++++++get+{
++++++++++++let+badCharacters+=+NSCharacterSet.decimalDigits.inverted
++++++++++++return+(self.rangeOfCharacter(from:+badCharacters)+==+nil)
++++++++}
++++}
}|code-block|syntax|javascript|892895|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Swift 3 solution could be like:

<pre><code>extension String {

 var doubleValue:Double? {
 return NumberFormatter().number(from:self)?.doubleValue
 }

 var integerValue:Int? {
 return NumberFormatter().number(from:self)?.intValue
 }

 var isNumber:Bool {
 get {
 let badCharacters = NSCharacterSet.decimalDigits.inverted
 return (self.rangeOfCharacter(from: badCharacters) == nil)
 }
 }
}
</code></pre>

blocks|key|892709|text|需要明确的是，此函数适用于字符串中的整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892710|以下是基于John上面的答案的一个小帮手类别：|892711|在.h文件中|892712|@interface+NSString+(NumberChecking)

%2B(bool)isNumber:(NSString+*)string;

@end|code-block|syntax|javascript|892713|在.m文件中|892714|#import+"NSString%2BNumberChecking.h"

@implementation+NSString+(NumberChecking)

%2B(bool)isNumber+{
++++if([self+rangeOfCharacterFromSet:[[NSCharacterSet+decimalDigitCharacterSet]+invertedSet]].location+==+NSNotFound)+{
++++++++return+YES;
++++}else+{
++++++++return+NO;
++++}
}

@end|892715|用法：|892716|#import+"NSString%2BNumberChecking.h"

if([someString+isNumber])+{
++++NSLog(@"is+a+number");
}else+{
++++NSLog(@"not+a+number");
}|892717|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|11|8|@]|9|@]|A|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

to be clear, this functions for integers in strings.

heres a little helper category based off of John's answer above:

in .h file

<pre><code>@interface NSString (NumberChecking)

+(bool)isNumber:(NSString *)string;

@end
</code></pre>

in .m file

<pre><code>#import "NSString+NumberChecking.h"

@implementation NSString (NumberChecking)

+(bool)isNumber {
 if([self rangeOfCharacterFromSet:[[NSCharacterSet decimalDigitCharacterSet] invertedSet]].location == NSNotFound) {
 return YES;
 }else {
 return NO;
 }
}

@end
</code></pre>

usage:

<pre><code>#import "NSString+NumberChecking.h"

if([someString isNumber]) {
 NSLog(@"is a number");
}else {
 NSLog(@"not a number");
}
</code></pre>

blocks|key|892926|text|还有另一种选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892927|-+(BOOL)isValidNumber:(NSString*)text+regex:(NSString*)regex+{
++++@try+{
++++++++NSPredicate+*predicate+=+[NSPredicate+predicateWithFormat:@"SELF+MATCHES+%25@",+regex];
++++++++return+[predicate+evaluateWithObject:text];
++++}
++++@catch+(NSException+*exception)+{
++++++++assert(false);+
++++++++return+NO;
++++}
}|code-block|syntax|javascript|892928|使用示例：|892929|BOOL+isValid+=+[self+isValidNumber:@"1234"+regex:@"%5E[0-9]%2B$"];|892930|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Yet another option:

<pre><code>- (BOOL)isValidNumber:(NSString*)text regex:(NSString*)regex {
 @try {
 NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", regex];
 return [predicate evaluateWithObject:text];
 }
 @catch (NSException *exception) {
 assert(false); 
 return NO;
 }
}
</code></pre>

Usage example:

<pre><code>BOOL isValid = [self isValidNumber:@"1234" regex:@"^[0-9]+$"];
</code></pre>

blocks|key|676045|text|测试字符串是否为数字可能会有所帮助|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676046|int+i+=+[@"12.3"+rangeOfCharacterFromSet:+[+[NSCharacterSet+characterSetWithCharactersInString:@"0123456789."]+invertedSet]+].location;

if+(i+==+NSNotFound)+{
+++++//is+a+number
}|code-block|syntax|javascript|676047|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Test if a string is a number Might be Helpful

<pre><code>int i = [@"12.3" rangeOfCharacterFromSet: [ [NSCharacterSet characterSetWithCharactersInString:@"0123456789."] invertedSet] ].location;

if (i == NSNotFound) {
 //is a number
}
</code></pre>

blocks|key|892838|text|Swift扩展：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|892839|extension+NSString+{
func+isNumString()+->+Bool+{
++++let+numbers+=+NSCharacterSet(charactersInString:+"0123456789.").invertedSet
++++let+range+=+self.rangeOfCharacterFromSet(numbers).location
++++if+range+==+NSNotFound+{
++++++++return+true
++++}
++++return+false
}++}|code-block|syntax|javascript|892840|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Swift extension :

<pre><code>extension NSString {
func isNumString() -&gt; Bool {
 let numbers = NSCharacterSet(charactersInString: "0123456789.").invertedSet
 let range = self.rangeOfCharacterFromSet(numbers).location
 if range == NSNotFound {
 return true
 }
 return false
} }
</code></pre>

blocks|key|893052|text|当你有来自mixed+languages的使用(或不使用)+0-9数字格式的数字时，你将需要运行一个正则表达式来查找任何数字，下一步是将所有数字转换为0-9格式(如果你需要实际的值)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|893053|//+Will+look+for+any+language+digits
let+regex+=+try+NSRegularExpression(pattern:+"[%5E[:digit:]]",+options:+.caseInsensitive)
let+digitsString+=+regex.stringByReplacingMatches(in:+string,
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++options:+NSRegularExpression.MatchingOptions(rawValue:+0),
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++range:+NSMakeRange(0,+string.count),+withTemplate:+"")
//+Converting+the+digits+to+be+0-9+format
let+numberFormatter+=+NumberFormatter()
numberFormatter.locale+=+Locale(identifier:+"EN")
let+finalValue+=+numberFormatter.number(from:+digitsString)

if+let+finalValue+=+finalValue+{
++let+actualValue+=+finalValue.doubleValue
}|code-block|syntax|javascript|893054|entityMap|0|LINK|mutability|MUTABLE|url|https://www.omniglot.com/language/numerals.htm^0|5|F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

When you have digits that are from <a href="https://www.omniglot.com/language/numerals.htm" rel="nofollow noreferrer">mixed languages</a>, that use (or don't) the 0-9 digits formats, you will need to run a regex that will look for any number, next thing is to convert all digits to be 0-9 format (if you need the actual value):

<pre><code>// Will look for any language digits
let regex = try NSRegularExpression(pattern: "[^[:digit:]]", options: .caseInsensitive)
let digitsString = regex.stringByReplacingMatches(in: string,
 options: NSRegularExpression.MatchingOptions(rawValue: 0),
 range: NSMakeRange(0, string.count), withTemplate: "")
// Converting the digits to be 0-9 format
let numberFormatter = NumberFormatter()
numberFormatter.locale = Locale(identifier: "EN")
let finalValue = numberFormatter.number(from: digitsString)

if let finalValue = finalValue {
 let actualValue = finalValue.doubleValue
}
</code></pre>

blocks|key|676419|text|最简单和最可靠的方法是尝试强制转换为Double，如果结果为nil+-它不能形成一个合法的数字。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|676420|let+strings+=+["test",+"123",+"123.2",+"-123",+"123-3",+"123..22",+".02"]
let+validNumbers+=+strings.compactMap(Double.init)

print(validNumbers)
//+prints+[123.0,+123.2,+-123.0,+0.02]|code-block|syntax|javascript|676421|文档中有更多信息：https://developer.apple.com/documentation/swift/double/2926277-init|676422|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/swift/double/2926277-init^0|I|6|U|3|0|0|9|1V|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

The easiest and most reliable way is trying to cast as <code>Double</code>, if the result is <code>nil</code> - it can't be formed into a legit number.

<pre class="lang-swift prettyprint-override"><code>let strings = ["test", "123", "123.2", "-123", "123-3", "123..22", ".02"]
let validNumbers = strings.compactMap(Double.init)

print(validNumbers)
// prints [123.0, 123.2, -123.0, 0.02]
</code></pre>

More info in the documentation:
<a href="https://developer.apple.com/documentation/swift/double/2926277-init" rel="nofollow noreferrer">https://developer.apple.com/documentation/swift/double/2926277-init</a>

blocks|key|676467|text|Simple编写了字符串扩展：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676468|extension+String+{
++++var+isNumeric:+Bool+{
++++++++return+self.rangeOfCharacter(from:+CharacterSet.decimalDigits.inverted)+==+nil
++++}
}|code-block|syntax|javascript|676469|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Simple wrote the string extension:
<pre><code>extension String {
 var isNumeric: Bool {
 return self.rangeOfCharacter(from: CharacterSet.decimalDigits.inverted) == nil
 }
}
</code></pre>

How can we check if a string is made up of numbers only. I am taking out a substring from a string and want to check if it is a numeric substring or not.

<pre><code>NSString *newString = [myString substringWithRange:NSMakeRange(2,3)];
</code></pre>

Finding out whether a string is numeric or not

Objective-C

Swift

我们如何检查一个字符串是否只由数字组成。我正在从字符串中取出一个子字符串，并希望检查它是否是数值子字符串。NSString *newString = [myString substringWithRange:NSMakeRange(2,3)];

问判断字符串是否为数字
EN

回答 14

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问判断字符串是否为数字EN

回答 14

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问判断字符串是否为数字
EN