objective-c 如何将 NSString 转换为 NSNumber

How to convert an NSString into an NSNumber

提问by Enyra

How can I convert a NSStringcontaining a number of any primitive data type (e.g. int, float, char, unsigned int, etc.)? The problem is, I don't know which number type the string will contain at runtime.

如何可以转换一个NSString包含许多任何原始数据类型（例如int，float，char，unsigned int，等等）？问题是，我不知道字符串在运行时将包含哪种数字类型。

I have an idea how to do it, but I'm not sure if this works with any type, also unsigned and floating point values:

我有一个想法如何做到这一点，但我不确定这是否适用于任何类型，也适用于无符号和浮点值：

long long scannedNumber;
NSScanner *scanner = [NSScanner scannerWithString:aString];
[scanner scanLongLong:&scannedNumber]; 
NSNumber *number = [NSNumber numberWithLongLong: scannedNumber];

Thanks for the help.

谢谢您的帮助。

回答by Dave DeLong

Use an NSNumberFormatter:

使用NSNumberFormatter：

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
NSNumber *myNumber = [f numberFromString:@"42"];

If the string is not a valid number, then myNumberwill be nil. If it is a valid number, then you now have all of the NSNumbergoodness to figure out what kind of number it actually is.

如果字符串不是有效数字，myNumber则将为nil。如果它是一个有效数字，那么您现在就NSNumber可以弄清楚它实际上是什么类型的数字。

回答by Barry Wark

You can use -[NSString integerValue], -[NSString floatValue], etc. However, the correct (locale-sensitive, etc.) way to do this is to use -[NSNumberFormatter numberFromString:]which will give you an NSNumber converted from the appropriate locale and given the settings of the NSNumberFormatter(including whether it will allow floating point values).

您可以使用-[NSString integerValue],-[NSString floatValue]等。但是，正确的（对区域设置敏感的等）方法是使用-[NSNumberFormatter numberFromString:]which 将为您提供从适当的区域设置转换而来的 NSNumber 并给出 的设置NSNumberFormatter（包括它是否允许浮动点值）。

回答by Kevin R

Objective-C

目标-C

(Note: this method doesn't play nice with difference locales, but is slightly faster than a NSNumberFormatter)

（注意：这种方法在不同的语言环境下效果不佳，但比 a 稍快NSNumberFormatter）

NSNumber *num1 = @([@"42" intValue]);
NSNumber *num2 = @([@"42.42" floatValue]);

---

---

Swift

迅速

Simple but dirty way

简单但肮脏的方式

// Swift 1.2
if let intValue = "42".toInt() {
    let number1 = NSNumber(integer:intValue)
}

// Swift 2.0
let number2 = Int("42')

// Swift 3.0
NSDecimalNumber(string: "42.42") 

// Using NSNumber
let number3 = NSNumber(float:("42.42" as NSString).floatValue)

The extension-wayThis is better, really, because it'll play nicely with locales and decimals.

扩展方式这更好，真的，因为它可以很好地处理语言环境和小数。

extension String {

    var numberValue:NSNumber? {
        let formatter = NSNumberFormatter()
        formatter.numberStyle = .DecimalStyle
        return formatter.number(from: self)
    }
}

Now you can simply do:

现在你可以简单地做：

let someFloat = "42.42".numberValue
let someInt = "42".numberValue

回答by ma11hew28

For strings starting with integers, e.g., @"123", @"456 ft", @"7.89", etc., use -[NSString integerValue].

对于开始整数，字符串如@"123"，@"456 ft"，@"7.89"，等使用-[NSString integerValue]。

So, @([@"12.8 lbs" integerValue])is like doing [NSNumber numberWithInteger:12].

所以，@([@"12.8 lbs" integerValue])就像做[NSNumber numberWithInteger:12].

回答by MacMark

You can also do this:

你也可以这样做：

NSNumber *number = @([dictionary[@"id"] intValue]]);

Have fun!

玩得开心！

回答by Ad1905

If you know that you receive integers, you could use:

如果您知道收到整数，则可以使用：

NSString* val = @"12";
[NSNumber numberWithInt:[val intValue]];

回答by miker

Here's a working sample of NSNumberFormatter reading localized number NSString (xCode 3.2.4, osX 10.6), to save others the hours I've just spent messing around. Beware: while it can handle trailing blanks ("8,765.4 " works), this cannot handle leading white space and this cannot handle stray text characters. (Bad input strings: " 8" and "8q" and "8 q".)

这是 NSNumberFormatter 读取本地化数字 NSString (xCode 3.2.4, osX 10.6) 的工作示例，以节省其他人我刚刚花费的时间。注意：虽然它可以处理尾随空白（“8,765.4”有效），但它不能处理前导空格，也不能处理杂散文本字符。（错误的输入字符串：“8”和“8q”和“8q”。）

NSString *tempStr = @"8,765.4";  
     // localization allows other thousands separators, also.
NSNumberFormatter * myNumFormatter = [[NSNumberFormatter alloc] init];
[myNumFormatter setLocale:[NSLocale currentLocale]]; // happen by default?
[myNumFormatter setFormatterBehavior:NSNumberFormatterBehavior10_4];
     // next line is very important!
[myNumFormatter setNumberStyle:NSNumberFormatterDecimalStyle]; // crucial

NSNumber *tempNum = [myNumFormatter numberFromString:tempStr];
NSLog(@"string '%@' gives NSNumber '%@' with intValue '%i'", 
    tempStr, tempNum, [tempNum intValue]);
[myNumFormatter release];  // good citizen

回答by Anthony

I wanted to convert a string to a double. This above answer didn't quite work for me. But this did: How to do string conversions in Objective-C?

我想将字符串转换为双精度。上面的答案对我来说不太适用。但这确实做到了：如何在 Objective-C 中进行字符串转换？

All I pretty much did was:

我几乎所做的就是：

double myDouble = [myString doubleValue];

回答by DNA App Lab

Thanks All! I am combined feedback and finally manage to convert from text input ( string ) to Integer. Plus it could tell me whether the input is integer :)

谢谢大家！我结合了反馈，最终设法从文本输入（字符串）转换为整数。另外它可以告诉我输入是否为整数:)

NSNumberFormatter * f = [[NSNumberFormatter alloc] init];
[f setNumberStyle:NSNumberFormatterDecimalStyle];
NSNumber * myNumber = [f numberFromString:thresholdInput.text];

int minThreshold = [myNumber intValue]; 

NSLog(@"Setting for minThreshold %i", minThreshold);

if ((int)minThreshold < 1 )
{
    NSLog(@"Not a number");
}
else
{
    NSLog(@"Setting for integer minThreshold %i", minThreshold);
}
[f release];

回答by JeanNicolas

I think NSDecimalNumber will do it:

我认为 NSDecimalNumber 会做到：

Example:

例子：

NSNumber *theNumber = [NSDecimalNumber decimalNumberWithString:[stringVariable text]]];

NSDecimalNumber is a subclass of NSNumber, so implicit casting allowed.

NSDecimalNumber 是 NSNumber 的子类，因此允许隐式转换。