blocks|key|935566|text|使用如下的Reguler表达式，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|935567|+NSString+*str+=@"000034234247236000049327428900000";
++++NSRange+range+=+[str+rangeOfString:@"%5E0*"+options:NSRegularExpressionSearch];
++++str=+[str+stringByReplacingCharactersInRange:range+withString:@""];
++++NSLog(@"str+%25@",str);|code-block|syntax|javascript|935568|O/P:-str+34234247236000049327428900000|offset|length|style|CODE|935569|entityMap^0|0|0|5|X|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use Reguler Expression like this,

<pre><code> NSString *str =@"000034234247236000049327428900000";
 NSRange range = [str rangeOfString:@"^0*" options:NSRegularExpressionSearch];
 str= [str stringByReplacingCharactersInRange:range withString:@""];
 NSLog(@"str %@",str);
</code></pre>

O/P:-<code>str 34234247236000049327428900000</code>

blocks|key|935332|text|这正是NSNumberFormatter的目标(而且它在这方面做得更好)。为什么要重新发明轮子呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|935333|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.apple.com/mac/library/documentation/Cocoa/Reference/Foundation/Classes/NSNumberFormatter_Class/Reference/Reference.html^0|3|H|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This is exactly the kind of thing <a href="http://developer.apple.com/mac/library/documentation/Cocoa/Reference/Foundation/Classes/NSNumberFormatter_Class/Reference/Reference.html" rel="noreferrer">NSNumberFormatter</a> is made for (and it's way better at it). Why reinvent the wheel?

blocks|key|1149742|text|您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1149743|NSString+*testStr+=+@"001234056";
testStr+=+[NSString+stringWithFormat:@"%25d",[testStr+intValue];|code-block|syntax|javascript|1149744|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|$]]

You can use this: 

<pre><code>NSString *testStr = @"001234056";
testStr = [NSString stringWithFormat:@"%d",[testStr intValue];
</code></pre>

blocks|key|935267|text|我在这里假设您只想删除前导零。即@"*00*1234**56“变成@"1234**56"，而不是@"123457”。要做到这一点，我会使用NSScanner。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|935268|//+String+to+strip
NSString+*test+=+@"001234056";

//+Skip+leading+zeros
NSScanner+*scanner+=+[NSScanner+scannerWithString:test];
NSCharacterSet+*zeros+=+[NSCharacterSet
++++++++++++++++++++++++++++characterSetWithCharactersInString:@"0"];
[scanner+scanCharactersFromSet:zeros+intoString:NULL];

//+Get+the+rest+of+the+string+and+log+it
NSString+*result+=+[test+substringFromIndex:[scanner+scanLocation]];
NSLog(@"%25@+reduced+to+%25@",+test,+result);|code-block|syntax|javascript|935269|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|$]]

I'm assuming here that you only want to remove the leading zeros. I.E. @"*00*1234*0*56" becomes @"1234*0*56", not @"123457". To do that, I'd use an NSScanner.

<pre><code>// String to strip
NSString *test = @"001234056";

// Skip leading zeros
NSScanner *scanner = [NSScanner scannerWithString:test];
NSCharacterSet *zeros = [NSCharacterSet
 characterSetWithCharactersInString:@"0"];
[scanner scanCharactersFromSet:zeros intoString:NULL];

// Get the rest of the string and log it
NSString *result = [test substringFromIndex:[scanner scanLocation]];
NSLog(@"%@ reduced to %@", test, result);
</code></pre>

blocks|key|935618|text|不使用正则表达式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|935619|func+trimLeadingZeroes(input:+String)+->+String+{
++++var+result+=+""
++++for+character+in+input.characters+{
++++++++if+result.isEmpty+&&+character+==+"0"+{+continue+}
++++++++result.append(character)
++++}
++++return+result
}|code-block|syntax|javascript|935620|使用正则表达式|935621|func+trimLeadingZeroes(input:+String)+->+String+{
++++return+input.stringByReplacingOccurrencesOfString(
++++++++"%5E0%2B(?!$)",
++++++++withString:+"",
++++++++options:+.RegularExpressionSearch,
++++++++range:+Range(start:+input.startIndex,+end:+input.endIndex)
++++)
}|935622|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|$]]

Without regex

<pre><code>func trimLeadingZeroes(input: String) -&gt; String {
 var result = ""
 for character in input.characters {
 if result.isEmpty &amp;&amp; character == "0" { continue }
 result.append(character)
 }
 return result
}
</code></pre>

With regex

<pre><code>func trimLeadingZeroes(input: String) -&gt; String {
 return input.stringByReplacingOccurrencesOfString(
 "^0+(?!$)",
 withString: "",
 options: .RegularExpressionSearch,
 range: Range(start: input.startIndex, end: input.endIndex)
 )
}
</code></pre>

blocks|key|935424|text|更好的方法是编写一个随时都可以使用的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|935425|-(NSString+*)trimZero:(NSString*)inputString+{

NSScanner+*scanner+=+[NSScanner+scannerWithString:inputString];
NSCharacterSet+*zeros+=+[NSCharacterSet
+++++++++++++++++++++++++characterSetWithCharactersInString:@"0"];
[scanner+scanCharactersFromSet:zeros+intoString:NULL];

return+[inputString+substringFromIndex:[scanner+scanLocation]];
}|code-block|syntax|javascript|935426|用法：|935427|NSString+*newString+=+[self+trimZero:yourString];|935428|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|$]]

Better way is to write a function that can be used anytime.

<pre><code>-(NSString *)trimZero:(NSString*)inputString {

NSScanner *scanner = [NSScanner scannerWithString:inputString];
NSCharacterSet *zeros = [NSCharacterSet
 characterSetWithCharactersInString:@"0"];
[scanner scanCharactersFromSet:zeros intoString:NULL];

return [inputString substringFromIndex:[scanner scanLocation]];
}
</code></pre>

Usage :

<pre><code>NSString *newString = [self trimZero:yourString];
</code></pre>

blocks|key|1149887|text|这是一个基于C-string的实现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1149888|char+*cstr+=+[string+UTF8String];
while+(*cstr+==+'0')
++++%2B%2Bcstr;
return+[NSString+stringWithUTF8String:cstr];|code-block|syntax|javascript|1149889|当然，这也不会跳过前导空格，而NSScanner+(默认情况下)会跳过。|1149890|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|$]]

Here's a C-string-based implementation.

<pre><code>char *cstr = [string UTF8String];
while (*cstr == '0')
 ++cstr;
return [NSString stringWithUTF8String:cstr];
</code></pre>

Of course, this won't also skip leading whitespace, whereas NSScanner (by default) will.

I have a string <code>0023525631</code> but what I want is the string <code>23525631</code> -- without the leading zeroes. How do I do this?

Removing leading zeroes from a string

我有一个字符串0023525631，但我想要的是字符串23525631 --没有前导零。我该怎么做呢？

问从字符串中删除前导零
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