blocks|key|183749|text|您是否正在尝试查找它是否至少包含(或)中的一个？您可以使用-rangeOfCharacterFromSet:|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|183750|NSCharacterSet+*cset+=+[NSCharacterSet+characterSetWithCharactersInString:@"()"];
NSRange+range+=+[mystr+rangeOfCharacterFromSet:cset];
if+(range.location+==+NSNotFound)+{
++++//+no+(+or+)+in+the+string
}+else+{
++++//+(+or+)+are+present
}|code-block|syntax|javascript|183751|entityMap^0|G|1|I|1|T|P|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Are you trying to find if it contains at least one of <code>(</code> or <code>)</code>? You can use <code>-rangeOfCharacterFromSet:</code>:

<pre><code>NSCharacterSet *cset = [NSCharacterSet characterSetWithCharactersInString:@"()"];
NSRange range = [mystr rangeOfCharacterFromSet:cset];
if (range.location == NSNotFound) {
 // no ( or ) in the string
} else {
 // ( or ) are present
}
</code></pre>

blocks|key|401872|text|如果给定的字符串包含给定的字符，下面的方法将返回Yes|type|unstyled|depth|inlineStyleRanges|entityRanges|data|401873|-(BOOL)doesString:(NSString+*)string+containCharacter:(char)character
{
++++return+[string+rangeOfString:[NSString+stringWithFormat:@"%25c",character]].location+!=+NSNotFound;
}|code-block|syntax|javascript|401874|您可以按如下方式使用它：|401875|+NSString+*s+=+@"abcdefg";

++++if+([self+doesString:s+containCharacter:'a'])
++++++++NSLog(@"'a'+found");
++++else
++++++++NSLog(@"No+'a'+found");


++++if+([self+doesString:s+containCharacter:'h'])
++++++++NSLog(@"'h'+found");
++++else
++++++++NSLog(@"No+'h'+found");|401876|输出：|401877|2013-01-11+11:15:03.830发现CharFinder17539:c07+'a‘|401878|2013-01-11+11:15:03.831+CharFinder17539:c07未找到'h‘|401879|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

The method below will return Yes if the given string contains the given char

<pre><code>-(BOOL)doesString:(NSString *)string containCharacter:(char)character
{
 return [string rangeOfString:[NSString stringWithFormat:@"%c",character]].location != NSNotFound;
}
</code></pre>

You can use it as follows:

<pre><code> NSString *s = @"abcdefg";

 if ([self doesString:s containCharacter:'a'])
 NSLog(@"'a' found");
 else
 NSLog(@"No 'a' found");


 if ([self doesString:s containCharacter:'h'])
 NSLog(@"'h' found");
 else
 NSLog(@"No 'h' found");
</code></pre>

Output:

2013-01-11 11:15:03.830 CharFinder[17539:c07] 'a' found

2013-01-11 11:15:03.831 CharFinder[17539:c07] No 'h' found

blocks|key|401897|text|-+(bool)+contains:+(NSString*)+substring+{
++++NSRange+range+=+[self+rangeOfString:substring];
++++return+range.location+!=+NSNotFound;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|401898|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>- (bool) contains: (NSString*) substring {
 NSRange range = [self rangeOfString:substring];
 return range.location != NSNotFound;
}
</code></pre>

blocks|key|183897|text|对于我在代码中使用的问题，我得到了一个通用的答案。此代码包括以下规则:+1.无特殊字符2.至少一个大写字母和一个小英文字母表3.至少一个数字|type|unstyled|depth|inlineStyleRanges|entityRanges|data|183898|+BOOL+lowerCaseLetter,upperCaseLetter,digit,specialCharacter;
++++int+asciiValue;
++++if([txtPassword.text+length]+>=+5)
++++{
++++++++for+(int+i+=+0;+i+<+[txtPassword.text+length];+i%2B%2B)
++++++++{
++++++++++++unichar+c+=+[txtPassword.text+characterAtIndex:i];
++++++++++++if(!lowerCaseLetter)
++++++++++++{
++++++++++++++++lowerCaseLetter+=+[[NSCharacterSet+lowercaseLetterCharacterSet]+characterIsMember:c];
++++++++++++}
++++++++++++if(!upperCaseLetter)
++++++++++++{
++++++++++++++++upperCaseLetter+=+[[NSCharacterSet+uppercaseLetterCharacterSet]+characterIsMember:c];
++++++++++++}
++++++++++++if(!digit)
++++++++++++{
++++++++++++++++digit+=+[[NSCharacterSet+decimalDigitCharacterSet]+characterIsMember:c];
++++++++++++}

++++++++++++asciiValue+=+[txtPassword.text+characterAtIndex:i];
++++++++++++NSLog(@"ascii+value---%25d",asciiValue);
++++++++++++if((asciiValue+>=33&&asciiValue+<+47)%7C%7C(asciiValue>=58+&&+asciiValue<=64)%7C%7C(asciiValue>=91+&&+asciiValue<=96)%7C%7C(asciiValue>=91+&&+asciiValue<=96))
++++++++++++{
++++++++++++++++specialCharacter=1;
++++++++++++}
++++++++++++else
++++++++++++{
++++++++++++++++specialCharacter=0;
++++++++++++}

++++++++}

++++++++if(specialCharacter==0+&&+digit+&&+lowerCaseLetter+&&+upperCaseLetter)
++++++++{
++++++++++++//do+what+u+want
++++++++++++NSLog(@"Valid+Password+%25d",specialCharacter);
++++++++}
++++++++else
++++++++{
++++++++++++NSLog(@"Invalid+Password+%25d",specialCharacter);
++++++++++++UIAlertView+*alert+=+[[UIAlertView+alloc]+initWithTitle:@"Error"
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++message:@"Please+Ensure+that+you+have+at+least+one+lower+case+letter,+one+upper+case+letter,+one+digit+and+No+Any+special+character"
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++delegate:nil+cancelButtonTitle:@"OK"+otherButtonTitles:nil];
++++++++++++[alert+show];
++++++++}|code-block|syntax|javascript|183899|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 got a generalized answer to your question which I was using in my code. This code includes the following rules:
1. No special characters 
2. At least one capital and one small English alphabet
3. At least one numeric digit

<pre><code> BOOL lowerCaseLetter,upperCaseLetter,digit,specialCharacter;
 int asciiValue;
 if([txtPassword.text length] &gt;= 5)
 {
 for (int i = 0; i &lt; [txtPassword.text length]; i++)
 {
 unichar c = [txtPassword.text characterAtIndex:i];
 if(!lowerCaseLetter)
 {
 lowerCaseLetter = [[NSCharacterSet lowercaseLetterCharacterSet] characterIsMember:c];
 }
 if(!upperCaseLetter)
 {
 upperCaseLetter = [[NSCharacterSet uppercaseLetterCharacterSet] characterIsMember:c];
 }
 if(!digit)
 {
 digit = [[NSCharacterSet decimalDigitCharacterSet] characterIsMember:c];
 }

 asciiValue = [txtPassword.text characterAtIndex:i];
 NSLog(@"ascii value---%d",asciiValue);
 if((asciiValue &gt;=33&amp;&amp;asciiValue &lt; 47)||(asciiValue&gt;=58 &amp;&amp; asciiValue&lt;=64)||(asciiValue&gt;=91 &amp;&amp; asciiValue&lt;=96)||(asciiValue&gt;=91 &amp;&amp; asciiValue&lt;=96))
 {
 specialCharacter=1;
 }
 else
 {
 specialCharacter=0;
 }

 }

 if(specialCharacter==0 &amp;&amp; digit &amp;&amp; lowerCaseLetter &amp;&amp; upperCaseLetter)
 {
 //do what u want
 NSLog(@"Valid Password %d",specialCharacter);
 }
 else
 {
 NSLog(@"Invalid Password %d",specialCharacter);
 UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error"
 message:@"Please Ensure that you have at least one lower case letter, one upper case letter, one digit and No Any special character"
 delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
 [alert show];
 }
</code></pre>

blocks|key|401982|text|为什么总是使用NSCharactorSet+?？这是一个简单而强大的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|401983|NSString+*textStr+=+@"This+is+String+Containing+/+Character";

if+([textStr+containsString:@"/"])
{
++++NSLog(@"Found!!");
}
else
{
+++NSLog(@"Not+Found!!");|code-block|syntax|javascript|401984|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|$]]

Why to always use for NSCharactorSet ?? this is simple and powerful solution

<pre><code>NSString *textStr = @"This is String Containing / Character";

if ([textStr containsString:@"/"])
{
 NSLog(@"Found!!");
}
else
{
 NSLog(@"Not Found!!");
</code></pre>

blocks|key|183704|text|您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|183705|NSString+rangeOfString:+(NSString+*)+string|code-block|syntax|javascript|183706|查看此处：https://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html%2523//apple_ref/occ/instm/NSString/rangeOfString|offset|length|183707|如果NSRange返回的属性'location‘等于NSNotFound，则字符串不包含传递的字符串(或字符)。|183708|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html%2523//apple_ref/occ/instm/NSString/rangeOfString^0|0|0|5|4V|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@$I|X|J|Y|1|Z]]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

You can use

<pre><code>NSString rangeOfString: (NSString *) string
</code></pre>

See here: <a href="https://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html%23//apple_ref/occ/instm/NSString/rangeOfString" rel="nofollow">https://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html%23//apple_ref/occ/instm/NSString/rangeOfString</a>:

If the NSRange comes back with property 'location' equal to NSNotFound, then the string does not contain the passed string (or character).

i have one NSString object , for ex:- ($45,0000)

Now i want to find if this string contains () or not

How can i do this?

How to detect if NSString contains a particular character or not?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个NSString对象，例如：- ($45,0000)现在我想看看这个字符串是否包含()我该怎么做呢？

问如何检测NSString中是否包含特定字符？
EN

回答 6

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问如何检测NSString中是否包含特定字符？
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