检查一个NSString是否只是由空格构成

Question

我想检查一个特定的字符串是否只是由空格组成。它可以是任意数量的空格，包括零。什么是最好的方式来确定？

Answer 1

NSString *str = @"   "; 
NSCharacterSet *set = [NSCharacterSet whitespaceCharacterSet]; 
if ([[str stringByTrimmingCharactersInSet: set] length] == 0) 
{ 
    // String contains only whitespace. 
} 

Answer 2

修整空间并检查剩余字符的数量。看看这个帖子here

Answer 3

尝试剥离它的空间，并将其与@ “”：

NSString *probablyEmpty = [myString stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]; 
BOOL wereOnlySpaces = [probablyEmpty isEqualToString:@""]; 

Answer 4

试试这个：

[mystring stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]; 

或

[mystring stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceAndNewlineCharacterSet]]; 

Answer 5

这里是基于@Alexander Akers的答案的NSString上的一个易于重用的类别，但这也是返回YES如果字符串包含 “新线” ......

@interface NSString (WhiteSpaceDetect) 
@property (readonly) BOOL isOnlyWhitespace; 
@end 
@implementation NSString (WhiteSpaceDetect) 
- (BOOL) isOnlyWhitespace { 
    return ![self stringByTrimmingCharactersInSet: 
      [NSCharacterSet whitespaceAndNewlineCharacterSet]].length; 
} 
@end 

，并为那些你不信任的灵魂在那里的..

#define WHITE_TEST(x) for (id z in x) printf("\"%s\" : %s\n",[z UTF8String], [z isOnlyWhitespace] ? "YES" :"NO") 

WHITE_TEST(({ @[ 

    @"Applebottom", 
    @"jeans\n", 
    @"\n", 
    @"" 
    "", 
    @" \ 
    \ 
    ", 
    @" " 
];})); 

➜

"Applebottom" : NO 
"jeans 
" : NO 
" 
" : YES 
"" : YES 
" " : YES 
" " : YES 

Answer 6

下面是一个简单的迅捷解决方案：

//Check if string contains only empty spaces and new line characters 
static func isStringEmpty(#text: String) -> Bool { 
    let characterSet = NSCharacterSet.whitespaceAndNewlineCharacterSet() 
    let newText = text.stringByTrimmingCharactersInSet(characterSet) 
    return newText.isEmpty 
} 

Answer 7

下面是相同的雨燕版本代码，

var str = "Hello World" 
if count(str.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())) == 0 { 
    // String is empty or contains only white spaces 
} 
else { 
    // String is not empty and contains other characters 
} 

或者你可以写一个简单的字符串扩展如下图所示，并使用相同的代码与在多个位置更好的可读性。

extension String { 
    func isEmptyOrContainsOnlySpaces() -> Bool { 
     return count(self.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())) == 0 
    } 
} 

，只是把它用这样的任意字符串，

var str1 = " " 
if str.isEmptyOrContainsOnlySpaces() { 
    // custom code e.g Show an alert 
} 

Answer 8

这是显著快检查的非空格字符，而不是微调整个字符串的范围。

NSCharacterSet *inverted = [[NSCharacterSet whitespaceAndNewlineCharacterSet] invertedSet]; 
NSRange range = [string rangeOfCharacterFromSet:inverted]; 
BOOL empty = (range.location == NSNotFound); 

请注意，“填充”可能是混合了空格和文本的最常见情况。

testSpeedOfSearchFilled - 0.012 sec 
testSpeedOfTrimFilled - 0.475 sec 
testSpeedOfSearchEmpty - 1.794 sec 
testSpeedOfTrimEmpty - 3.032 sec 

测试在我的iPhone 6+上运行。 代码here。粘贴到任何XCTestCase子类中。

Answer 9

不得不使用雨燕3验证码：

func isEmptyOrContainsOnlySpaces() -> Bool { 

    return self.trimmingCharacters(in: .whitespaces).characters.count == 0 
} 

Answer 10

我真的很惊讶，我是第一个谁提出正则表达式

NSString *string = @"   "; 
NSString *pattern = @"^\\s*$"; 
NSRegularExpression *expression = [[NSRegularExpression alloc] initWithPattern:pattern options:0 error:nil]; 
NSArray *matches = [expression matchesInString:string options:0 range:NSMakeRange(0, string.length)]; 
BOOL isOnlyWhitespace = matches.count; 

还是在斯威夫特：

let string = "   " 
let pattern = "^\\s*$" 
let expression = try! NSRegularExpression(pattern:pattern, options:[]) 
let matches = expression.matches(in: string, options: [], range:NSRange(location: 0, length: string.utf16.count)) 
let isOnlyWhitespace = !matches.isEmpty