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關(guān)于以下這個(gè)問(wèn)題,ios-IOS SWIFT 判斷字符串是否含有字母?

關(guān)于以下這個(gè)問(wèn)題,ios-IOS SWIFT 判斷字符串是否含有字母?

函數(shù)式編程 2019-11-19 19:14:44
ios-IOS SWIFT 判斷字符串是否含有字母
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3 回答

?
守著星空守著你

TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超8個(gè)贊

rangeOfString
如果不夠用的話,可以查詢一下range開(kāi)頭的函數(shù)
NSString* str = @"1232343453453";
NSRange range = [str rangeOfString:@"a"];
if (range.length > 0)
{
}

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人到中年有點(diǎn)甜

TA貢獻(xiàn)1895條經(jīng)驗(yàn) 獲得超7個(gè)贊

使用遍歷

用 OC 實(shí)現(xiàn)的擴(kuò)展:

@implementation NSString (Emoji)
+ (BOOL)stringContainsEmoji:(NSString *)string
{
__block BOOL returnValue = NO;
[string enumerateSubstringsInRange:NSMakeRange(0, [string length])
options:NSStringEnumerationByComposedCharacterSequences
usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
const unichar hs = [substring characterAtIndex:0];
if (0xd800 <= hs && hs <= 0xdbff) {
if (substring.length > 1) {
const unichar ls = [substring characterAtIndex:1];
const int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;
if (0x1d000 <= uc && uc <= 0x1f77f) {
returnValue = YES;
}
}
} else if (substring.length > 1) {
const unichar ls = [substring characterAtIndex:1];
if (ls == 0x20e3) {
returnValue = YES;
}
} else {
if (0x2100 <= hs && hs <= 0x27ff) {
returnValue = YES;
} else if (0x2B05 <= hs && hs <= 0x2b07) {
returnValue = YES;
} else if (0x2934 <= hs && hs <= 0x2935) {
returnValue = YES;
} else if (0x3297 <= hs && hs <= 0x3299) {
returnValue = YES;
} else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50) {
returnValue = YES;
}
}
}];
return returnValue;
}
其實(shí)它的原理還是去匹配 Emoji 所在區(qū)間,所以,如果我們應(yīng)用上 Swift 語(yǔ)句以及其特性,那么 Swift 版本就是這樣:

extension String {

var containsEmoji: Bool {
for scalar in unicodeScalars {
switch scalar.value {
case
0x00A0...0x00AF,
0x2030...0x204F,
0x2120...0x213F,
0x2190...0x21AF,
0x2310...0x329F,
0x1F000...0x1F9CF:
return true
default:
continue
}
}
return false
}

}



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