关于以下这个问题，ios-IOS SWIFT 判断字符串是否含有字母？

Question

ios-IOS SWIFT 判断字符串是否含有字母

Answer 1

rangeOfString
如果不够用的话，可以查询一下range开头的函数
NSString* str = @"1232343453453";
NSRange range = [str rangeOfString:@"a"];
if (range.length > 0)
{
}

Answer 2

使用遍历

用 OC 实现的扩展：

@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;
}
其实它的原理还是去匹配 Emoji 所在区间，所以，如果我们应用上 Swift 语句以及其特性，那么 Swift 版本就是这样：

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
}

}