ios 阻止用户输入表情

最新推荐文章于 2023-05-10 17:52:42 发布

原创最新推荐文章于 2023-05-10 17:52:42 发布 · 2.8k 阅读

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CC 4.0 BY-SA版权

本文介绍了一个iOS应用程序中用于禁止用户在UITextView中输入表情符号的方法。通过检测文本输入模式的语言来判断是否为表情输入，并阻止其显示。此外，还提供了一个字符串检查函数，能够扫描并确定一个字符串中是否包含表情符号。

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- (BOOL)textView:(UITextView *)textView shouldChangeTextInRange:(NSRange)range replacementText:(NSString *)text
 2 {
 3     // 不让输入表情
 4     if ([textView isFirstResponder]) {
 5         if ([[[textView textInputMode] primaryLanguage] isEqualToString:@"emoji"] || ![[textView textInputMode] primaryLanguage]) {
 6             return NO;
 7         }
 8     }
 9     
10     return YES;
11 }
12 
13 // 过滤所有表情  https://gist.github.com/cihancimen/4146056
14 - (BOOL)stringContainsEmoji:(NSString *)string {
15     __block BOOL returnValue = NO;
16     [string enumerateSubstringsInRange:NSMakeRange(0, [string length]) options:NSStringEnumerationByComposedCharacterSequences usingBlock:
17      ^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
18          
19          const unichar hs = [substring characterAtIndex:0];
20          // surrogate pair
21          if (0xd800 <= hs && hs <= 0xdbff) {
22              if (substring.length > 1) {
23                  const unichar ls = [substring characterAtIndex:1];
24                  const int uc = ((hs - 0xd800) * 0x400) + (ls - 0xdc00) + 0x10000;
25                  if (0x1d000 <= uc && uc <= 0x1f77f) {
26                      returnValue = YES;
27                  }
28              }
29          } else if (substring.length > 1) {
30              const unichar ls = [substring characterAtIndex:1];
31              if (ls == 0x20e3) {
32                  returnValue = YES;
33              }
34              
35          } else {
36              // non surrogate
37              if (0x2100 <= hs && hs <= 0x27ff) {
38                  returnValue = YES;
39              } else if (0x2B05 <= hs && hs <= 0x2b07) {
40                  returnValue = YES;
41              } else if (0x2934 <= hs && hs <= 0x2935) {
42                  returnValue = YES;
43              } else if (0x3297 <= hs && hs <= 0x3299) {
44                  returnValue = YES;
45              } else if (hs == 0xa9 || hs == 0xae || hs == 0x303d || hs == 0x3030 || hs == 0x2b55 || hs == 0x2b1c || hs == 0x2b1b || hs == 0x2b50) {
46                  returnValue = YES;
47              }
48          }
49      }];
50     
51     return returnValue;
52 }