iOS 字符串处理 为大于4位的字符打码 (就是emoji害的)

意外发现微信获取的用户名 带有emoji 然后与后台交互就出问题了

emoji的问题就是字符编码超过了4位 导致后台数据库不支持 后台试了下来说出问题的地方太多不能改 就只好前端过滤emoji了

因为是四位编码出问题所以 就写了如下 把四位编码的字符改为""的方法

+ (NSString *)filterMoreThanFourString:(NSString *)string {

string = [string stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

NSString *newStr = @"";

for(int i =0; i < [string length]; ){

NSString *temp = nil;

for (int j = 1; j <= string.length - i; j++) {

temp = [string substringWithRange:NSMakeRange(i,j)];

temp = [temp stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

if (temp) {

i += j;

//                NSLog(@"占%d位的字符:%@",j,temp);

break;

}

}

//        NSLog(@"%d位最后的字是:%@",i,temp);

if ([temp lengthOfBytesUsingEncoding:NSUTF8StringEncoding] > 3) {

temp = @"";

}

newStr = [NSString stringWithFormat:@"%@%@",newStr,temp];

}

return newStr;

}

稍微改动成 多位字符替换的成指定字符的方法:

#pragma mark  过滤max位以上字符(不包括

+ (NSString *)filterMoreThanNumber:(NSInteger)maxNum String:(NSString *)string replaceString:(NSString *)replaceString UsingEncoding:(NSStringEncoding)usingEncoding{

string = [string stringByAddingPercentEscapesUsingEncoding:usingEncoding];

NSString *newStr = @"";

for(int i =0; i < [string length]; ){

NSString *temp = nil;

for (int j = 1; j <= string.length - i; j++) {

temp = [string substringWithRange:NSMakeRange(i,j)];

temp = [temp stringByReplacingPercentEscapesUsingEncoding:usingEncoding];

//只验证了NSUTF8StringEncoding 2017.04.01之前的编码格式下 上一句如果获取的字符不正常temp == nil

if (temp) {

i += j;

 NSLog(@"占%d位的字符:%@",j,temp);

break;

}

}

NSLog(@"%d位最后的字是:%@",i,temp);

if ([temp lengthOfBytesUsingEncoding:usingEncoding] > maxNum) {

temp = replaceString;

}

newStr = [NSString stringWithFormat:@"%@%@",newStr,temp];

}

return newStr;

}