iOS 应用层

仅支持X位整数Y位小数 的处理

2018-12-04  本文已影响0人  介和

仅支持X位整数Y位小数 的处理:

-(void) handleInPut:(UITextField*)textfield  IntBitCount:(NSInteger) intBitCount dotBitCount:(NSInteger)dotBitCount toastMsg:(NSString*) msg{

    BOOLisHaveDot =YES;

    if([textfield.textrangeOfString:@"."].location==NSNotFound) {

        isHaveDot =NO;

    }

    if(isHaveDot) {//存在小数点

        NSRangerange = [textfield.textrangeOfString:@"."];

        if(textfield.text.length- range.location> (dotBitCount +1)) {

            [WAGToastUtil showToastInRootViewWithMsg:msg];

            NSUIntegerlen = textfield.text.length;

            textfield.text= [textfield.textsubstringToIndex:len -1];

        }

        if(range.location> intBitCount) {

            [WAGToastUtil showToastInRootViewWithMsg:msg];

            textfield.text= [textfield.textsubstringFromIndex:range.location- intBitCount];

        }

    }else{

        NSUIntegerlen = textfield.text.length;

        if(len > intBitCount) {

            [WAGToastUtil showToastInRootViewWithMsg:msg];

            textfield.text= [textfield.textsubstringToIndex:intBitCount];

        }

    }

}

调用:

        [self handleInPut:textfield IntBitCount:6 dotBitCount:2 toastMsg:@"仅支持6位整数2位小数"];

判断是空字符串

- (BOOL)isBlankString:(NSString*)str {

    NSString *noBlankStr = [str stringByReplacingOccurrencesOfString:@" " withString:@""];

    noBlankStr = [noBlankStrstringByReplacingOccurrencesOfString:@"\n" withString:@""];

    noBlankStr = [noBlankStrstringByReplacingOccurrencesOfString:@"\r" withString:@""];

    return(noBlankStr.length==0);

}

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