Question

有人可以告诉我我需要做什么来解决这个问题

我有一个对象UserPrefs * userpref
有一个名为userpref.userid的变量，它是一个NSNumber
我正在解析XML，这里的一切都很好，直到我将“myxmluseridnumber”元素分配给userpref.userid

这是错误消息。

将'NSMutableString * _strong'传递给'NSNumber

    - (void) parser:(NSXMLParser *)parser didStartElement:(NSString *)elementName namespaceURI:(NSString *)namespaceURI qualifiedName:(NSString *)qName attributes:(NSDictionary *)attributeDict {

        if ([elementName isEqualToString:@"User"]) {
            inItemElement = YES;
        }

        if (inItemElement && [elementName isEqualToString:@"myxmluseridnumber"]) {
            capturedCharacters = [[NSMutableString alloc] initWithCapacity:100];
        }
    }

    - (void)parser:(NSXMLParser *)parser foundCharacters:(NSString *)string {
       if(capturedCharacters != nil) {
           [capturedCharacters appendString:string];
       }
    }

    - (void)parser:(NSXMLParser *)parser didEndElement:(NSString *)elementName namespaceURI:(NSString *)namespaceURI qualifiedName:(NSString *)qName {

        if (inItemElement && [elementName isEqualToString:@"myxmluseridnumber"]) {
            NSLog(@"%@ - myxmlnumber", capturedCharacters);
            //This is what is producing my error
            userpref.userid = capturedCharacters;
            capturedCharacters = nil;
        }

        if ([elementName isEqualToString:@"User"]) {
            inItemElement = NO;
        }

    }

由于

Answer 1

由于用户ID是NSNumber，您应该从字符串NSNumber

创建capturedCharacters

userpref.userid = [NSNumber numberWithInt：[capturedCharacters 的intValue]];

Answer 2

用以下代码替换错误产生线：

userpref.userid = [capturedCharacters intValue];

（假设您的capturedCharacters是s NSString）

XMLParsing - 将字符串转换为NSNumber

2 个答案: