我已经构建了一个XML Parse应用程序,它从PHP(XML Content)文件中检索我的数据库中的一些值,其中一个值我有一个URL,问题是,每次我尝试加载webView使用此URL没有任何反应!我有以下代码:
-(void)parser:(NSXMLParser *)parser didEndElement:(NSString *)elementName namespaceURI:(NSString *)namespaceURI qualifiedName:(NSString *)qName {
if ( [elementName isEqualToString:@"users"] ) {
[messages addObject:[NSDictionary dictionaryWithObjectsAndKeys:msgIDEN,@"id",msgAdded,@"added",msgUsername,@"username",msgEmail,@"email",msgMainCode,@"maincode",msgPictureAddress,@"userpictureaddress",nil]];
[[messages reverseObjectEnumerator] allObjects];
lastId = msgId;
[msgAdded release];
[msgUsername release];
[msgEmail release];
[msgMainCode release];
[msgPictureAddress release];
NSLog(@"UserData:Success");
NSString *urlAddress = msgPictureAddress;
NSURL *url = [NSURL URLWithString:urlAddress];
NSURLRequest *requestObj = [NSURLRequest requestWithURL:url];
[webViewMateus loadRequest:requestObj];
}
if ( [elementName isEqualToString:@"id"] ) { inIDEN = NO; }
if ( [elementName isEqualToString:@"username"] ) { inUsername = NO; }
if ( [elementName isEqualToString:@"email"] ) { inEmail = NO; }
if ( [elementName isEqualToString:@"usermaincode"] ) { inMainCode = NO; }
if ( [elementName isEqualToString:@"userpictureaddress"] ) { inPictureAddress = NO; }
}
如何让UIWebView地址与我的NSMutableString = userpictureaddress
相同?
答案 0 :(得分:0)
如果查看代码,只需将msgPictureAddress丢弃(释放),然后再将其分配给urlAddress。你确定你当时想发布那些东西吗?