我试图将一个变量附加到一个字符串,但它显示一个错误。我在这里错过了一些非常容易的事情,但我的头脑被击中了。
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSURL *urla = [NSURL URLWithString:@"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=",eID];
我试图将一个变量附加到一个字符串,但它显示一个错误。我在这里错过了一些非常容易的事情,但我的头脑被击中了。
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSURL *urla = [NSURL URLWithString:@"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=",eID];
您不能仅通过在它们之间放置逗号来连接 2 个字符串变量。试试这个:
NSURL *urla = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=%@",eID]];
如果您所做的只是追加,您有几个选择:
A. 使用 NSString 和 stringWithFormat:
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSString *urlString = [NSString stringWithFormat:@"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=%@", eID];
NSURL *urla = [NSURL URLWithString:urlString];
B. 使用 NSString & stringByAppendingString:
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSString *baseUrl = @"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=";
NSString *urlString = [baseUrl stringByAppendingString:eID];
NSURL *urla = [NSURL URLWithString:urlString];
C. 使用 NSMutableString & appendString:
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSString *baseUrl = @"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=";
NSMutableString *urlString = [NSMutableString stringWithString:baseUrl];
[urlString appendString:eID];
NSURL *urla = [NSURL URLWithString:urlString];
试试这样: -
NSString *eID = [entertainmentArticle objectForKey:@"eID"];
NSString *url=@"http://www.mydomain.com/iostest/appPHPs/schedule.php?eID=";
NSURL *urla = [NSURL URLWithString:[url stringByAppendingString:eID];