Question

我从我的网站获取数据：

NSString *website = [NSString stringWithFormat:@"http://www.mysite.com/fbconnect.php?email=%@&name=%@&pass=***", nameTrimmmed, [jsonObject objectForKey:@"email"]];
NSLog(@"%@", website);
NSError *error = nil;
NSString *contents = [[NSString alloc] initWithContentsOfURL:[NSURL URLWithString:website] encoding:NSUTF8StringEncoding error:&error];

内容有Cocoa错误256.我哪里错了？

Answer 1

问题在于希伯来字符，你应该html-escape他们，也可以尝试用英文字符请求，看它是否有效

- (void)yourMethod
{
    NSString *name = @"שימרגוליס";
    name = AFURLEncodedStringFromStringWithEncoding(name, NSUTF8StringEncoding);
    NSString *website = [NSString stringWithFormat:@"http://www.ba-cafe.com/fbconnect.php?email=%@&name=%@&pass=SwHyK17!",@"email@mail.com",name];
    NSLog(@"%@", website);
    NSError *error = nil;
    NSString *contents = [[NSString alloc] initWithContentsOfURL:[NSURL URLWithString:website] encoding:NSUTF8StringEncoding error:&error];
}

其中AFURLEncodedStringFromStringWithEncoding是来自AFNetworking framework

的函数

Answer 2

检查控制台日志中的NSLog(@"%@", website);

你会看到这样的事情：

http://www.mysite.com/fbconnect.php?email=thetrimmedname&name=emailaddress&pass=***

这样做：

NSString *website = [NSString stringWithFormat:@"http://www.mysite.com/fbconnect.php?email=%@&name=%@&pass=***", [jsonObject objectForKey:@"email"], nameTrimmmed ];

而不是：

NSString *website = [NSString stringWithFormat:@"http://www.mysite.com/fbconnect.php?email=%@&name=%@&pass=***", nameTrimmmed, [jsonObject objectForKey:@"email"]];

Answer 3

这是因为电子邮件地址中的点。

看看这里： Error while trying to access Google translate with NSURL

Cocoa错误256，使用initWithContentsOfURL时：

3 个答案: