Question

我在视图控制器上有一个按钮，我想点击按钮，然后进入一个网站。该网站在parse.com上举行。

代码如下

- (IBAction)WebAddressBtn:(id)sender {
    NSString *url = [self.exam objectForKey:@"Website"];
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:url]];
    NSLog(@"website: %@",url);
}

NSLog显示值为url

的Null

但数据保存在

self.exam objectForKey:@"Website"

NSLog确认其存在

这可行并将转到谷歌

- (IBAction)WebAddressBtn:(id)sender {


    NSString *url = @"http://www.google.com";
    //[self.exam objectForKey:@"Website"];
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:url]];
    NSLog(@"website: %@",url);


}

如果我尝试

NSURL *url = [self.exam objectForKey:@"Website"];

url仍显示为Null 但我知道数据在self.exam objectForKey：@＆＃34;网站

数据的NSLog输出

    name = "DMK Media & Photography Ltd";
    phone1 = 01993835148;
    phone2 = 07795966848;
    postcode = "OX28 4BT";
    products = "<PFRelation: 0x10dc75320>(<00000000 00000000>.(null) -> products)";
    website = "http://www.dmkmedia.co.uk";

Answer 1

[UIApplication sharedApplication]如果没有格式正确的url，openURL将无法正常工作。请检查您的网址是否为＆＃34; http：//＆＃34;

IOS IBAction按钮不使用URL访问网站

1 个答案: