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In my app I have a UITableView displaying tweets from a downloaded json file. Many of the tweets contain a link (to Instagram for example). How would I go about making these links open in Safari? What code would I need?

Patrick
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rexr
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2 Answers2

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[[UIApplication sharedApplication] openURL:myurl];

Or is your problem elsewhere?

If that doesn't solve the problem, some code would be helpful to show what your table view cells look like.

johnyu
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  • i havent attempted to actually make my links open in safari as i have no idea how, where would i imput this code? – rexr Feb 08 '13 at 11:30
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  1. First of all you will need to detect your links. Is the link part of the tweet string or is it a separate object in an array?

    NSString *yourLink = @""; //we will need more detail in your question for this.

  2. Once you have your link you need to store it as an NSURL like so:

    NSURL *url = [[NSURL alloc] initWithString:yourLink];

  3. And then you can open this in a browser using:

    [[UIApplication sharedApplication] openURL:url];

It would be useful to see what you are displaying in your table. Is it an NSArray from your JSON, or an NSDictionary?

Patrick
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  • Based on the above question, I think you should look this (http://stackoverflow.com/a/2620632/1359306) and maybe lecture 6 of http://thenewboston.org/list.php?cat=28 – Patrick Feb 08 '13 at 14:30
  • and i dont actually have a url to input as it comes from the tweets which are always changing?? – rexr Feb 08 '13 at 16:06
  • I think you need to expand your question. Post an example of the data coming in, how you store it and how you currently display it. I think you need to go through each tweet, detect if it has a link and store results in an NSMutableArray where each object contains another array of the tweet and the link (If there is no link it will store a null value). You will then display each object of the NSMutableArray in the table. – Patrick Feb 08 '13 at 16:15