Create NSURL using AssetPath

Question

Using UIImagePickerController, I'm able to get the Asset Path for a Media like assets-library://asset/asset.MOV?id=2A2CE6C9-C178-4395-977B-E6F159BF6D5E&ext=MOV.

NSURL *instagramURL = [NSURL URLWithString:[NSString stringWithFormat:@"instagram://library?AssetPath=%@&InstagramCaption=%@",[referenceURL absoluteString], strInstagramCaption]];

Now, I want to convert this Asset Path into NSURL to pass it into following code block :

if ([[UIApplication sharedApplication] canOpenURL:instagramURL])
{
    [[UIApplication sharedApplication] openURL:instagramURL];
}

Can I ask you a question. When you do this, do you find that the Instagram app sometimes goes to the image/video which was added before? So in other words it opens up the second to latest one for some reason? — , Jun 01 '15 at 09:21
Well I have been testing it and it opens the Photo/Video uploader. — , Jun 01 '15 at 09:26
I am basically using the code from this post: http://stackoverflow.com/a/27702493/4657588 — , Jun 01 '15 at 09:29
I want `assets-library://asset/asset.JPG?id=B202E621-FCCB-48BA-817C-8DF9605ED799&ext=JPG` path from `/var/mobile/Containers/Data/Application/ABADEDFA-FDEE-47BA-A7F5-575AAA40F937/Documents/InstaImage.png` path. Please help — Vaibhav Jhaveri, Jun 02 '15 at 05:29

score 1 · Accepted Answer · edited May 23 '17 at 12:33

1

This is the way how you should convert asset path to valid NSUrl:-

NSURL *instagramURL = [NSURL URLWithString:[[NSString stringWithFormat:@"instagram://library?AssetPath=%@&InstagramCaption=%@",[assetURL absoluteString],caption]stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
        [[UIApplication sharedApplication] openURL:instagramURL];
    }

Referred from this

edited May 23 '17 at 12:33

Community

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answered Jun 01 '15 at 06:20

Vizllx

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`.percentEscape ` is custom property ? – Vaibhav Jhaveri Jun 01 '15 at 06:22
Yes... it is. Updated the answer. – Vizllx Jun 01 '15 at 06:25
1

Can you please share the whole code to share to share a Video to Instagram? – Vaibhav Jhaveri Jun 01 '15 at 12:30
No, that's against the rule of SF. Your objective was to Create NSURL using AssetPath which is already answered here.Try yourself, then if still you find problem then share your code here or raise another question , Will surely help you then. – Vizllx Jun 01 '15 at 12:34

score 0 · Answer 2 · answered Jun 02 '15 at 07:49

I presume you would have used UIImagePickerController to select a photo from the library or to bring up the camera? Either way you must be making use of the following delegate method:

-(void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info {

}

Well in that method, you can get the URL of the asset and then pass that to AVAsset framework for the asset URL and then put that in the Instagram URL.

(NSURL *)[info valueForKey:UIImagePickerControllerReferenceURL];

Use the above code to obtain a URL for the image in the imagePickerController delegate method. You can then pass that to AVAsset.

Update This is also a useful link to look at, its explains how to use UIImagePickerController - http://www.appcoda.com/ios-programming-camera-iphone-app/.

Hope this helps :)

Create NSURL using AssetPath

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