How can I differentiate between fingerprints with TOUCH ID [duplicate] - ios

This question already has an answer here:
Know which user logging into app depending on touchid in iOS
(1 answer)
Closed 6 years ago.
Hello I would like to know how I can differentiate between two fingerprints to relate them to a user and password, because if I can not know who owns that footprint I could not relate them to a user within my app.
Thank you

I'm afraid that's not currently possible. The result is authenticated or not, not authenticated by X.

Related

Detect current number calling - iPhone [duplicate]

This question already has answers here:
How to get phone number of incoming call in iOS Swift? [closed]
(2 answers)
Closed 5 years ago.
I am currently working on an app which should show informations about the current person calling when it is opened during an ongoing phone call.
So, I need to get the phone number of the person that is currently calling me when launching the app. I have read that it is not possible to do that. Is that right, because that would suck.
Thanks in advance for your help.
Because of the app sandbox, it's not possible to track the current caller with a third-party application, as it could be a security issue.

Can i get Unique value of each fingerprint while using TouchID in ios? [duplicate]

This question already has an answer here:
Is it possible to get TouchID information and compare to a fingerprint database?
(1 answer)
Closed 5 years ago.
I wanted to get unique value of each fingerprint while using TouchID. I'm doing that in swift3.1. Can anyone help me for this?
You can't get any unique information about fingerprint while using touchID. Apple doesn't expose that details as of now.

Get call history in iOS [duplicate]

This question already has answers here:
iPhone call log / history
(3 answers)
Closed 7 years ago.
I want details of dial calls,received calls and missed calls in a form of Array or List.
So, That I can display that records in UITableView of my app.
Please suggest me code for button touch up event instead of suggesting any links.
Not possible.
Luckily that is not possible. That kind of data and information is far too private to be accessible for everyone.
Something like that can only be achieved via private APIs or on jailbroken devices - both will not get you into the app store.
There is no API to access such private information

Authentication using ios keychain - touch id (is this possible) [duplicate]

This question already has an answer here:
iPhone 5s Touch ID sensor API [closed]
(1 answer)
Closed 8 years ago.
Instead of using the user object and password from ios keychain to authenticate in an app do developers have access to the user object's touchId. I do not want to store the fingerprint (and I know that apple will not allow this) I just want to authenticate using the fingerprint (touchid) that is encrypted on the hardware. I can't seem to find any subclasses or methods in the documentation that would allow me to perform the authentication using touchId rather than user and pass. Any insight would be great.
Thanks!
No, (at the time of writing) developers have no access to the touch sensor or any information from it.
It is possible only if you jailbreak.

Can I get user's phone number by requesting for permission in ios 6? [duplicate]

This question already has answers here:
Programmatically get own phone number in iOS
(9 answers)
Closed 9 years ago.
I am building an IOS6 app which requires the user's phone number,
Is there a way to ask the user for permission and get the number in programmatic way?
Or he has to manually type it in?
I am using Xcode 4.5 if it matters..
You can't get the phone via any easy API within iOS. There might be a way to do it via the technique described in this related question, but this is a few years old and Apple may have closed this hole (which uses an undocumented key). I also wonder if Apple wouldn't allow the app on the App Store for privacy reasons. Probably not.
You'll have to trust the user to type in the correct phone number if you prompt him/her to.

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