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[WM6.1]How to query a contact's call records? RRS feed

  • Question

  • Hello,
    How to query a contact's call records? Could you please give me some advice? sample code? thanks.

    Thursday, March 12, 2009 1:32 AM

Answers

  • HI Zeng_Zeng,

    In addition to Paul reply, since we can enumerate all the call log, we can wrap a method to filter a contact call log.  In the filter method, we enumerate all call log and check whether it is equal to the contact you want to filter, if it is, add it to an array. At last, the array is  the contact's all call log.


    Best regards,
    Guang-Ming Bian - MSFT
    Please remember to mark the replies as answers if they help and unmark them if they provide no help
    • Marked as answer by Zeng_Zeng Monday, March 16, 2009 5:46 AM
    Monday, March 16, 2009 5:32 AM

All replies

  • Hi,

    The following article might help you :-

    http://www.codeproject.com/KB/mobile/wm_callhistory.aspx

    Hope this helps.

    Paul Diston
    http://www.smartmobiledevice.co.uk/
    • Marked as answer by Zeng_Zeng Monday, March 16, 2009 5:46 AM
    • Unmarked as answer by Zeng_Zeng Monday, March 16, 2009 5:46 AM
    Thursday, March 12, 2009 7:24 AM
  • Thanks for your reply.

    It seems to be able to get all the call records, I just want to query a contact's call records, I heard that can query the EDB database, but i cannot implement it successfully. please feel free to help me, Thanks a lot.
    Thursday, March 12, 2009 8:30 AM
  • HI Zeng_Zeng,

    In addition to Paul reply, since we can enumerate all the call log, we can wrap a method to filter a contact call log.  In the filter method, we enumerate all call log and check whether it is equal to the contact you want to filter, if it is, add it to an array. At last, the array is  the contact's all call log.


    Best regards,
    Guang-Ming Bian - MSFT
    Please remember to mark the replies as answers if they help and unmark them if they provide no help
    • Marked as answer by Zeng_Zeng Monday, March 16, 2009 5:46 AM
    Monday, March 16, 2009 5:32 AM
  • Thanks in advance.

    Good idea. I'm also ready to do so.
    • Marked as answer by Zeng_Zeng Monday, March 16, 2009 5:45 AM
    • Unmarked as answer by Zeng_Zeng Monday, March 16, 2009 5:46 AM
    Monday, March 16, 2009 5:45 AM