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Problem with displaying the selected record in the form's field

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    Problem with displaying the selected record in the form's field

    Hi,

    I am trying to load the found record number in the Main Form (mf)'s field (File_No), but somehow it keeps giving me the same record number no matter what I put in... What am I doing wrong? I have used several ways to do this: activate, tabbed1, show, etc. but none have done the job. I appreciate your feedbacks. Below is the code (I left the mf.activate() alone for now):

    " dim global mf as p
    dim vfileno as c
    dim tbl as p
    dim ix as p
    dim qfind as n
    vfileno=ui_get_text("File Number","Enter the file
    number","","00000000")
    if vfileno=""
    end
    end if
    tbl=table.open("Claim")
    query.order="File_No"
    query.filter="File_No= "+quote(var->vfileno)
    query.options=""
    ix=tbl.query_create()
    tbl.Fetch_First()
    qfind=ix.records_get()
    if (qfind = 0) then
    mf.hide()
    ui_beep()
    ui_msg_box("Message","Sorry! There is no matching file
    number.")
    else
    mf.show()
    mf.activate()
    end if "

    #2
    Ali,

    Is there some particular reason you're not using Action Scripting? It can be a great way to get to learn Alpha Five.

    The table.open() statement opens a new instance of the table. Your script then runs a query against that instance. This is ok but is probably not what you want. Tell us about the context in which this script is running. Is it running from a button on a form? What's the form's name? What table or set of tables is supporting that form? (If set, give us the set structure).

    In this situation are you trying to "find" a particular record, or are you trying to find all records in the table that match your file_no ? i.e. is the file_no field value in each record unique?

    -- tom

    Comment


      #3
      Hi Tom,

      Thank you for your feedback.
      Answer to your question: It's a button on the form called Main Menu. The table that's supporting this form is Claim.dbf. I tried "tbl=table.current() but it never seemed to be able to find the right record... Let's say, when I put in a record # that does not match the records in File_No field, it gives me the message:"Sorry..." just like it's supposed to, but when there is a match it open the right Main Form (as it is supposed to) but not the right record.
      I am trying to find a particular record that match the File_No. The File_No field value in each record is unique. BTW. I also tried with 'find' command but if there is no match, it finds the "closest match" as it implies.

      Comment


        #4
        Sorry for some confusion:
        "I tried "tbl=table.current() but it never seemed to be able to find the right record": I meant it didn't open the table I had in mind, which is "Claim.dbf".
        "Let's say, when I put in a record # that does not match the records in File_No field, it gives me the message:"Sorry..." just like it's supposed to, but when there is a match it open the right Main Form (as it is supposed to) but not the right record.": This is when I open the database and click on the button in the Main Menu, if I select to find an existing file by File number.
        "BTW. I also tried with 'find' command but if there is no match, it finds the "closest match" as it implies.": I meant using the line "qfind=mf.find(vfileno) instead of "qfind=ix.records_get()"... the errors seem to be inconsistent... earlier when I tried "qfind=mf.find(vfileno), it was working( but not what I wanted)... after I did some debugging and ran the database, it gave me errors like "variable mismatch", "Conflict...", etc.

        Comment


          #5
          Originally posted by lemow
          "I tried "tbl=table.current() but it never seemed to be able to find the right record": I meant it didn't open the table I had in mind, which is "Claim.dbf".
          If you are working with a form based on a set you need tbl = table.current(n) where n is the relative position in the set hierarchy.

          1 for parent
          2 for first child shown when you view the set structure in the edit set dialog
          3 for second child ....
          There can be only one.

          Comment


            #6
            Tom,

            I am trying to look more into action scripting...
            Do you think there is any hope with the existing way? I am trying to avoid major changes since the database was already developed in A5V4... Thanks.

            Comment

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