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Thread: new to table joins?

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  1. #1
    Join Date
    Jun 2005
    Location
    coronado
    Posts
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    new to table joins?

    I have an activity table which has A_ID (PK), C_NUMBER (FK), A_DATE, A_DESC, and a MEDIVAS field. When I perform a search by date, all the dates in that column show up in a combo box, when a cerrtain date is selected, I'm trying to output all the info for that activity (description, medivas) as well as the coresponding company information (C_NUMBER) would be its unique identifier.

    how do I do this using table joins?
    thanks

  2. #2
    Join Date
    Feb 2003
    Posts
    1,048
    Select *
    From CompanyTable As C
    Inner Join ActivityTable AS A On A.C_NUMBER = C.C_NUMBER
    Where C.A_DATE = 'Date Value'

  3. #3
    Join Date
    Jun 2005
    Location
    coronado
    Posts
    76

    Talking k

    Thank you rawhide,
    I did this
    Select *
    From COMPANY As C
    Inner Join ACTIVITY AS A On A.C_NUMBER = C.C_ID
    Where C.A_DATE = '$date'

    COMPANY is the name of my company table
    ACTIVITY is the name of my activity table
    C_ID is the primary key of my company table
    $date is the phhp variable of the date choosen.

    How should I display the results, like this?

    while ($result = mysql_fetch_array($query)){
    print "<TD>".$result['C_NAME']."</TD>";
    print "<TD>".$result['C_CITY'].", ".$result['C_STATE']."</TD>";
    print "<TD>".$result['C_DESC']."</TD>";
    print "<TD>".$result['RATIONALE']."</TD>";
    print "<td>".$result['contact_NAME']."</TD>";
    print "<td>".$result['contact_PHONE']."</td>";
    print "<TD>".$result['A_DATE']."</TD>";
    print "<TD>".$result['A_DESC']."</TD>";
    print "<td>".$result['NEXT']."</TD>";
    print "<td>".$result['MEDIVAS']."</td>";
    print "</TR>\n";
    }

    How would I differentiate between the 2 tables if thats the case?

  4. #4
    Join Date
    Feb 2003
    Posts
    1,048
    If there are no fields by the same name in the two tables, so all you need to know are the field names. If there are any by the same name, you'll need to alias one of the tables with a different name.

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