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Thread: Help needed with php forms

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  1. #1
    Join Date
    Jan 2013
    Posts
    20

    Help needed with php forms

    Hello,
    I am totally new to php and need a help with solving the followinf problem:
    I've got a database file creating 3 tables:
    <?

    $db=sqlite_open("../paintings.db");



    @sqlite_query($db, "DROP TABLE Picture");
    @sqlite_query($db, "DROP TABLE Artist");
    @sqlite_query($db, "DROP TABLE Sales");


    @sqlite_query($db,"CREATE TABLE Picture( ID integer PRIMARY KEY , name VARCHAR[200], year integer, artistID integer)",$sqliteerror);
    @sqlite_query($db,"CREATE TABLE Artist( ID integer PRIMARY KEY , name VARCHAR[200], DOB integer, DOD integer, nationality VARCHAR[200])",$sqliteerror);
    @sqlite_query($db,"CREATE TABLE Sales( pictureID integer , date VARCHAR[20], price REAL)",$sqliteerror);


    //NB use of ' and "
    sqlite_query($db,'INSERT INTO Picture (name, year, artistID) VALUES ( "The Haywain", 1821, 1)');
    sqlite_query($db,"INSERT INTO Picture (name, year, artistID) VALUES ( 'Salisbury Cathedral from the Meadows', 1831, 1)");
    sqlite_query($db,"INSERT INTO Picture (name, year, artistID) VALUES ( 'A Wheatfield, with Cypresses', 1889, 2)");

    sqlite_query($db,"INSERT INTO Artist (name, DOB, DOD, nationality) VALUES ( 'John Constable', 1776, 1837, 'British')");
    sqlite_query($db,"INSERT INTO Artist (name, DOB, DOD, nationality) VALUES ( 'Vincent Van Gogh', 1753, 1890, 'Dutch')");


    sqlite_query($db,"INSERT INTO Sales VALUES ( 1, '1/9/2008', 23.56)");
    sqlite_query($db,"INSERT INTO Sales VALUES ( 1, '9/1/2009', 500230.00)");
    sqlite_query($db,"INSERT INTO Sales VALUES ( 3, '15/12/2009', 87.56)");


    sqlite_close($db);

    ?>


    What I have to do is:
    Create a form that allows the user to input information about another picture for a given artist and then handles this in a different page. The second page shows all the artists in the database.
    For the first page try to use a drop down list for the choice of artist for ones that are currently in the database hint you will need to have a query on the database to select all the artists, then use a for loop to add these to items on a drop down list.

    I am so confused, do not really know what I have to do.

    I would appreciate any help.

  2. #2
    Join Date
    Jan 2013
    Posts
    20
    To begin with, I've done a drop down list that would display artist names in it. Something is wrong here as names are not displayed. Is there anything obvious I am missing here?

    Code:
    <html>
    <body>
    <?php
    
    
    
    $db=sqlite_open("../paintings.db");
    
    $result=sqlite_query($db,"SELECT ID, name from Artist");
    
    $options="";
    
    
    
    while($row=sqlite_fetch_array($result,SQLITE_ASSOC ))
    {
    $id=$row["ID"];
    $name=$row["name"];
    $option.="<OPTION VALUE=\"$id\">".$name.'</option>';
    }
    
    sqlite_close($db);
    
    	
    ?>
    
    <SELECT NAME=id>
    <OPTION VALUE=0>Choose
    <?php echo $options ?>
    </SELECT>
    
    </body>
    </html>

  3. #3
    Join Date
    Jan 2013
    Posts
    20
    Oh I just noticed my mistake which was the following:
    I had ending 's' missing in $option in this line:

    $option.="<OPTION VALUE=\"$id\">".$name.'</option>';

  4. #4
    Join Date
    Jan 2013
    Posts
    20
    So now I need to have some code that would allow to insert information about new picture for a given artist. Any ideas?

  5. #5
    Join Date
    Jan 2013
    Posts
    20
    So now I need to have some code that would allow to insert information about new picture for a given artist. Any ideas?

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