Member Avatar for tip

I am a second year IT student who is working on a project using PHP(frontend) and MySQL(backend). I am brand new to these softwares which are not being touch in class. I have 2 questions:

  1. How do I populate fields (textboxes) in a html form from a selected item in a droplist. The droplist and fields in the form are from the same table. I have spend hours on this but seem to get no where.

  2. How do I integrate the the above query into my code where I have add, update, and delete button on one page. The name of the page is add_event2.php

Table: events

event_code
title
duration
grade
type
desc_short
desc_long

<html>
<head><title>(Event Form)</title></head>
<body>
<?php
// db connectivity works
include 'db.php';
$tablename = 'events';

$value = "";
if (isset($_POST['submit'])){
    $value = $_POST['submit'];
    }

// create SQL statement 
$sql = "SELECT event_code FROM $tablename ORDER BY event_code ASC"; 

// execute SQL query and get result 
$sql_result = mysql_query($sql,$connection) 
    or die("Couldn't execute query."); 

// put data into drop-down list box 
while ($row = mysql_fetch_array($sql_result)) { 
    $event_code  = $row["event_code"]; 
    // line gives an error but still copiles the list. 
    $option_block .= "<OPTION value=\"$event_code\"> $event_code</OPTION>"; 
    } 

switch($value){

        case "Add":
            print "Added..........<P>";
            break;

        case "Update":
            print "Updated..........<P>";
            break;

        case "Delete":
            print "Deleted..........<P>";
            break;

        case "": //first time to open this form. 
            //should populate fields in the form with the last record in  the table event
            print "First time to open this form .......<p>";
            break;                  
}
?>

<form method="post" action="add_event2.php" ID="event_form">
            <P><STRONG><U>Events</U></STRONG></P>
            <P>Event Code: <br> 
                <SELECT  name="$event_code"> 
                <? echo "$option_block"; ?> 
                </SELECT> 

            <P>Title: <INPUT type="text" name="title"></P>
            <P>Duration: <INPUT type="text" name="duration"></P>   
            <P>Grade: <INPUT type="text" name="grade"></P>
            <P>Type: <INPUT type="text" name="type"></P>
            <P>Short Description: <INPUT type="text" name="desc_short"></P>
            <P>Long Description: <INPUT type="text" name="desc_long"></P>
            <P>Road End Number: <INPUT type="text" name="road_end_id"></P>
            <p><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Add"
            <p><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Update" 
            <p><INPUT TYPE="SUBMIT" NAME="submit" VALUE="Delete"
        </form>
<?
// mysql_close($connection);
?>
</body>
</html>

Note that the above codes produce this error meassage but still seems to work:
Notice: Undefined variable: option_block in :\inetpub\wwwroot\add_event2.php on line 25

I would appreciate help urgently. If there is a better way of writing the codes, pls feel free to show me.

Thanks
tip

Edited by Dani because: Formatting fixed
  • 2 Contributors
  • 5 Replies
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  • 19 Hours Discussion Span
  • Latest Post Latest Post by tip
Member Avatar for tip

I have solve the following error by declaring $option_block = ''; at the top.

Note that the above codes produce this error meassage but still seems to work:
Notice: Undefined variable: option_block in :\inetpub\wwwroot\add_event2.php on line 25

Member Avatar for Gary King

To turn off notices, you should set error_reporting() to something else; check www.php.net/error_reporting for more info :) Your level is probably way too low; you should NEVER see Notice messages, they are only if you really want to debug things but about literally 95% of the time, notices are worthless.

Member Avatar for tip

Thanks Gary. Any solutions for my queries. Query 1 is vital at this state.

ta

tip

Member Avatar for tip

These parts of my code produce a droplist which is populated by event_code field from the events table. But how do I populate the rest of the data on the form which an event_code is selected from droplist?


// put data into drop-down list box
while ($row = mysql_fetch_array($sql_result)) {
$event_code = $row["event_code"];
// line gives an error but still copiles the list.
$option_block .= "<OPTION value=\"$event_code\"> $event_code</OPTION>";
}
.
.
.
<FORM...............>
.
.
<P>Event Code: <br>
<SELECT name="$event_code">
<? echo "$option_block"; ?>
</SELECT>
.
.
</FORM>

SOS ta

tip

Member Avatar for tip

help! help! help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! anyone

tip

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