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Views: 5682 | Replies: 5
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Join Date: Sep 2004
Posts: 13
Reputation: 
Rep Power: 4
Solved Threads: 0
Newbie Poster
Populating a fields on a form base on a selected item from a droplist. 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
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
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Join Date: Sep 2004
Posts: 13
Reputation:
Rep Power: 4
Solved Threads: 0
Newbie Poster
I have solve the following error by declaring $option_block = ''; at the top.
Note that the above codes produce this error meassage but still seems to work:
Notice: Undefined variable: option_block in :\inetpub\wwwroot\add_event2.php on line 25
Note that the above codes produce this error meassage but still seems to work:
Notice: Undefined variable: option_block in :\inetpub\wwwroot\add_event2.php on line 25
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Join Date: Nov 2003
Location: Toronto, Canada
Posts: 354
Reputation:
Rep Power: 6
Solved Threads: 5
PHP/vBulletin Guru
To turn off notices, you should set error_reporting() to something else; check www.php.net/error_reporting for more info 
Your level is probably way too low; you should NEVER see Notice messages, they are only if you really want to debug things but about literally 95% of the time, notices are worthless.
Your level is probably way too low; you should NEVER see Notice messages, they are only if you really want to debug things but about literally 95% of the time, notices are worthless. •
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Join Date: Sep 2004
Posts: 13
Reputation:
Rep Power: 4
Solved Threads: 0
Newbie Poster
Thanks Gary. Any solutions for my queries. Query 1 is vital at this state.
ta
tip
ta
tip
•
•
Join Date: Sep 2004
Posts: 13
Reputation:
Rep Power: 4
Solved Threads: 0
Newbie Poster
These parts of my code produce a droplist which is populated by event_code field from the events table. But how do I populate the rest of the data on the form which an event_code is selected from droplist?
// put data into drop-down list box
while ($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 "$option_block"; ?>
</SELECT>
.
.
</FORM>
SOS ta
tip
// put data into drop-down list box
while ($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
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