I'm busy trying to create a website for my football team. The thing I'm having problems with is creating a web form with drop down boxes to select and insert the match data. I'm already able to add a match in phpmyadmin where I can just select team_home and team_away, so the relational database seems to work.
I've got the following 2 tables:
So how can I make a web form with drop down boxes so I can add matches into my database?
UPDATE:
I've got the form working with drop down boxes, but I'm getting the following error when I'm submitting the form:
Error: Cannot add or update a child row: a foreign key constraint fails (
roflz.matches, CONSTRAINTmatches_ibfk_1FOREIGN KEY (team_home) REFERENCESteams(name))I've posted my submit form code and insertmatch.php code
$sql="SELECT id, name FROM Teams";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["id"];
$name=$row["name"];
$optionshometeam.="<OPTION VALUE=\"$id\">".$name;
$optionsawayteam.="<OPTION VALUE=\"$id\">".$name;
}
?>
<form action="insertmatch.php" method="post">
<SELECT NAME=Teams>
<OPTION VALUE=0>Home Team
<?=$optionshometeam?>
</SELECT>
<SELECT NAME=Teams>
<OPTION VALUE=0>Away team
<?=$optionsawayteam?>
</SELECT>
Score Home team: <input type="text" name="score_home" />
Score Away team: <input type="text" name="score_away" />
Match Date: <input type="text" name="score_away" />
<input type="submit" />
</form>
mysql_select_db("roflz", $con);
$sql="INSERT INTO matches (team_home, team_away, score_home, score_away, date)
VALUES
('$_POST[team_home]','
$_POST[team_away]','
$_POST[score_home]','
$_POST[score_away]'
$_POST[date]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "Match added";
mysql_close($con);
?>
So what's causing this error?
Error: Cannot add or update a child row: a foreign key constraint fails (roflz.matches, CONSTRAINT matches_ibfk_1 FOREIGN KEY (team_home) REFERENCES teams (name))
okay you have very basic problem. i don't know much about php but i can suggest you some logical thing which you can perform.
<select>
<option value="t1">Team 1</option>
<option value="t2">Team 2</option>
<option value="t3">Team 3</option>
<option value="t4">Team 4</option>
</select>
this will create drop down box. What you need to do is to set your teams id(using php) in "value" and team name between "option" tag. The "value" of the particular selected team will be passed in request when you submit your form.
ok try out this..
<?
...
mysql cnx code
...
$sql="SELECT id, name FROM Teams";
$result=mysql_query($sql);
$options="";
while ($row=mysql_fetch_array($result)) {
$id=$row["id"];
$name=$row["name"];
$options.="<OPTION VALUE=\"$id\">".$name;
}
?>
...
html code
...
<SELECT NAME=Teams>
<OPTION VALUE=0>Choose
<?=$options?>
</SELECT>
but don't forget to wrap it with in the "form" tag.
This question is a bit broad and alot has to be cover prior to nailing the result, its almost asking how to build a website and that in itself could take years to learn. That said, these 2 links should give you what you need to get going.
W3C MySql introduction and tutorials
Please note how much code is behind phpmyadmin (just go through its source code) and you will find there is no simple 1 click way off doing it.
you may try something like
<?php
$db = "database_name";
$con = mysql_connect("localhost","username","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db, $con);
$query = "SELECT * FROM YOUR_TABLE";
$result = mysql_query($query);
?>
Also, in your html body part, inside your form, you may use this for generating a dropdown box
<select>
<?php
while($info = mysql_fetch_array($result)){
$name = $info["table_column_name"];
echo '<option value="'.$name.'">'.$name.'</option>';
}
?>
</select>
Hope this helps.. :)
Error: Cannot add or update a child row: a foreign key constraint fails (roflz.matches, CONSTRAINT matches_ibfk_1 FOREIGN KEY (team_home) REFERENCES teams (name))
This is all you need to know, blah blah....foreign key constraint fails blah bla....REFERENCES teams
You are updating matches, but since matches uses foreign keys to teams you cannot add something in matches that is not in teams
Error: Cannot add or update a child row: a foreign key constraint fails (roflz.matches, CONSTRAINT matches_ibfk_1 FOREIGN KEY (team_home) REFERENCES teams (team_home)). Weird thing is that it's not giving any error when I add a match in phpmyadmin, it just works - Johan 2012-04-05 10:41