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Kristy24
Mar 11th 2008, 7:43 am
Hello,

I have 3 databases, one with names of people I know, another one with names of tools I own and another one for "events" that involve those people and those tools.
I'm using forms (drop list) to put it all together. Right now I'm having trouble with the "events" part. I have a form where I choose a person and a tool and then I want it to send that info to the "events" database. Everything seems to work fine, I don't get any errors but the info is not printed on the site. I was wondering if someone could analyze my code and help me out.

Thanks in advance.

Here's the form code:

<form action="add_event.php" method="post">
<?php
include("config.php");
$sql_tools = "SELECT * FROM list_tools";
$sql_people = "SELECT * FROM list_people";
$result_tools = mysql_query($sql_tools);
$result_people = mysql_query($sql_people);
$endselect = "</SELECT>";
$select_tools = "<SELECT name=\"list_tools\">";
$select_people = "<SELECT name=\"lista_people\">";
echo ("".$select_tools."");
$submit = "<BR><input type=\"submit\" value=\"Add Event\" style=\"
color: #fff;
font-weight:bold;
background-color:#000000;
border: 2px solid #fff;
padding: 2px 5px;
font-family: Arial, ..., serif;
font-size: 11px;
height: 30px;
display: block;
width: 350px;\">";
while ($row_tools = mysql_fetch_array($result_tools))
{
$option_tools = "<option value=\"".$row_tools[toolsid]."\" />".$row_tools[toolsname]."";
echo ("".$option_tools."");
}
echo ("".$endselect."");
echo ("".$select_people."");
while ($row_people = mysql_fetch_array($result_people))
{
$option_people = "<option value=\"".$row_people[peopleid]."\" />".$row_people[peoplename]."";
echo ("".$option_people."");
}
echo ("".$endselect."".$submit."");
?>
</form>

And here's the code that inserts those 2 values in the "events" database:

<?php

include("config.php");
$result = mysql_query("INSERT INTO events (tools, people)
VALUES ('".$_POST['toolsid']."', '".$_POST['peopleid']."')",$connect) or die("Could not insert data because ".mysql_error());
echo "<meta http-equiv=Refresh content=1;url=index.php>";
?>

Altari
Mar 11th 2008, 8:02 am
It looks like you're trying to insert a value that doesn't exist? I think those two items should match name='foo' and insert $_POST['foo']

$select_tools = "<SELECT name=\"list_tools\">";
$select_people = "<SELECT name=\"lista_people\">";


$result = mysql_query("INSERT INTO events (tools, people)
VALUES ('".$_POST['toolsid']."', '".$_POST['peopleid']."')",$connect) or die("Could not insert data because ".mysql_error());
echo "<meta http-equiv=Refresh content=1;url=index.php>";
?>

Kristy24
Mar 11th 2008, 9:02 am
Problem solved. I googled all evening yesterday after a solution for this!
Thanks a lot Altari. :)