Inserting from form

topcat

I'm still having trouble trying to get info from a form and a table entered into a second table.

Here's how I want it to work.
-people enter business info into a form
-that info is entered into table1
-a business key is selected from table1
-the business key and zip codes are entered into table2
(one record for each zip code)

I'm able to enter all the business info into table1 and select the business key.
The problem comes when I try to enter the zip codes. All I get is a new record with one zip code and a 0 for the BusinessKey.

For the INSERT statement I've tried using ".$row['BusinessKey']." or '.$row['BusinessKey'].' instead of '$BusinessKey' and echo $row['BusinessKey'];

And, I know I need a while statement on the INSERT but can't figure out how to write one that works.

Any help would be greatly appreciated as really having a tough time with this.

Thanks,
TC

FIRST PAGE
(This is actually the second page...after the form where people enter their business info)

<?php
session_start();
?>

<html>
<head>
<title>Untitled Document</title>
<?php
session_register ("BusinessName");
session_register ("ContactName");
session_register ("Phone1");
?>

</head>
<body bgcolor="#FFFFFF" text="#000000">
<form name="form1" method="post" action="insert_zipcodes.php">
<input type="text" name="zipcode">
<input type="text" name="zipcode">
<input type="text" name="zipcode">
<input type="text" name="zipcode">
<input type="text" name="zipcode">
<input type="submit" name="Submit" value="Submit">
</form>
</body>
</html>

SECOND PAGE

<?php
session_start();
session_unregister("BusinessName");
session_unregister ("ContactName");
session_unregister ("Phone1");
?>

<html>
<head>
<title>insert_zipcodes.php</title>
</head>
<body bgcolor="#FFFFFF" text="#000000">

<?php
echo "$BusinessName<br>
$ContactName<br>
$Phone1<br>";


@ $db = mysql_connect("X", "X", "X");

if (!$db)
{
echo "Error: Could not connect. Please try again later.";
exit;
}

@mysql_select_db("X");


$query = "SELECT BusinessKey FROM tblMain WHERE BusinessName='$BusinessName' AND ContactName='$ContactName' AND Phone1='$Phone1'";
$resultID = mysql_query($query) or die("Invalid query query 1: " . mysql_error());
while ($row = mysql_fetch_assoc($resultID))

{
echo $row['BusinessKey'];
}

$BusinessKey = "$row['BusinessKey']";

$query = "INSERT INTO tblZipCodes (BusinessKey, Zip) VALUES('$BusinessKey', '$Zip')";
$resultID = mysql_query($query) or die("Invalid query query 2: " . mysql_error());

{  

if ($result)
echo mysql_affected_rows(). "<center><font face=Arial, Helvetica, sans-serif size=3><b>Your information has been submitted.
<br>
<br>

<a href='http://www.mywebsite.com'>Continue</center>";
}

mysql_close($linkID);

?>

hermand

The text boxes in the form have all the same names. That wont work!

hermand

I think I may be being dumb here, but you need to have different names for the textboxes. THen in the next page, you can call the value from $testboxname or $_POST["textboxname"] if you want to be strict!

Bunkermaster

Try to echo your SQL statement and run it in phpmyadmin to see if insert works.

$BusinessKey = $row['BusinessKey'];

will work BUT it's outside of the loop so if you have more than one tuple (😃) to update it'll only update the last one.

for multipe zip code you can create multiple form fields named as follow

<input type="text" name="zipcode"> 
<input type="text" name="zipcode[0]"> 
<input type="text" name="zipcode[1]"> 
<input type="text" name="zipcode[2]"> 
<input type="text" name="zipcode[3]">

you can then access each one using

$_POST['zipcode']['0'];
$_POST['zipcode']['1'];
$_POST['zipcode']['2'];
$_POST['zipcode']['3'];

use a foreach loop and voila.

If I am off topic try to explain more clearly for your post is HUGE.

Hope it helped.

topcat

Thanks for responding. I'm not quite sure I understand how to do what you're saying but I'll give it a shot.

I'm sorry for the length of my post . I'm just learning to how to use PHP and I wasn't sure how much information people would need to understand what I'm trying to do. :rolleyes:

jib

also (and i may be wrong) but I think that session_register needs to be used right at the top of the page before any HTML headers are used...

you are also then unregistering the variables before they are used by unregistering them at the top (but in the correct palce) of the second page...

Bunkermaster

you can use predefined variables anywhere you want them.

topcat

I'm a bit lost. I'm sure I'm looking in the wrong places but I'm having trouble trying to find some examples of what you're talking about in your last post. I'm not sure where this goes in the code:

$_POST['zipcode']['0'];
$_POST['zipcode']['1'];
$_POST['zipcode']['2'];
$_POST['zipcode']['3'];

Bunkermaster

this goes where you need to gather the zipcodes...

topcat

I think (hope) I'm getting it but now I've got an error that I can't figure out:

Parse error: parse error, unexpected $ in /www/.../insert_zipcodes.php on line 75

line 75 is this:

</html>

Any ideas?

hermand

Give us the lines around it, sounds like someone didn't escape something or whatever

topcat

According the the error statement, the </html> line is supposedly the one that has the $ in it...but it doesn't....

{

if ($result)
echo mysql_affected_rows(). "<center><font face=Arial, Helvetica, sans-serif size=3><b>Your information has been submitted.
<br>
<br>

<a href="http://www.mywebsite.com">Continue</center>";
}

mysql_close($linkID);
?>

</body>
</html>

hermand

The start loop { is before the IF

hermand

Generally speaking, any parse error that claims to be after the PHP closes (?>) is usually a loop error.

topcat

I just changed it and it's still saying there's a $ on the </html> line...there must be something else wrong with my loop then.

hmmmm....

hermand

Hmm, try and count all the open loops ,{, and then count all the close loops, }. Make sure they equal each other.

topcat

Thanks...I counted and the number of { = }, but I now get a different kind of error. I think it's time I walk away from the computer for a while before my head blows up! I'll be back...

hermand

Hmm, wanna post the whole code here? It's sometimes easy to overlook something, because you assume its there, kinda thing. Like when you automatically add missing words to a sentance, because you "know" it'll be there

Bunkermaster

Originally posted by topcat
{

if ($result)
echo mysql_affected_rows(). "<center><font face=Arial, Helvetica, sans-serif size=3><b>Your information has been submitted.
<br>
<br>

<a href="http://www.mywebsite.com">Continue</center>";
}

mysql_close($linkID);
?>

</body>
</html>

{  

if ($result){
echo mysql_affected_rows();
?>
<center><font face=Arial, Helvetica, sans-serif size=3><b>Your information has been submitted.
<br>
<br>

<a href="http://www.mywebsite.com">Continue</center>
<?
}
}

mysql_close($linkID);
?>

</body>
</html>

or something 😃

topcat

I think I'm making progress with this thing lol Hopefully I'll get it soon.

I don't get any error messages but when I check to see what's entered into the database it only enters one record with a 0 for the Zip and a 0 for the BusinessKey.

Here's the code now, after I've played with it many times today....

<?php

echo "$BusinessName<br>
$ContactName<br>
$Phone1<br>";


@ $db = mysql_connect("X", "X", "X");

if (!$db)
{
echo "Error: Could not connect. Please try again later.";
exit;
}

@mysql_select_db("X");


$query = "SELECT BusinessKey FROM tblMain WHERE BusinessName='$BusinessName' AND ContactName='$ContactName' AND Phone1='$Phone1'";
$resultID = mysql_query($query) or die("Invalid query query 1: " . mysql_error());
while ($row = mysql_fetch_assoc($resultID))

{
echo $row['BusinessKey'];
}


$_POST['zipcode']['0'];
$_POST['zipcode']['1'];
$_POST['zipcode']['2'];
$_POST['zipcode']['3'];

foreach ($_POST['zipcode'] as $zipcode)
 {

 $query = "INSERT INTO tblZipCodes (BusinessKey, Zip) VALUES('".$_POST['zipcode']."', '.$row['BusinessKey'].')";
 $resultID = mysql_query($query) or die('Invalid query query 2: ' . mysql_error());

 exit();
 }


if ($result)
{
echo mysql_affected_rows();

mysql_close($linkID);
}
?>