displaying details

[harlequin2k5]

so I have a summary list page and what I'd like to do is when you click on a particular vendor the rest of their information displays

<?php
$conn = doconnect();
$query = "SELECT `VendorName` FROM `tblVendorInfo` where VendorName LIKE 'A%'";
$result = mysql_query($query);
echo "<ul>\n";
while($row = mysql_fetch_assoc($result))
{
echo "<li><a href=\"#\">" . $row['VendorName'] . "</a></li>";
}
echo "</ul>\n";
mysql_free_result($result);
?>

VendorName is the link

I'm in the process of adding VendorID because there are a few vendors with the same name located in different parts of the state

and for what it's worth...PHP and MySQL For Dynamic Web Sites by Larry Ullman is a huuuuuuuuge disappointment :(

thanks in advance! :)

[Brandoe85]

Hi,
So you do have vendorID in the table? Pass the vendorID to a page, and grab it with $_GET, then in your query, select fields from table where vendorID = $_GET['vendorID'].


// i like to use single quotes and concatenat on the variable you need...this way you can use double quotes for your html and not have all the escaping.
echo '<li><a href="somepage.php?vendorID=' . $row['vendorID'] . '">' . $row['VendorName'] . '</a></li>';


Now, you should have the link set up. Next, create somepage.php(or whatever you're going to call it) and do your query:

$id = (isset($_GET['vendorID'])) ? $_GET['vendorID'] : false;
if($id)
{
// vendor id was passed
$query = 'select fields from table where vendorID = ' . $id;
$result = mysql_query($query) or die(mysql_error()); // add mysql_error() for debugging.
while($row = mysql_fetch_assoc($result))
{
// display your info
}
}


Hopefully that'll help some.
Good luck;

[ns1987]

Don't forget to protect your SQL by using intval(), typecasting or some other method on $id.

[harlequin2k5]

ok I used your code sample and the subdetail.php comes up just fine except that it doesn't display any data - I've been at this a couple of hours now so I'm sure I'm just not seeing something real simple

<?php
$id = (isset($_GET['VendorID'])) ? $_GET['VendorID'] : false;
if($id)
{
// vendor id was passed
$query = 'SELECT VendorID, VendorName, VendorAddress1, VendorAddress2, VendorCity, VendorState, VendorZip, VendorPhone, VendorFax, VendorURL, VendorEmail from tblVendorInfo where VendorID =' . $id;
//select fields from table where VendorID = ' . $id;
$result = mysql_query($query) or die(mysql_error()); // add mysql_error() for debugging.
while($row = mysql_fetch_assoc($result))
{
// display your info
echo $row['VendorName'];
}
echo "</ul>\n";

} ?>

Don't forget to protect your SQL by using intval(), typecasting or some other method on $id.
I don't understand what you mean

[Brandoe85]

Does it make it passed the if statement? Add this on:

<?php
$id = (isset($_GET['VendorID'])) ? $_GET['VendorID'] : false;
if($id)
{
echo 'The id is: ' . $id;
// vendor id was passed
$query = 'SELECT VendorID, VendorName, VendorAddress1, VendorAddress2, VendorCity, VendorState, VendorZip, VendorPhone, VendorFax, VendorURL, VendorEmail from tblVendorInfo where VendorID =' . $id;
//select fields from table where VendorID = ' . $id;
$result = mysql_query($query) or die(mysql_error()); // add mysql_error() for debugging.
while($row = mysql_fetch_assoc($result))
{
// display your info
echo $row['VendorName'];
}
echo "</ul>\n";

} ?>

[harlequin2k5]

I tried that too...I'd show you a screen shot but I can't seem to figger out how *chuckles*

The address bar shows that it has selected vendorid 540 which is the vendorid for abbey carpet - but it doesn't show even the vendor name on the detail page

I think you can get to the page from here (http://67.78.185.42:1985/subinfo.php)

[Brandoe85]

Heh, link doesn't work :(

But, did it display "The id is 540"? If not, make sure the querystring "VendorID" is the same case as the $_GET['VendorID']

Good luck :)

[harlequin2k5]

the address bar displays...
http://67.78.185.42:1985/subdetail.php?vendorID=1459
I even added some dummy text after and that displays so I think it's at least passing the "if"

I checked to make sure I was using the proper case for everything and I am - I've gotten caught by that before

I tried adding the vendorid to the results echo and that didn't come up either - and yes there are details to be displayed - if only the vendor name - this data is taken from an existing database

[Brandoe85]

Hmm..I see it's a small case v in the querystring and a capital V in the assignment...so, not sure what you have but make sure it's:

$id = (isset($_GET['vendorID'])) ? $_GET['vendorID'] : false;

Note the small v in vendorID. Can you verify that's what you have?

[harlequin2k5]

when you had originally given me the code you wrote it as vendorID - I changed it to VendorID because that's the proper field name - should I change it back to the lowercase?

[harlequin2k5]

before you answer that - it displays the id now and then says no database selected

[Brandoe85]

Whichever you'd like. Just make sure they are the same case. Just make sure the URL:

http://67.78.185.42:1985/subdetail.php?vendorID=1459


Is the same as that statement:

$id = (isset($_GET['vendorID'])) ? $_GET['vendorID'] : false;


That should work, now..if you'd like them to be a capital V just make sure you change them in both places.

[harlequin2k5]

ok so I'm a dope - I didn't have my connect() in there and changing it to lower case worked

I'm just curious why? if the field name is VendorID why wouldn't I look for it instead of vendorID?

[harlequin2k5]

I got it - that makes perfect sense

I'll get what I ask for

thanks for your patience brandoe85!

[Brandoe85]

The reason is because that's not the name from your DB...it's just a name we made up when we built the URL here:

echo '<li><a href="somepage.php?vendorID=' . $row['vendorID'] . '">' . $row['VendorName'] . '</a></li>';


So we have to make sure it's vendorID in $_GET. We could've named it harlequin2k5=55 :D

But now i'm just blabing on...i'm glad it works :)

[harlequin2k5]

thanks brandoe85

but for your edification...harlequin2k5=37 :p

may I be so bold as to ask one more question? I can't seem to make a linebreak

I've tried "\n", \"n", \n, '\n', \'n' - it is currently "\n"
echo $row['VendorName'] . "\n" . $row['VendorAddress1'] . "\n" . $row['VendorAddress2'] . "\n" . $row['VendorCity'] . ", " . $row['VendorState'] . " " . $row['VendorZip'] . "\n" . $row['VendorEmail'] . "\n" . $row['VendorURL'];

does the line break have to go into its own echo?

[Brandoe85]

I stand corrected! :p

"\n" will put a line feed in your source, so you can format your white space of the generated html. You'll want to use <br>.

Good luck;

[harlequin2k5]

that's absolutely perfect!

I shoulda realized about the < br >

ok go to bed now - you've earned it :)

[Brandoe85]

Great!

I didn't even know you guys got a chance to be on a computer in Florida...are you on the beach with your laptop? :D