[SOLVED] Can someone spot the error?

[englishcodemonkey]

When i run this code nothing shows in the table...Well i guess the table doesn't even show up!  I was told this was because of the SQL connection or query?? I know my connection is correct because I have other PHP scripts running correctly from it??  Here is how my table is set up.  This is confusing me because I had it working perfect on localhost.

 

CREATE TABLE HOSTA (
hosta_id INT AUTO_INCREMENT PRIMARY KEY,
hosta_name VARCHAR(50) NOT NULL,
hybridizer VARCHAR(50),
size VARCHAR(10),
description VARCHAR (200),
price DECIMAL(3,2));

 

 

 

<?php
require_once('conn.php');

$hosta = $_REQUEST['hosta'];
$hosta1 = $_REQUEST['hosta'];

$q = "SELECT hybridizer, size, price, description from hostajess where hosta_name='$hosta'";

$r = @mysqli_query ($dbc, $q);


if ($r) {


echo '
	<table align="center" cellspacing="3" cellpadding="3" width="75%">
		<tr>
			<td align="left" width="33.3%">Hybridizer</td>
			<td align="left" width="33.3%">Size</td>
			<td align="left" >Price</td>
		</tr>';

while ($row = mysqli_fetch_array($r,MYSQLI_ASSOC)) {
	echo '
		<tr>
			<td align="left">' .$row['hybridizer'] . '</td>
			<td align="left">' .$row['size'] . '</td>
			<td align="left">' .$row['price']. '</td>
		</tr>';
}
}

echo '</table>';

$q = "SELECT description from hostajess where hosta_name='$hosta1'";
$r = mysqli_query ($dbc, $q);

if ($r) {


echo '
	<table align="center" cellspacing="3" cellpadding="3" width="75%">
		<tr>
			<td align="left">Description</td>
		</tr>';

while ($row = mysqli_fetch_array($r,MYSQLI_ASSOC)) {
	echo '
		<tr>
			<td align="justify">' .$row['description'] . '</td>
		</tr>';
}
}

echo '</table>';



mysqli_close($dbc);


?>

 

This is really frustrating me! thanks for all your help! 

[dannyb785]

your query is selecting from a table "hostajess" but you've created a table named "HOSTA".

[englishcodemonkey]

I tried that already if you need more info the problem is on this page...http://www.wadeandgattonnurseries.com/hosta_new.html

[dannyb785]

Tried what already? If your table's name is HOSTA, you have to say "SELECT * FROM HOSTA" it can't be anything FROM hostajess... hostajess doesn't exist(as afar as I know)

[englishcodemonkey]

The table on my local MySQL is called 'hosta'. The table I created on my host is called 'hostajess'.  I posted the code that I used to set up the table because that has the fields in, forgetting that I had changed the name.  The code for the table should have been posted as:

 

CREATE TABLE hostajess (
hosta_id INT AUTO_INCREMENT PRIMARY KEY,
hosta_name VARCHAR(50) NOT NULL,
hybridizer VARCHAR(50),
size VARCHAR(10),
description VARCHAR (200),
price DECIMAL(3,2));

 

Sorry for the confusion!

[revraz]

Use mysqli_error after your query to see if it's failing.

[englishcodemonkey]

Ok i added:

mysqli_error($dbc);

 

but when i select the option from the drop down box and click the submit button i still get just a blank page????

 

Any ideas?

[dannyb785]

you need to do the error function as:

 

$result = mysql_query("QUERY") or die(mysql_error());

that should show the error

 

 

also, when you get the blank page, view the source and see if anything is printing. Like the </table> tag that  you have set to output whether $r is true or not. If it does display, then there's an error grabbing the data. If it's blank, then you have error in the code.

[englishcodemonkey]

Ok so i'm not quiet sure what the problem was but it's working ok now!

 

Thanks everyone though