[SOLVED] Simple Query

[iceblox]

Hi Guy this is probably going to be the blondest post of the day but i cant for the life of me get this simple query right. Ive change the format ive changed everything.

 

Can anyone see anything wrong with this code?

 

<?php

$connection = mysql_connect('localhost','user','pass');

mysql_select_db('name');

$query = "SELECT * FROM Cars WHERE CarID = '$CarID'";

$result = mysql_query($query);

if (mysql_num_rows($result) > 0)
{

while($row = mysql_fetch_row($result))
{

echo '>  <a href=' . str_replace(" ","-",$row['1'] ) . '-Cars-' . str_replace(" ","-",$row['3'] ) . '.html><font class="traillinks">' . $row['2'] . '</a>  >  <font class="traillinks">' . $row['3'] . '</font>';

}

}
else
{
echo 'No rows found!';
}


?>

[Dragen]

are you getting any errors?

[iceblox]

Just No Rows found but ive checked the CarID the name everything. its a bit weird

[Dragen]

try this:

<?php
$connection = mysql_connect('localhost','user','pass');

mysql_select_db('name');

$query = "SELECT * FROM `Cars` WHERE `CarID` = '$CarID'";

if($result = mysql_query($query)){
	if(mysql_num_rows($result) > 0){
		while($row = mysql_fetch_row($result)){*/
			echo '>  <a href="' . str_replace(" ","-",$row['1'] ) . '-Cars-' . str_replace(" ","-",$row['3'] ) . '.html"><font class="traillinks">' . $row['2'] . '</a>  >  <font class="traillinks">' . $row['3'] . '</font>';
		}
	}else{
		echo 'No rows found!';
	}
}else{
	echo mysql_error();
}
?>

 

you didn't have any error handling the the actual mysql_query itself. Also, you should surround table name etc in a query with `.

Another small error with the output was that you had no " around the url, so it may have caused problems when outputting.

[iceblox]

Hmmm,

 

Still no rows found. I really dont get it

[DeadEvil]

or you can try this..

 

<?php
$connection = mysql_connect('localhost','user','pass');
mysql_select_db('name');

$result = mysql_query("SELECT * FROM Cars WHERE CarID = '$CarID'");
$num = mysql_num_rows($result);
if ($num > 0){
$content = '';
while($row = mysql_fetch_array($result)){
	$content .= '><a href=';
	$content .= str_replace(" ","-",$row['column_name']).'-Cars-'; // changed "column_name" to real column name
	$content .= str_replace(" ","-",$row['column_name']).'.html>';
	$content .= '<font class="traillinks">'.$row['column_name'].'</a>>';
	$content .= '<font class="traillinks">'.$row['column_name'].'</font>';
	echo $content;
}
}
else{
echo 'No rows found!';
}
?>

[vbnullchar]

may be there is something wrong with the query...

try this things to help you find what is wrong

 

$connection = mysql_connect('localhost','user','pass') or die('connection failed');

 

mysql_select_db('name') or die('DB selection failed');

 

if(!mysql_query($query)) {

die('Query Failed');

}

 

[vbnullchar]

or you can try this..

 

<?php
$connection = mysql_connect('localhost','user','pass');
mysql_select_db('name');

$result = mysql_query("SELECT * FROM Cars WHERE CarID = '$CarID'");
$num = mysql_num_rows($result);
if ($num > 0){
$content = '';
while($row = mysql_fetch_array($result)){
	$content .= '><a href=';
	$content .= str_replace(" ","-",$row['column_name']).'-Cars-'; // changed "column_name" to real column name
	$content .= str_replace(" ","-",$row['column_name']).'.html>';
	$content .= '<font class="traillinks">'.$row['column_name'].'</a>>';
	$content .= '<font class="traillinks">'.$row['column_name'].'</font>';
	echo $content;
}
}
else{
echo 'No rows found!';
}
?>

 

if he's getting "No rows found" as output then there must be something wrong with the query

[Dragen]

try changing your query to this:

$query = "SELECT * FROM `Cars` WHERE `CarID` = '" . $CarID . "'";

just to check it's not reading the variable as it's name.

If it's not dinding the rows, than your query must be incorrect. Are you sure 'Cars' isn't 'cars' or 'CarID' isn't 'carid'?

 

or do this:

echo "SELECT * FROM `Cars` WHERE `CarID` = '" . $CarID . "'";

and see what it outputs

[vbnullchar]

try to output the value of $CarID..

[iceblox]

Ive tried all your ideas and nothing. it just says no rows found..

 

this is my table

 

CREATE TABLE `Cars` (
  `CarID` int(10) NOT NULL auto_increment,
  `MakeName` varchar(255) NOT NULL,
  `CarName` varchar(255) NOT NULL,
  `MakeID` varchar(255) NOT NULL,
  `CarType` varchar(255) NOT NULL,
  `Text` text NOT NULL,
  `Img1` varchar(255) NOT NULL,
  `Img2` varchar(255) NOT NULL,
  `Img3` varchar(255) NOT NULL,
  `Img4` varchar(255) NOT NULL,
  `Img5` varchar(255) NOT NULL,
  `Img6` varchar(255) NOT NULL,
  `ActiveImg` varchar(255) NOT NULL,
  PRIMARY KEY  (`CarID`)
) ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=utf8 AUTO_INCREMENT=2 ;

-- 
-- Dumping data for table `Cars`
-- 

INSERT INTO `Cars` (`CarID`, `MakeName`, `CarName`, `MakeID`, `CarType`, `Text`, `Img1`, `Img2`, `Img3`, `Img4`, `Img5`, `Img6`, `ActiveImg`) VALUES (1, 'Ford', 'Fiesta', '2', 'Supermini', 'Displayed below are links to find your perfect car! Simply click on a logo and it will take you to a page full of cars by that manufacturer. We currently have X cars stored in the database. If there is one that is missing and you feel deserves a place on the site then just send us an email via this link and we will add it asap!', 'Ford-Fiesta-1.jpg', 'Ford-Fiesta-2.jpg', 'Ford-Fiesta-3.jpg', 'Ford-Fiesta-4.jpg', 'Ford-Fiesta-5.jpg', 'Ford-Fiesta-6.jpg', 'Ford-Fiesta-1.jpg');

[vbnullchar]

try to output your $query

 

$query;

then check the value of $CarID then see your table if you really have  this record.

 

your sql dump says you only have 1 row

 

then your query should look like this to print 1 row

 

### SELECT * FROM Cars WHERE CarID = '1');

 

[iceblox]

OK, we are getting there when i changed this

 

SELECT * FROM Cars WHERE CarID = '1');

 

it outputting the query but wont allow me to add a $vairable. Im not sure why

[DeadEvil]

it means no problem with the connection.. huh? I think you should print the query first.

 

$query = "SELECT * FROM Cars WHERE CarID = '$CarID'";
print $query; exit;

 

.. copy the output of the above statement then paste it on your pma query window.

[vbnullchar]

therefore there is a problem with $CarID, try print its value

[iceblox]

I excuted this script

 

 

$query = "SELECT * FROM Cars WHERE CarID = '$CarID'";
print $query; exit;

 

and got this as an output

 

SELECT * FROM Cars WHERE CarID = ''

[vbnullchar]

that is why your are getting 0 rows $CarID has no value

[ViN86]

where is $CarID declared?

[iceblox]

Just found out from my hosts that they moved my server and is no running PHP5 wih register globals switched off?

So i need to structure the vairalbe like this.

 

$_GET['carid']

 

but im getting this error

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /usr/local/psa/home/vhosts/freecarreviews.com/httpdocs/cars.php on line 28

[Dragen]

okay.

having globals off is an incredibly good thing!

 

How are you setting the $CarID variable? you simply need to carry it for use in your table

[ViN86]

Just found out from my hosts that they moved my server and is no running PHP5 wih register globals switched off?

So i need to structure the vairalbe like this.

 

$_GET['carid']

 

but im getting this error

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /usr/local/psa/home/vhosts/freecarreviews.com/httpdocs/cars.php on line 28

 

first and foremost,  where are you declaring $CarID?

 

secondly, $_GET['carid'] wont work unless your URL looks like pagename.php?carid=value

in which case $_GET['carid'] would return a value of "value"

[iceblox]

carid is specified in the url so its cars.php?carid=1

 

 

[Dragen]

what's line 28?

[ViN86]

carid is specified in the url so its cars.php?carid=1

 

 

so where do you declare

$CarID = $_GET['carid']   

???

 

just cause its in the url doesnt mean the value is automatically set.  you still need to declare the $CarID variable and set it equal to the $_GET['carid'] value thats passed.

[Dragen]

unless he's set his query as this:

$query = "SELECT * FROM Cars WHERE CarID = '" . $_GET['carid'] . "'";