showing results from database

[Porkie]

<?php
$con = mysql_connect("localhost","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$sql = mysql_query("SELECT * FROM Bands ");

while($row = mysql_fetch_array($sql))

 

 

how can i echo the results so they are put in a table going from right to left instead of going down to up ?

 

Example

I want it going like this,

Apple Avacado Bannana .........

 

Cheers

 

[Ken2k7]

Well it's not rocket science. Learn a bit of HTML.

 

<?php
echo '<table><tr>';
while ($row = mysql_fetch_assoc($sql)) {
     echo '<td>';
     // print your stuff here
     echo '</td>';
}
echo '</tr></table>';

[Porkie]

	<?php
echo '<table><tr>';
while ($row = mysql_fetch_array($sql)) {
     echo '<td>';
     echo 'name';
     echo '</td>';
}
echo '</tr></table>';
$con = mysql_connect("localhost","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);


$sql = mysql_query("SELECT * FROM gg_video ");


?>

 

with that code i get this error,

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/myuklive/public_html/New directory/taster.php on line 108

 

on this line

 

while ($row = mysql_fetch_array($sql)) {

 

also i only want it to show  five results so how would i do that ?

 

cheers

 

[wildteen88]

Umm, you put your code in then wrong order. Your while loop needs be placed after this line

$sql = mysql_query("SELECT * FROM gg_video ");

 

If you only want your query to return 5 results then use the LIMIT clause in your query

$sql = mysql_query("SELECT * FROM gg_video LIMIT 5");

[Porkie]

cheers mate,

 

echo '<table><tr>';
while ($row = mysql_fetch_array($sql)) {
     echo '<td>';
     echo $row['name'];
 echo 'http://img.youtube.com/vi/['videoid']/default.jpg';
     echo '</td>';
}
echo '</tr></table>';

 

how would i get this line to work?

 

echo 'http://img.youtube.com/vi/['videoid']/default.jpg';

 

videoid getting the data from my database?

 

cheers

[wildteen88]

You'd use $row['videoid']

 

echo 'http://img.youtube.com/vi/'.$row['videoid'].'/default.jpg';

[chmpdog]

You would have to put it in an array before you can echo it like that

 

$row = mysql_fetch_array($result, MYSQL_BOTH);

[wildteen88]

You would have to put it in an array before you can echo it like that

 

$row = mysql_fetch_array($result, MYSQL_BOTH);

MySQL_fetch_array returns both a numerical and associative array.

[Porkie]

ok so where would i have to put that line?

 

and how could i make that url it creates become an actual image?

 

thanks for all your help

[corbin]

An image tag with an src attribute....

 

 

 

echo '<img src="'.$image_path.'" />';

 

 

 

Hopefully this doesn't offend you, but before trying to use PHP to output HTML, perhaps you should learn HTML.

[chmpdog]

 

 

Hopefully this doesn't offend you, but before trying to use PHP to output HTML, perhaps you should learn HTML.

 

I second that notion, php basically wraps itself around html

[Ken2k7]

I second that notion, php basically wraps itself around html

It does? Hmm... never knew that.

[chmpdog]

I second that notion, php basically wraps itself around html

It does? Hmm... never knew that.

 

well if your creating a webpage, which Im assuming he is