List everything that meets criteria

[Deacon Frost]

Ok, I've selected the rows from my database, and assigned it to variables.

Now I need to list everything that falls under the "Where row=value" criteria. Basically a giant list of everything that is in it.

My problem understanding how it would work is, i don't see how the variables would change less i changed the extraction method...


I also need an advanced MySQL query, how would I set more than one criteria. Would it be like...

SELECT * FROM table WHERE row=value && row=value ?

And...

After all of the above, where I get everything..

How would I set it so it displays, for instance 30, random results that match the criteria?

[Andrew Johnson]

Originally Posted by Deacon Frost

Now I need to list everything that falls under the "Where row=value" criteria. Basically a giant list of everything that is in it.

PHP Code:

<table border="0" cellpadding="3" cellspacing="0">
   <tr>
      <th>Col1</th>
      <th>Col2</th>
      <th>Col3</th>
   </tr>
<?php

$query = mysql_query("SELECT * FROM table");
while ($results = mysql_fetch_array($query))
{
?>
   <tr>
      <td><?php echo $results["Col1"]; ?></td>
      <td><?php echo $results["Col2"]; ?></td>
      <td><?php echo $results["Col3"]; ?></td>
   </tr>
<?php
}

?>

Originally Posted by Deacon Frost

I also need an advanced MySQL query, how would I set more than one criteria.

PHP Code:

SELECT * FROM table WHERE col='value' AND col2='value2' 


Originally Posted by Deacon Frost

How would I set it so it displays, for instance 30, random results that match the criteria?

PHP Code:

SELECT * FROM table LIMIT 30 


Unfortunately there is no random command for MySQL

Hopefully this helps, you'll need to change the names of everything though since you didn't post your schema...

[Deacon Frost]

thanks! I'ma update right now to see if I can alter it correctly to get it to work.

Unfortunately, everything above is gonna be used on one page XD!

EDIT:

But wait, I need to display 30 of the results using variabls. saying col1, col2, and col3 will only display the same results...

like so:

http://downstage.tv/test.php

:S

[Deacon Frost]

PHP Code:

<?php
include("/home/stage/public_html/css/inc/theaterconn.php"); 

$sql = "SELECT id,filmid,name FROM theater WHERE status='2' LIMIT 2 ORDER BY id";
$result = mysql_query( $sql ) or die('mysql_error()' . 'Error:  ' . mysql_errno() );
$row = mysql_fetch_assoc($result);
$num = mysql_num_rows($result);

$id = $row['id'];
$filmid = $row['filmid'];
$name = $row['name'];

?>



<table>
<tr>
<td>
<form action="theater.php" method="get">
<input type="hidden" name="id" value ="<? echo $id ?>">
<input type="image" src="http://images.stage6.com/video_images/<? echo $filmid ?>t.jpg" value="Submit" alt="Go to Theater">
<br />
<center><? echo $name ?></center>
<tr>
<td>
<input type="hidden" name="id" value ="<? echo $id ?>">
<input type="image" src="http://images.stage6.com/video_images/<? echo $filmid ?>t.jpg" value="Submit" alt="Go to Theater">
<br />
<center><? echo $name ?></center>
</form>
</tr>
</table>

mysql_error()Error: 1064

Basically... I need it to behave much like that of comments. It needs to show the data in a table listed by the ID number, displaying only those that meet the criteria of the SQL database, and I want it to display 30 of them.

EDIT: (if I take out the LIMIT 2, then it shows TWO of the same image w/ the same database data. Instead of retrieving the other row's data and putting it in there...)


After trying something else:

PHP Code:

<?php
include("/home/stage/public_html/css/inc/theaterconn.php"); 

$sql = "SELECT id,filmid,name FROM theater WHERE status='2' ORDER BY id";
$result = mysql_query( $sql ) or die('mysql_error()' . 'Error:  ' . mysql_errno() );

while($row = mysql_fetch_array($result));


$id = $row['id'];
$filmid = $row['filmid'];
$name = $row['name'];
 
{

echo "<table>";
echo "<tr>";
echo "<td>";
echo "<form action='theater.php' method='get'>";
echo "<input type='hidden' name='id' value="; 
echo $id . ">";
echo "<input type='image' src='http://images.stage6.com/video_images/";
echo $filmid . "t.jpg' value='Submit' alt='Go to Theater'>";
echo "<center>"; 
echo $name . "</center>";
echo "</form>";
echo "</tr>";
echo "</table>";
}
?>

It kept some things, but some things didn't work, experimenting w/ that to see what I get... >.>


(It seems when I do this, it doesn't display the image =/.)