[Resolved] [Resolved] mySQL Array Help

Reality15

Hello,

I have an output into an array, but I don't know how I'd go about displaying the items I have in the array. I want my list of things to look like this:

http://www.pointbuzz.com/mmrides.htm

But with 3 across instead of 4. There is still a link and another line of text under each 100x75 image.

What would the echo be for this?

Thanks,

~ Reality15
www.mirollercoast.com

planetsim

Ok well here is the logic behind it ill let you do the formatting.

$length = sizeof($myArray);
for ($i = 0; $i < $length; $i++)
{
    if (($i%3)==0)
    {
        //new row
    }
    //info here
}

Reality15

So where would I place the code for the image/text?

robert_w_ark

Hey -- you can always try this...


$count = 0;
$query = mysql_query("SELECT * FROM table") or die(mysql_error());
echo "<table>";
while($row = mysql_fetch_array($query))
{
  if($count == 3)
  {
    echo "\n</tr>\n<tr>";
    echo "\n  <td>\n    <img src=\"someimage.jpg\" border=\"0\" alt=\"\" /><br/>Image Description\n  </td>";
    $count = 0;
  }
  else
    echo "\n  <td>\n    <img src=\"someimage.jpg\" border=\"0\" alt=\"\" /><br/>Image Description\n  </td>";
}

if($count < 3) echo "</tr>";
echo "</table>";

Matt_Kindig

I think robert_w_ark meant to include an increment statement for $count. The corrected code is here:

$count = 0; 
$query = mysql_query("SELECT * FROM table") or die(mysql_error()); 
echo "<table>"; 
while($row = mysql_fetch_array($query)) 
{ 
  if($count == 3) 
  { 
    echo "\n</tr>\n<tr>"; 
    echo "\n  <td>\n    <img src=\"someimage.jpg\" border=\"0\" alt=\"\" /><br/>Image Description\n  </td>"; 
    $count = 0; 
  } 
  else 
    {
    echo "\n  <td>\n    <img src=\"someimage.jpg\" border=\"0\" alt=\"\" /><br/>Image Description\n  </td>"; 
   $count++;
   }
} 

if($count < 3) echo "</tr>"; 
echo "</table>";

Reality15

Thank you very much!

Reality15

Just came across a problem. It doesn't like either way that you guys wrote up the code for some problem. Here is my page, can anyone tell me what's wrong?

<?php
//CONNECT TO THE DATABASE
$link=mysql_connect('HOSTNAME','USERNAME','PASSWORD')
or die('Could not connect: ' . mysql_error());
mysql_select_db('DATABASENAME') or die('Could not select database');

//GET ALL OF THE ENTRIES FROM THE DATABASE
$result = mysql_query("SELECT `id` , `name` , `opened` , `manufacturer` , `image` FROM `rollercoasters` ORDER BY `name` DESC LIMIT 0 , 50");
if (!$result) { 
echo mysql_error(); 
exit; 
}
while ($array = mysql_fetch_array($result)) {
$id = $array['id'];
$name = $array['name'];
$opened = $array['opened'];
$manufacturer = $array['manufacturer'];
$image = $array['image'];
}

//DISCONNECT FROM THE DATABASE
mysql_close($link);

include "skin_header.php";
?>
                <tr>
                  <td width="100%" height="100%" valign="top">
                  <table border="0" width="100%" cellspacing="0" cellpadding="5" height="189">
                    <tr>
                      <td width="100%" valign="top" height="176">
                        <p align="center"><b><font face="Arial" size="3">Roller
                        Coasters</font></b></p>
                        <table border="0" width="100%" cellspacing="0" cellpadding="5">

<?php
$length = sizeof($array); 
for ($i = 0; $i < $length; $i++) 
{ 
if (($i%3)==0) 
{ 
    echo "</tr><tr><td width='33%'><img border='0' src='attractions/rollercoasters/$image.gif' width='102' height='77'><br><font face='Arial' size='2'><u><a href='viewcoaster.php?id=$id'><font color='#000000'>$name</font></a></u><br></font><font face='Arial' size='1'>$opened - $manufacturer</font></td>"; 
} 
else
{
    echo "<td width='33%'><img border='0' src='attractions/rollercoasters/$image.gif' width='102' height='77'><br><font face='Arial' size='2'><u><a href='viewcoaster.php?id=$id'><font color='#000000'>$name</font></a></u><br></font><font face='Arial' size='1'>$opened - $manufacturer</font></td>";
} 
?>

                    </table>
                  </td>
<?php
	include "skin_footer.php";
?>

And I did try it without the things you guys gave me and it at least showed my site's skin.

Any help is greatly appreciated.

Thanks,

~ Reality15

Matt_Kindig

A few things I've noticed....
1. Each time the while() loop iterates, $array is reassigned to the next row in your MySQL result set, so that $array is always just the last row.
2. The length of that array will be the number of columns in the row, which would be id+name+opened+manufacturer+image=5, which is NOT the total number of rows in the result set (up to 50).
3. This length isn't useful anyway, because what you want is the number of rows in the result set, not the number of columns.

Instead of trying to make this code work, I've rewritten it below.

<?php 
//CONNECT TO THE DATABASE 
 $link=mysql_connect('HOSTNAME','USERNAME','PASSWOR
D') 
or die('Could not connect: ' . mysql_error()); 
mysql_select_db('DATABASENAME') or die('Could not select database'); 

//GET ALL OF THE ENTRIES FROM THE DATABASE 
$result = mysql_query("SELECT `id` , `name` , `opened` , `manufacturer` , `image` FROM `rollercoasters` ORDER BY `name` DESC LIMIT 0 , 50"); 
if (!$result) {  

echo mysql_error();  

exit;  

} 
//DISCONNECT FROM THE DATABASE 
mysql_close($link); 

include "skin_header.php"; 
?> 
                <tr> 
                  <td width="100%" height="100%" valign="top"> 
                  <table border="0" width="100%" cellspacing="0" cellpadding="5" height="189"> 
                    <tr> 
                      <td width="100%" valign="top" height="176"> 
                        <p align="center"><b><font face="Arial" size="3">Roller 
                        Coasters</font></b></p> 
                        <table border="0" width="100%" cellspacing="0" cellpadding="5"> 

<?php 
$i=0;
while ($array = mysql_fetch_array($result)) { 
extract($array);
if($i==3) { 
echo '</tr><tr>';
$i=0;
} else {
$i++;
} // end if
    echo "<td width='33%'><img border='0' src='attractions/rollercoasters/$image.gif' width='102' height='77'><br><font face='Arial' size='2'><u><a href='viewcoaster.php?id=$id'><font color='#000000'>$name</font></a></u><br></font><font face='Arial' size='1'>$opened - $manufacturer</font></td>";  

} //end while
?> 

                    </table> 
                  </td> 
<?php 
    include "skin_footer.php"; 
?>

Note the use of extract. It is a quick way to create variables from the key names of an array, i.e., it does the same thing as

$id = $array['id']; 
$name = $array['name']; 
$opened = $array['opened']; 
$manufacturer = $array['manufacturer']; 
$image = $array['image'];

EDITS: Some grammar edits at the beginning of this post. I have to learn to stop trying to write coherently late at night!

Reality15

Thank you very much!

You can see this (finally) in action at http://www.mirollercoast.com/rollercoasters.php

Thanks again!