PHP - Displaying Results From A Mysql Database

Hi,
This has been baffling me for a couple hours now and i cant seem to figure it out.
I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site.
Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath.
To help explain i would be looking at a layout as follows:
Code: [Select]
line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25

line 1      line 1      line 1
line 2      line 2      line 2
...      ...      ...
line 25      line 25      line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct?

Thanks,
Alex

Hey all,

First off, if this is in the wrong section I apologize. I wasn't sure if it should be here or the mySQL section.

What's going on is, I'm in the process of learning the Prepared Statement way of doing things and am changing / updating my code to reflect the changes. Everything was going fine until I attempted to do what I could do using old MySQL methods and that is display the queried results on the same page.


I can place a query and display the results as they should be displayed if I only use one block of code. However, if I try to do any additional queries on the same page, they get killed and do not display anything even though I know the query is fine because I can test the exact same syntax below one a different page and it works.

Here's a code snippet for an example:
Code: [Select]

Code:
<table>
<tr>
<td>
// The below code will display a selection box containing various strings such as "hello world", "great to be here", "Wowserz", "this is mind blowing" etc. that are stored in the database.

<?php

   echo "<select = \"SpecialConditions\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
   
       echo "<option value=\"$specialId\">$specialcondition</option>";
   }
}
   echo "</select>";
?>
</td>
<td>
// If I place another fetch query below the above fetch() query, this one will not show up. This one is supposed to display values 1 - 20 that have been stored in the DB.

<?php
echo "<select = \"NumberSets\">";
   if($stmt->num_rows == NULL){
      echo "No results found.";
   }else{
      while($stmt->fetch()){
      
          echo "<option value=\"".$numbers."\">".$numbers."</option>";
      }
   }
echo "</select>";

?>

</td>
</tr>
</table>


What am I doing wrong with this? When I use regular SQL queries I can display multiple results on the same page.

The results are being pulled from two separate joined tables but I don't think that's the issue.

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

Hey guys,
Having a slight problem with part of the code in my index.php file

Code: [Select]
mysql_select_db('db_name', $con);

$result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25");
$resulto = mysql_query("SELECT * FROM spy ORDER BY id desc");
$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))
  {
?>
<div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for  "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div>
<?}?>
</div>
  <div id="login"></div>
          <? include("footer.php"); ?>
</div>
</body>
</html>
I'm getting the following error when viewing the file

Code: [Select]
Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71

where lines 70 and 71 are

$count = mysql_num_rows($resulto);
while($row = mysql_fetch_array($result))

Any ideas on how to fix this?