foreach

[raptor30506090]

Hi can any one help please im still very new but hard trying

 

The problem is that i only get 1 result back from the database i need it to loop any ideas please

	$query = mysql_query("SELECT * FROM categorys, sub_categorys WHERE categorys.cat_id = sub_categorys.cat_id");
        
	echo "<ul>";
	$row[] = mysql_fetch_array($query);
				  foreach($row as $category){
                  echo "<li>";
	              echo $category['category'];

			      foreach($row as $sub_category){
		          if($category['cat_id'] == $sub_category["cat_id"]){
			      echo "<ul><li>".$sub_category["sub_category"]."</li></ul>";
		          }
		       }
            echo "</li>";
                }
              echo "</ul>";

[xyph]

http://www.php.net/manual/en/function.mysql-fetch-assoc.php#90030

[raptor30506090]

Sorry no errors

its getting back

 

•Home

◦our staff

 

thats it need it to loop get rest of the menu

 

<?php 

	$query = mysql_query("SELECT * FROM categorys, sub_categorys WHERE categorys.cat_id = sub_categorys.cat_id")or die(mysql_error());
        
	echo "<ul>";
	$row[] = mysql_fetch_array($query);
				  foreach($row as $category){
                  echo "<li>";
	              echo $category['category'];

			      foreach($row as $sub_category){
		          if($category['cat_id'] == $sub_category["cat_id"]){
			      echo "<ul><li>".$sub_category["sub_category"]."</li></ul>";
		          }
		       }
            echo "</li>";
                }
              echo "</ul>";
?>

[xyph]

Because you're only fetching one result. You have to use mysql_fetch_xxx in a loop to grab every row returned.

[raptor30506090]

ok changed that to this now and i now get 2 submenu results but still not doing the job

<?php 

	$query = mysql_query("SELECT * FROM categorys, sub_categorys WHERE categorys.cat_id = sub_categorys.cat_id")or die(mysql_error());
        
	echo "<ul>";
	$row[] = mysql_fetch_array($query);
				  foreach($row as $category){
                  echo "<li>";
	              echo $category['category'];

	$row[] = mysql_fetch_array($query);
			      foreach($row as $sub_category){
		          if($category['cat_id'] == $sub_category["cat_id"]){
			      echo "<ul><li>".$sub_category["sub_category"]."</li></ul>";
		          }
		       }
            echo "</li>";
                }
              echo "</ul>";
?>

[ManiacDan]

$row[] = mysql_fetch_array($query);

 

 

You need to use mysql_fetch_array IN THE LOOP.

 

The manual page you were linked to contains examples on how to do it properly. 

[raptor30506090]

whats the manual page?

 

Thank for your help

[xyph]

Search this page for the words 'manual'

 

One of those results will be a link

[raptor30506090]

Yes had a look and now it returns this as you can see it nown returns two home links there should only be one and also a service link as well below the about driving me mad  thanks for this help

Home
        ◦our history
        ◦our staff
Home
        ◦our history
        ◦our staff
        ◦plastering
About
        ◦why we started
        ◦about the company
Contact
        ◦company policy

 

<?php 

	$query = mysql_query("SELECT * FROM categorys, sub_categorys WHERE categorys.cat_id = sub_categorys.cat_id")or die(mysql_error());
        

	echo "<ul>";
	while($row[] = mysql_fetch_array($query)){
				  foreach($row as $category){ 
				 }
                  echo "<li>";
	              echo $category['category'];
				  




	$row[] = mysql_fetch_array($query);
			      foreach($row as $sub_category){
		          if($category['cat_id'] == $sub_category["cat_id"]){
			      echo "<ul><li>".$sub_category["sub_category"]."</li></ul>";
		          }
		       }
            echo "</li>";
                }
              echo "</ul>";
		  
?>

[ManiacDan]

Ignore the specific example that you were linked to, since it's a more advanced tactic than you can handle at the moment.  The example from the top of the page is easier to follow, look:

while ($row = mysql_fetch_assoc($result)) {
    echo $row["userid"];
    echo $row["fullname"];
    echo $row["userstatus"];
}

You don't need the $row[] syntax or your inner foreach.

[raptor30506090]

:'( mm sorry cant see how that is going to help im trying to make a drop down dymanic menu if i do two querys how do i compare the id fields?

 

Many thanks Daz

[ManiacDan]

Because this is wrong:

 

		while($row[] = mysql_fetch_array($query)){
				  foreach($row as $category){ 

 

And this is right:

 

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

 

[raptor30506090]

Still having probs with this any more help would be great this works but when i try to get it from database thats when it go's wrong.

 

<?php

$conn = mysql_connect("localhost","root","");
$db = mysql_select_db("array", $conn);

// array.categorys
$categorys = array(
  array('cat_id'=>1,'category'=>'Home'),
  array('cat_id'=>2,'category'=>'About'),
  array('cat_id'=>3,'category'=>'Services'),
  array('cat_id'=>4,'category'=>'Contact')
);

// array.sub_categorys
$sub_categorys = array(
  array('sub_cat_id'=>1,'cat_id'=>3,'sub_category'=>'our history'),
  array('sub_cat_id'=>2,'cat_id'=>3,'sub_category'=>'why we started'),
  array('sub_cat_id'=>3,'cat_id'=>3,'sub_category'=>'about the company'),
  array('sub_cat_id'=>4,'cat_id'=>3,'sub_category'=>'our staff'),
  array('sub_cat_id'=>5,'cat_id'=>3,'sub_category'=>'plastering'),
  array('sub_cat_id'=>6,'cat_id'=>3,'sub_category'=>'plumbing'),
  array('sub_cat_id'=>7,'cat_id'=>3,'sub_category'=>'company policy')
);

echo "<ul>";
foreach($categorys as $category){
echo "<li>";
	echo $category['category'];

	foreach($sub_categorys as $sub_category){
		if($category['cat_id'] == $sub_category["cat_id"]){
			echo "<ul><li>".$sub_category["sub_category"]."</li></ul>";
		}
	}
echo "</li>";
}
echo "</ul>";
?>

 

 

 

Database side this is how fare i have got

 

<?php $sql = "SELECT * FROM categorys INNER JOIN sub_categorys ON categorys.cat_id = sub_categorys.cat_id ORDER BY categorys.cat_id, sub_categorys.sub_cat_id";
      $query = mysql_query($sql) or die(mysql_error()." $sql");$Category = '';

echo "<ul>";
while($row = mysql_fetch_array($query)){	
if ($row['category'] != $Category){		
if ($Category != ''){			
echo "</ul>";		
}		
$Category = $row['category'];		
echo "<li>" . $row['category'] . "</li>";		
echo "<ul>";}	
echo "<li>".$row['sub_category']."</li>";}

if ($Category != ''){	
echo "</ul>";}
echo "</ul>"; 
?>

17163_.gz

[ManiacDan]

Please indent your code properly and put braces on their own lines so we can see where your control structures begin and end.

 

what does the result set look like for this sql query?

[raptor30506090]

Hi the problem is that i want to show all the menu as well as sub menu this will show all if i put sub menu to each.

 

Home

        submenu

 

About

      submenu

 

but id i dont add a submenu then you cant see the above Home, About, Contact

 

Thank you for any help

 

Darren

[ManiacDan]

Then maybe you want a left join.

 

Or you could provide the two things I asked for last time.

 

I'm going to go ahead and ignore this thread from now on.  It really isn't that hard to copy/paste the results of your query and I'm not getting paid to beg you for the information I need to help you.

[raptor30506090]

Hi Guys

 

Any ideas to get this dusted

 

Thanks Daz

[Pikachu2000]

Please indent your code properly and put braces on their own lines so we can see where your control structures begin and end.

 

what does the result set look like for this sql query?