Populating an array with MySQL data

[Rulkster]

Here is the sql statement I am using to grab the data:

$sql2 = "Select scategoryid,scategory_name from subcategory where mcategoryid =\"stationery\" ";
$result2 = @mysql_query($sql2,$db)or die("Couldn't execute query2.");

while ($row = mysql_fetch_array($result2)){
	$scategory_name = $row['scategory_name'];

}

Here is the array, the array I need to populate is within another array

$structure['greetings'] = array(

array(
   "menu"   => "",
   "label"  => "$scategory_name",
   "link"   => "index.php",
   "target" => "",
  ),

);

I tried this but it doesn't work:

$sql2 = "Select scategoryid,scategory_name from subcategory where mcategoryid =\"stationery\" ";
$result2 = @mysql_query($sql2,$db)or die("Couldn't execute query2.");

while ($row = mysql_fetch_array($result2)){
	$scategory_name = $row['scategory_name'];


$display_array .= "
	array(
   \"menu\"   => \"\",
   \"label\"  => \"$scategory_name\",
   \"link\"   => \"index.php\",
   \"target\" => \"\",
  ),
  ";




}


$structure['greetings'] = array(

  $display_array

);

How should I be handling this instead?

[Shrike]

Are you trying to print out an array-like structure, or actually create an array?

[sidney]

 $display_array .= " 
    array( 
   \"menu\"   => \"\", 
   \"label\"  => \"$scategory_name\", 
   \"link\"   => \"index.php\", 
   \"target\" => \"\", 
  ), 
  "; 

//should look more like this i think

$display_array[] = array( 
   "menu"   => "", 
   "label"  => "$scategory_name", 
   "link"   => "index.php", 
   "target" => "" 
  ); 

[Rulkster]

Sidney: I am using the variable $display_array because I may have more than 1 record pulled from the table so that is why I am using .=" "

And therefore using the slash before the quotes to ignore them.

Create the array Shrike

[Rulkster]

Sidney,

I tried that and it worked, but only for the first record. I did this:

$structure['greetings'] = array(

  $display_array[0]

);

but like I said above, i will not know the number of records, and I do not think I can put a for loop with the $structure array.

[sidney]

using[] tell php to auto increment the arry so in a loop ever time
it goes round $display_array[]
becomes
$display_array[0]
next
$display_array[1]
etc

[Rulkster]

I tried that and unfortunetly it does not work. The array: structure['greetings'] only displays the last $display_array[whatever number is here]

I think the problem lies in the final );

$display_array[] = array(
"menu" => "",
"label" => "$scategory_name",
"link" => "http://www.the-petset.com/DEV/MainPage.php?Nav=stationery&Category=$scategory_name",
"target" => ""
);

It should be ), because I am not ready to finish the array, but if I put ), instead of ); I get a parser error on that line.

UGH!!!!!

[sidney]

can i see more of your code

[Rulkster]

SURE,

Here is the whole thing:

<?php
$structure = array();


// main cats

$structure['root'] = array(
  array(
   "menu"   => "greetings",
   "image"  => "sitehelm_menus",
   "label"  => "Greetings",
   "link"   => "MainPage?Nav=greetings",
   "target" => "",
  ),
  array(
   "menu"   => "invitations",
   "label"  => "Invitations",
   "link"   => "MainPage?Nav=invitations",
   "target" => "_blank",
  ),

 array(
   "menu"   => "stationery",
   "label"  => "Stationery",
   "link"   => "MainPage?Nav=stationery",
   "target" => "_blank",
  ),

);

// sub cats


$sql2 = "Select scategoryid,scategory_name from mim_subcategory where mcategoryid =\"stationery\" ";
$result2 = @mysql_query($sql2,$db)or die("Couldn't execute query2.");
$i2=0;
while ($row = mysql_fetch_array($result2)){
	$scategory_name = $row['scategory_name'];

$display_array[] = array( 
   "menu"   => "", 
   "label"  => "$scategory_name", 
   "link"   => "MainPage.php?Nav=stationery&Category=$scategory_name", 
   "target" => "" 
  );	


}

$structure['stationery'] = array(

$display_array[0]

);
?>

Now if I do this, simulating a for loop at the end:

$structure['stationery'] = array(

$display_array[0]

);

$structure['stationery'] = array(

$display_array[1]

);

Only the $display_array[1] record appears.

Thanks for all help so far.

[sidney]

try this

while ($row = mysql_fetch_array($result2)){ 
    $scategory_name = $row['scategory_name']; 

$display_array[] = array(  

   "menu"   => "",  

   "label"  => "$scategory_name",  

   "link"   => "MainPage.php?Nav=stationery&Category=$scategory_name",  

   "target" => ""  

  );     


} 

$structure['stationery'] = array( 
$display_array //edited here
); 
?>   

[Rulkster]

FYI, i also tried this and It did not work:

$structure['stationery'] = array(

$display_array[0]

$display_array[1]
);

Even if that did work above, It would be useless since it do not know before hand how many $display_array[] I need, and you can not put a for loop inside the array.

[Rulkster]

Okay, I tried that and nothing appeared.

Tried:

$structure['stationery'] = array(
$display_array
);

I hate arrays..............

[Rulkster]

Sidney,

Thanks again for your help. I am going to take a break (sometimes it helps) If you can think of anything else, please let me know.

Thanks

[sidney]

then i dont think you need th array asignment

$structure['stationery'] = $display_array;

[Rulkster]

Sidney,

OH MY GOD!!!!!!!!!

that was it, that was the code I needed!!

I can NOT believe I did not think of that first and test it.

Here I am thinking about nested arrrays and how to do a great FOR LOOP and then save the outcome to a STRING and grab data from that STRING, Damn!

TY , TY, TY, TY sidney.. wow!!! it works

Thanks again

Brian

[sidney]

$display_array is an array not a string

i think you need to reead up on arrays