What im i doing wrong??

[Vinnepin]

Hey guys,

I have 2 almost equal function but one of then gives me an error and i really dont know why 🙁

Here are the 2 functions :

This one he does ok :

function cato() {
global $description;

$sql = mysql_query("select * from file_cat order by cat_id asc"); # De query word gemaakt

while($row = mysql_fetch_array($sql))
  {    # Hier worden alle waardes aan de variablen gekoppeld
        $cat_id      = $row['cat_id'];
    	$cat_name    = $row['cat_name'];
        $cat_desc    = $row['cat_desc'];

    $sql2 = mysql_query("select * from file_specs where cat_id='$cat_id'");
    $aantalfiles = mysql_num_rows($sql2);

    $catogories = ("<BR>&nbsp;<strong><img src=\"images/arrow.gif\" width=\"9\" height=\"11\"><a href=\"index.php?event=cat&cat=$cat_id\">$cat_name</strong><br>
    &nbsp;&nbsp;&nbsp;- $cat_desc<BR>
    &nbsp;&nbsp;&nbsp;- aantal files : $aantalfiles<br></a>");

    $description["catogories"] .= $catogories;

 } # Einde while loop

display_template('catogories'); # Shows the catagories

unset ($description["catogories"]);
} # einde function

And this one gives me an error :

function nieuwstefiles() {
global $description;

$sql = mysql_query("select * from file_specs order by fid asc"); # De query word gemaakt

while($row = mysql_fetch_array($sql))
  {
  # Hier worden alle waardes aan de variablen gekoppeld
  $fid         = $row['fid'];
  $size        = $row['size'];
  $name        = $row['name'];
  $description = $row['description'];
  $date        = $row['date'];
  $hits        = $row['hits'];
  $rating      = 75;

  $files = "<strong><img src=\"images/arrow.gif\" width=\"9\" height=\"11\">
  <a href=\"$PHP_SELF?event=file&id=$fid\" onMouseOver=\"drc('<strong>Desctiption : </strong>$description<BR><strong>Size : </strong>$size<BR><strong>Date : </strong>$date<BR><strong>Hits : </strong>$hits<BR><strong>Rating : </strong>$rating <BR>','$name','<img src=images/dl_button2.gif>');\" onMouseOut=\"nd();\">
  $name<br></strong>&nbsp;&nbsp;- posted on : $date</a><br><BR>";

  $description["files"] .= $files;

 } # Einde while loop

display_template('newfiles'); # Shows the catagories

unset ($description["files"]);
} # einde function

This is the error i get fot his function :
Fatal error: Cannot use assign-op operators with overloaded objects nor string offsets in c:\apache\util\download_system\functions.php on line 145

This error reffers to this line in the second funtion :
$description["files"] .= $files;

Do u guys know what i am doing wrong???

Any help appreciated 😃

Vinnie

[su1d]

$description = $row['description'];

I'm not sure whether you really wanted to use $description here. It's like a global variable. Try to use something like $desc instead of $description in this line.

[Vinnepin]

No that not it.

$description["files"] .= $files; <- Here im putting all the info in $files into the array $description
for to use it with templates u see

And the $row u use is an query var to pull data out the database

Vinnie

[su1d]

$description = $row['description'];

here you store a string into the variable $description (which is BTW has been declared as global in the begining)

$description["files"] .= $files;

and now you treat $description as an array (read the line above - it is a string now! string !== array).

Do me (yourself?) a favor: change "$description = $row['description'] to $desc = $row['description'];
and change $description to $desc in the line with $files = ...

Try that out first, and tell me if the error has disappeared.

[128kb_com]

su1d is correct, you are asinging $description a string value (php will automatically do the type conversions for you) and the you use it as an array. Maybe you want to try declaring

$description = array();

or casting the array type
code=php $description;[/code]
or using the []'s to have php do the type conversion for you:

$description[] = $row['description'];