[SOLVED] How do I do an if statement for current page?

[TechMistress]

I have a list that shows on a page from a mysql database.  I've made the queries and I've created the following function:

 

        $current = $_GET['cat'];

 

How do I write the "if" statement  that if cat = $current then echo blah blah blah else echo yadda yadda yadda

 

Here's a snippet:

 

         $uSql = "SELECT cat_name, cat_id FROM categories WHERE parent_id ='0' ORDER by cat_name";
         $uResult = mysql_query($uSql);
         if(!$uResult){
         echo 'no data found';
         }else{
         $i = 1;
         while($uRow = mysql_fetch_array($uResult)){
         $pid = $uRow["cat_id"];
         $cat_name = $uRow['cat_name'];
         $current = $_GET['cat'];
         //$keys = array_keys ($cat_name);
                     ?>
  <tr>
    <td>
   <img src="/img/down_arrow.gif"> <a href="#" onclick="toggle_visibility('<? echo "$i"; ?>');"><? echo "$cat_name"; ?></a>
   
                  <? $sql2  = "SELECT cat_name, cat_id FROM categories WHERE parent_id='$pid'";
                     $result = mysql_query($sql2);
                     if(!$result){
                     echo 'no data found';
                     }else{
                     echo "<table id=" . $i ." style='display:none'>";
                     while($row = mysql_fetch_array($result)){
                     $cname = $row['cat_name'];
                     $cid = $row['cat_id'];
                     ?>

[trq]

How do I write the "if" statement  that if cat = $current then echo blah blah blah else echo yadda yadda yadda

 

Exactly as you said it.

 

if ($cat == $current) {
  echo "blah blah blah";
} else {
  echo "yadda yadda yadda";
}

 

I'm not exactly sure what your talking about when you refer to cat but you get the idea I'm sure.

[TechMistress]

That's the syntax I was looking for - it's perfect. Unfortunately it doesn't work with my code.  I just had an idea to try. Not the fault of your code, of course, but of mine.

 

I just wanted to find a way figure out how to determine what category the person is currently viewing, and the page url looks something like "projects.php?cat=55".  So then on the menu, cat_id 55 would be bold or something like that.

 

I was looking at an include for the page that says $page->SetParameter('CURRENT_CAT', $_GET['cat']); so I tried to rework it.

 

I was trying too hard

 

thanks!

[justinh]

 
<?php
if(isset($_GET['cat'])){ // checks to see if cat is set 

       $catid = $_GET['cat']; 

       echo "<b>".$catid."</b>"; 
?>

}

 

If you would want to display the name of the category instead of the id: (assuming you have a table with catid and catname)

 

 
<?php
if(isset($_GET['cat'])){ // checks to see if cat is set 

       $catid = $_GET['cat']; 
       $query = "SELECT * FROM tbl_cats WHERE catid = $catid"; 
       $doit = mysql_query($query) or die("Error:" . mysql_error()); 
       if(!mysql_num_rows($doit)){ 
       echo "Category Not found"; 
       }else{ 
       while($getcatname = mysql_fetch_array($doit)){ 
       $catname = $getcatname['catname'];    
       } 
       echo "<b>" . $catname . "</b>"; 
       }
     }
?>

 

this isn't tested, but should give you some idea.

 

[TechMistress]

Well that will work great on the top of the page.

 

This menu is a show/hide menu:

 

<script type="text/javascript">
<!--
    function toggle_visibility(id) {
       var e = document.getElementById(id);
       if(e.style.display == 'block')
          e.style.display = 'none';
       else
          e.style.display = 'block';
    }
//-->
</script>

 

What I'd like to do is if the cat is set, to open the toggle, but I'm not even sure if I understand how the toggle open/close is working, since I was given the code by a colleague. As it is, all main items are closed - when clicked they open:

 

<a href="#" onclick="toggle_visibility('<? echo "$i"; ?>');"><? echo "$cat_name"; ?></a>

 

I was hoping to at least bold the current cat being viewed - at most to open it up.

 

What do you think?

 

[TechMistress]

Or How about this:  I can display the current category and it's subs in one place, and I'll leave my previously mentioned menu where it is.

 

This is the code you wrote, which I modified to suit my table:

 

<?php
if(isset($_GET['cat'])){ // checks to see if cat is set 

       $catid = $_GET['cat']; 
       $query = "SELECT * FROM categories WHERE cat_id = $catid"; 
       $doit = mysql_query($query) or die("Error:" . mysql_error()); 
       if(!mysql_num_rows($doit)){ 
       echo "Category Not found"; 
       }else{ 
       while($getcatname = mysql_fetch_array($doit)){ 
       $catname = $getcatname['cat_name'];
       } 
       echo "<span class='title1'>" . $catname . " Projects</span>";
       }
     }
?>

 

That currently displays, for example, this html output:

 

Manufacturing Projects

 

What if we can add to your code above to make it display this output:

 

Manufacturing Projects

sub1 | sub2 | sub3 | sub4 | sub5 | sub6 | sub7 | sub8

 

Then the menus stay where they are, but the current cat and it's subs are displayed by themselves where I want them at the top.

 

[Maq]

Assuming the sub-categories are "cat_name" from the "categories" table with a parent_id of $catid (the one from the URL).

 

if(isset($_GET['cat'])){ // checks to see if cat is set 
   $catid = $_GET['cat']; 
   $query = "SELECT * FROM categories WHERE cat_id = $catid"; 
   $query2 = "SELECT cat_name FROM categories WHERE parent_id = '$catid'";
   $doit = mysql_query($query) or die("Error:" . mysql_error());
   $doit2 = mysql_query($query2) or die(mysql_error());
   if(!mysql_num_rows($doit)){ 
      echo "Category Not found"; 
   } else { 
      while($getcatname = mysql_fetch_array($doit)){ 
      $catname = $getcatname['cat_name'];
      } 
         echo "" . $catname . " Projects
";
	 if(!mysql_num_rows($doit2)) {
            echo 'no data found';
         } else {
            while($row = mysql_fetch_array($result)){
		echo "" . $row['cat_name'] . " ";
         }
      }
   }
}	 
?>

[TechMistress]

Ahhh, that looks like it's the one!  I changed $result to $doit2 (oops).

 

I want to make the subs output as a link, so would I just add a new element for the cat id of the subs?  Like this:

 

            while($row = mysql_fetch_array($doit2)){
            $subid= $doit2['cat_id'];
         echo "<strong>" . $row['cat_name'] . " </strong>";

 

and then put my link in the echo?

[TechMistress]

Aha!  Got it!

 

Here's the final code:

<?php
if(isset($_GET['cat'])){ // checks to see if cat is set 
   $catid = $_GET['cat']; 
   $query = "SELECT * FROM categories WHERE cat_id = $catid"; 
   $query2 = "SELECT cat_name,cat_id FROM categories WHERE parent_id = '$catid'";
   $doit = mysql_query($query) or die("Error:" . mysql_error());
   $doit2 = mysql_query($query2) or die(mysql_error());
   if(!mysql_num_rows($doit)){ 
      echo "Category Not found"; 
   } else { 
      while($getcatname = mysql_fetch_array($doit)){ 
      $catname = $getcatname['cat_name'];
      } 
         echo "<span class='title1'>" . $catname . " Projects</span><br />";
       if(!mysql_num_rows($doit2)) {
            echo 'no data found';
         } else {
	 echo "Sub-Categories: ";
            while($row = mysql_fetch_array($doit2)){
		$subid = $row[cat_id];
         echo "<strong><a href='projects.php?cat=$subid'>" . $row['cat_name'] . " </a></strong> | ";
         }
      }
   }
}    
?>

 

Thank You!  I've actually learned alot from this exercise. You're the best!

 

[Maq]

Ok I see.  You wanted to make the sub categories actual links.  Glad you figured it out.  The code is a little sloppy and lacks documentation (my fault) but if it works that's all that matters right now, but I would strongly urge that you document so you understand what's going on in case you have to refer to that code again.