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string variable not working

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  #1  
Old June 5th, 2003, 06:06 AM
tansan tansan is offline
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string variable not working
hi!

need help to resolve this...

i'm doing a popup menu in javascript and php. i noticed that if i do a

php Code:
Original - php Code 
  1.  
  2. <script language="Javascript">
  3.   x = computer;
  4.   category.addMenuItem(x)
  5. </script>


then the x is added to the menu, but everytime i do

php Code:
Original - php Code 
  1.  
  2. <script language="Javascript">
  3.   x = <?php echo $row['prod_category'] ?>;
  4.   category.addMenuItem(x)
  5. </script>


then nothing happens.
php Code:
Original - php Code 
  1. <?php echo $row['prod_category'] ?>
returns a value of 'computer', though. note that 'category' is a class object.

any idea here?

thanks!
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  #2  
Old June 5th, 2003, 07:20 AM
crisp crisp is offline
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RE: string variable not working
Like in PHP strings should be enclosed between quotes:

<script type="text/javascript" language="Javascript">
x = '<?php echo $row['prod_category'] ?>';
category.addMenuItem(x);
</script>
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  #3  
Old June 5th, 2003, 08:41 AM
tansan tansan is offline
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RE: string variable not working
thanks crisp!

by the way, sorry for not emphasizing this in my previous post but doing a

x = 'computer'

doesn't work either so it seems that assigning the variable x with a string value causes the object category not to accept x. and perhaps javascript treats <?php echo $row['prod_category'] ?> as string. this is what i actually mean in this post's title...
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  #4  
Old June 5th, 2003, 10:01 AM
crisp crisp is offline
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RE: string variable not working
javascript will only treat a value as string when it is enclosed in quotes. Else it will treat it as a number or an object reference.
The fact that you write your script with PHP shouldn't matter, what matters is what is sent to the browser, which you can easily see by using view-source in your browserscreen.

Could you post some more of this script? I would assume that you need to sent a string to your addMenuItem method..
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  #5  
Old June 5th, 2003, 11:03 AM
tansan tansan is offline
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RE: string variable not working
here's the code. it's a bit long:

php Code:
Original - php Code 
  1.  
  2. <head>
  3. <title>Document1</title>
  4. <script language="JavaScript">
  5. function mmLoadMenus() {
  6.     if (window.category) return;        
  7.     
  8.     <!-- create objects for categories. database driven
  9.     <?php while ($rowCategory = $rsCategory->fetchRow()) { ?>      
  10.         var cat = '<?php echo $rowCategory[1] ?>';
  11.         window.cat = new Menu("cat",0,17,"Verdana, Arial, Helvetica, sans-serif",11,"#000000","#FFFFFF","#CCCCCC","#000084","left","middle",3,0,1000,-5,7,true,true,true,0,true,true);  
  12.       <!-- place submenus. should be database driven also      
  13.         cat.addMenuItem(subcat1);
  14.         cat.addMenuItem(subcat2);      
  15.     <?php } ?>
  16.     
  17.     <!-- create the root menu
  18.     window.category = new Menu("root",0,17,"Verdana, Arial, Helvetica, sans-serif",11,"#000000","#FFFFFF","#CCCCCC","#000084","left","middle",3,0,1000,-5,7,true,true,true,0,true,true);
  19.   
  20.     <!-- populate the root menu with created objects above. database driven
  21.     <?php while ($rowCategory = $rsCategory->fetchRow()) { ?>  
  22.         var i = '<?php echo $rowCategory[1] ?>';    
  23.         category.addMenuItem(i);    
  24.     <?php } ?>
  25.     category.writeMenus();
  26. } 
  27. </script>
  28. <script language="JavaScript" src="mm_menu.js"></script>
  29. </head>


i manually coded the the script then use the data from dbase and it ran properly. i just notice that doing a category.addMenuItem(prod1); works fine but category.addMenuItem("prod1"); didn't.
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  #6  
Old June 5th, 2003, 12:38 PM
crisp crisp is offline
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RE: string variable not working
try changing

var i = '<?php echo $rowCategory[1] ?>';
category.addMenuItem(i);

into

category.addMenuItem(<?php echo $rowCategory[1] ?>);
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