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Populating a Drop-down List

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Join Date: Dec 2006

Posts: 71

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pygmalion

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Junior Poster in Training

Populating a Drop-down List

Apr 18th, 2007

hi

i have the following issue. i'd much appreciate someone helping me out here.

i'm trying to let a drop-down list (which will be placed, and vary on many pages) with the contents of an array. how can i get that done?

plus, once a certain element has been selected, i'd like that a link appears that links to a specific (filename with certain naming, according to the selected element) file. how could that be implmented?

again, i'd much apprecite your help
pygmalion

Microsoft Employee: "Hey, it compiles! Ship it."

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Join Date: Apr 2007

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landae

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Re: Populating a Drop-down List

Apr 18th, 2007

Below is a basic example of how I wrote an array drop down for a choice of messengers. $msgr is simply a post result for msgr, so it retains what you selected.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
<fieldset><label for="msgr">Primary Messenger:</label>
<?php
	echo "<select name=\"msgr\" id=\"msgr\">\n";
	$msgrArr = array();
	$msgrArr[] = "MSN";
	$msgrArr[] = "AIM";
	$msgrArr[] = "YIM";
	$msgrArr[] = "ICQ";
	$msgrArr[] = "Jabber/Gtalk";
	foreach($msgrArr as $v){
		if ($msgr == $v) {
			echo "<option value=\"$v\" selected=\"selected\">$v</option>\n";
		} else {
			echo "<option value=\"$v\">$v</option>\n";
		}
	}
	echo "</select>";
?>
</fieldset>
<fieldset><label for="msgr">Primary Messenger:</label>
<?php
	echo "<select name=\"msgr\" id=\"msgr\">\n";
	$msgrArr = array();
	$msgrArr[] = "MSN";
	$msgrArr[] = "AIM";
	$msgrArr[] = "YIM";
	$msgrArr[] = "ICQ";
	$msgrArr[] = "Jabber/Gtalk";
	foreach($msgrArr as $v){
		if ($msgr == $v) {
			echo "<option value=\"$v\" selected=\"selected\">$v</option>\n";
		} else {
			echo "<option value=\"$v\">$v</option>\n";
		}
	}
	echo "</select>";
?>
</fieldset>

As for the other thing your on about, perhaps you want to write some javascript thats populated then just shows the data. Alternatively you could use AJAX or simply reload the page based on the selection. Perhaps you can explain what your after more in depth.

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Join Date: Apr 2007

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dr4g

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Re: Populating a Drop-down List

Apr 20th, 2007

Another solution would be, instead of using a foreach() you could use a for() loop.

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$arr = array("cat", "dog", "horse", "frog");
 
echo "<select name=dropdown_list><option value="bleh">Choose Me</option>";
 
// loop while $i is less than count(the amout of elements) of $arr
for($i=0; $i < count($arr); $i++)
   echo <option value=". $arr[$i] .">" . $arr[$i] . "</option>";
 
echo "</select>";
 
$arr = array("cat", "dog", "horse", "frog");
 
echo "<select name=dropdown_list><option value="bleh">Choose Me</option>";
 
// loop while $i is less than count(the amout of elements) of $arr
for($i=0; $i < count($arr); $i++)
   echo <option value=". $arr[$i] .">" . $arr[$i] . "</option>";
 
echo "</select>";

Enjoy.

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Join Date: Dec 2006

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pygmalion

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Junior Poster in Training

Re: Populating a Drop-down List

May 1st, 2007

thanks a lot for your elaborate replies.
the populating of the drop-down list now works fine.

the case is as follows, i'm trying to populate the dropdown-list via a multidimensional array (because they will be audio recordings of parts of each chapter of a book).
now, first i'd like to declare the multi-dimensional array, which i have done as follows.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$book1 = array("chapter1"    =>    array(),
                     "chapter2"    =>    array(),
                     "chapter3"    =>    array());
$book1 = array("chapter1"    =>    array(),
                     "chapter2"    =>    array(),
                     "chapter3"    =>    array());

now, i'd like to populate the second dimension of this array by using for-loops, as follows:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
    $chapter1_amount = 4;
    $chapter2_amount = 6;
    $chapter3_amount = 8;
 
    for($i=1; $i <= $chapter1_amount; $i++)
    {
        $book1["chapter1"][] =  $i;
    }
 
    for($i=1; $i <= $chapter2_amount; $i++)
    {
        $book1["chapter2"][] =  $i;
    }
 
    for($i=1; $i <= $chapter3_amount; $i++)
    {
        $book1["chapter3"][] =  $i;
    }
    $chapter1_amount = 4;
    $chapter2_amount = 6;
    $chapter3_amount = 8;

    for($i=1; $i <= $chapter1_amount; $i++)
    {
        $book1["chapter1"][] =  $i;
    }
    
    for($i=1; $i <= $chapter2_amount; $i++)
    {
        $book1["chapter2"][] =  $i;
    }

    for($i=1; $i <= $chapter3_amount; $i++)
    {
        $book1["chapter3"][] =  $i;
    }

i think herein lies the mistake, at declaring the variables at the 2nd dimension. am i doing this correctly?

afterwards, i show the drop-down list as you guys have suggested:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
echo "<select name=\"book1\"><option value=\"bleh\">Choose Part</option>";
 
    foreach($book1 as $itemindex[] => $item[])
    {
        echo "<option value=\"$item\">$item</option>\n";
    }
 
    echo "</select>";
echo "<select name=\"book1\"><option value=\"bleh\">Choose Part</option>";
 
    foreach($book1 as $itemindex[] => $item[])
    {
        echo "<option value=\"$item\">$item</option>\n";
    }
 
    echo "</select>";

when doing it this way, i get a drop-down list which just has three identical elements, all entitled 'Array'.

once this is done, i was just wondering how i could for example manipulate a link next to it to link to a file according to the chosen element in the list.

i'd very much appreciate your help, really
pygmalion

Microsoft Employee: "Hey, it compiles! Ship it."

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Join Date: Dec 2006

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pygmalion

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Junior Poster in Training

Re: Populating a Drop-down List

May 2nd, 2007

hi again.

i'd much appreciate anyone's help. this is quite of an importance to me.

thanks in advance
pygmalion

Microsoft Employee: "Hey, it compiles! Ship it."

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Join Date: Apr 2007

Posts: 136

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dr4g

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Junior Poster

Re: Populating a Drop-down List

May 2nd, 2007

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
     /* init */
     $arr = array(
                       "chapter1" => array(),
                       "chapter2" =>  array(),
                       "chapter3" => array()
      );
 
    $chapter1_amount = 4;
    $chapter2_amount = 6;
    $chapter3_amount = 8;
 
      for($i = 1; $i <= $chapter1_amount; $i++)
         array_push($arr['chapter1'], $i);
 
      for($i = 1; $i <= $chapter2_amount; $i++)
         array_push($arr['chapter2'], $i);  
 
      for($i = 1; $i <= $chapter3_amount; $i++)
         array_push($arr['chapter3'], $i);
     /* init */
     $arr = array(
                       "chapter1" => array(),
                       "chapter2" =>  array(),
                       "chapter3" => array()
      );

    $chapter1_amount = 4;
    $chapter2_amount = 6;
    $chapter3_amount = 8;

      for($i = 1; $i <= $chapter1_amount; $i++)
         array_push($arr['chapter1'], $i);

      for($i = 1; $i <= $chapter2_amount; $i++)
         array_push($arr['chapter2'], $i);  

      for($i = 1; $i <= $chapter3_amount; $i++)
         array_push($arr['chapter3'], $i);

This theoretically works. It's 3am and i didnt compile it, but looks fine to me.

For manipulating the link acording to the dropdown list.
You'd need to call a javascript function to generate the link.
When you choose an option in the list, you'd call the javascript function to update the link.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
      /* HTML */
      <select onChange="updateLink(this)">....</select>
 
      /* JavaScript */
     function updateLink(list) {
        if(list.value='bleh')
            link = "bleh";
 
       ......
 
       return link;
 
     }
      /* HTML */
      <select onChange="updateLink(this)">....</select>

      /* JavaScript */
     function updateLink(list) {
        if(list.value='bleh')
            link = "bleh";

       ......
  
       return link;

     }