Auto populate multiple list box..

[techker]

hey guys i can't seem to get my second list box to work?

 

the first one is good it shows the brands but the second one does not show at all?

 

i have 2 DB Brands and others products

 

so in brands i have id and name ..

 

second i have product_id-product_Nand Brand

 

this is my query(select the brand then it list the products that are associated with the brand)

 

<?php 
$Brand_N = $Product_N = null; //declare vars 

$conn =  mysql_connect("localhost", "2", "2"); 
$db = mysql_select_db('2',$conn); 

if(isset($_POST["Brand_N"]) && is_numeric($_POST["Brand_N"])) 
{ 
$Brand_N = $_POST["Brand_N"]; 
} 

if(isset($_POST["Product_N"]) && is_numeric($_POST["Product_N"])) 
{ 
$Product_N = $_POST["Product_N"]; 
} 
?> 

<script language="JavaScript"> 
function autoSubmit() 
{ 
var formObject = document.forms['theForm']; 
formObject.submit(); 
} 
</script> 

<form name="theForm" method="post"> 


<select name="brand" onChange="autoSubmit();"> 
<option value="Brand">Brand</option> 
<?php 
$sql = "SELECT * FROM Brand"; 
$Brand = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Brand)) 
{ 
echo ("<option value=\"$row[brand_id]\" " . ($Brand_N == $row["Brand_N"]? " selected" : "") . ">$row[brand_N]</option>"); 
} 
?> 
</select> 

<?php 
if($Brand_N!= null && is_numeric($Brand_N)) 
{ 
?> 

<select name="Product" onChange="autoSubmit();"> 
<option value="Product">Product</option> 

<?php 
$sql = "SELECT * FROM products WHERE Brand = $Brand_N "; 
$Products = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Products)) 
{ 
echo ("<option value=\"$row[Product_id]\" " . ($Product_N == $row["Product_N"]? " selected" : "") . ">$row[Products_N]</option>"); 
} 
?> 
</select> </form>
<?php 
} 
?>

[awjudd]

Your select for brand's name is 'brand' not 'Brand_N', so the value will never be viewed as posted.

 

~juddster

[techker]

so it's just this part?i modified it but still..

 

<?php 
$sql = "SELECT * FROM Brand"; 
$Brand = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Brand)) 
{ 
echo ("<option value=\"$row[brand_id]\" " . ($Brand_N == $row["Brand_id"]? " selected" : "") . ">$row[brand_N]</option>"); 
} 
?> 
</select> 

<?php 
if($Brand_N!= null && is_numeric($Brand_N)) 
{ 
?> 

<select name="Product" onChange="autoSubmit();"> 
<option value="Product">Product</option> 

<?php 
$sql = "SELECT * FROM products WHERE Brand_id = $Brand_N "; 
$Products = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Products)) 
{ 
echo ("<option value=\"$row[Product_id]\" " . ($Product_N == $row["Product_N"]? " selected" : "") . ">$row[Products_N]</option>"); 
} 
?> 

[awjudd]

Now you aren't setting the variables.  All you needed to change in your last post was the posted value you were looking for (i.e. 'Brand_N' to 'brand').  To get that value.

 

~juddster

[techker]

sorry im rusted out..i don't get it..brand is the db name.

 

echo ("<option value=\"$row[brand_id]\" " . ($Brand_N == $row["Brand_id"]? " selected" : "") . ">$row[brand_N]</option>"); ???

[awjudd]

In your previous code you had:

 

<select name="brand" onChange="autoSubmit();"> 

 

PHP uses this name when posted in the $_POST array.  Because of that, you should be looking for $_POST['brand'] instead of $_POST['Brand_N'].

 

~juddster

[techker]

ok so im started to get it..lol

 

if(isset($_POST["Brand_N"]) && is_numeric($_POST["Brand_N"])) 
{ 
$Brand_N = $_POST["Brand"]; 
} 

if(isset($_POST["Product_N"]) && is_numeric($_POST["Product_N"])) 
{ 
$Product_N = $_POST["Product"]; 
}

 

so i need to change the product post to..

[awjudd]

$_POST [ 'Brand_N' ] should be referenced as $_POST [ 'brand' ] since that is the name of the drop down list being submitted.

 

~juddster

[techker]

ok so i changed it all to the select name..but still nothing?is there a way to echo the results so that i can see when i select a brand the output?

 

if(isset($_POST["Brand"]) && is_numeric($_POST["Brand"])) 
{ 
$Brand_N = $_POST["Brand"]; 
} 

if(isset($_POST["Product"]) && is_numeric($_POST["Product"])) 
{ 
$Product_N = $_POST["Product"]; 
} 

[awjudd]

add

print_r ( $_POST );

to your code and you can see everything being posted.

 

Edit: I said $_POST [ 'brand' ] not $_POST [ 'Brand' ] ... please note ... the array is case sensitive.

 

~juddster

[techker]

odd

 

select box (Brand) Array ( [brand] => 1 )

[awjudd]

<select name="brand" onChange="autoSubmit();"> 

 

Please Note: lowercase "brand" as in the name attribute of the select box.

 

~juddster

[techker]

typo error

 

in the select box by default it is writen brand

 

when i select a brand it just resets to brand

[awjudd]

Huh?

[techker]

ok i got your point..lol

 

i changed it to uppercase like all of it.now it shows the second drop down but empty

 

 

Array ( [brand] => 1 [Product] => Product )

[awjudd]

Post updated code?

 

~juddster

[techker]

<?php 
$Brand_N = $Product_N = null; //declare vars 

$conn =  mysql_connect("localhost", "1", "2"); 
$db = mysql_select_db('3',$conn); 

if(isset($_POST["Brand"]) && is_numeric($_POST["Brand"])) 
{ 
$Brand_N = $_POST["Brand"]; 
} 

if(isset($_POST["Product"]) && is_numeric($_POST["Product"])) 
{ 
$Product_N = $_POST["Product"]; 
} 
?> 

<script language="JavaScript"> 
function autoSubmit() 
{ 
var formObject = document.forms['theForm']; 
formObject.submit(); 
} 
</script> 

<form name="theForm" method="post"> 


<select name="Brand" onChange="autoSubmit();"> 
<option value="Brand">Brand</option> 
<?php 
$sql = "SELECT * FROM Brand"; 
$Brand = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Brand)) 
{ 
echo ("<option value=\"$row[brand_id]\" " . ($Brand_N == $row["Brand_N"]? " selected" : "") . ">$row[brand_N]</option>"); 
} 
?> 
</select> 

<?php 
if($Brand_N!= null && is_numeric($Brand_N)) 
{ 
?> 

<select name="Product" onChange="autoSubmit();"> 
<option value="Product">Product</option> 

<?php 
$sql = "SELECT * FROM products WHERE Brand_id = $Brand_id "; 
$Products = mysql_query($sql,$conn); 
while($row = mysql_fetch_array($Products)) 
{ 
echo ("<option value=\"$row[Product_id]\" " . ($Product_N == $row["Product_N"]? " selected" : "") . ">$row[Products_N]</option>"); 
} 
?> 
</select> 

<?php 
} 
?>
</form>
<? print_r ( $_POST );?>

[awjudd]

$sql = "SELECT * FROM products WHERE Brand_id = $Brand_id ";

 

That $Brand_id should be $Brand_N.

 

~juddster

[techker]

oK

 

so now it works i had another typo Products_Nneeded to be product_N

 

echo ("<option value=\"$row[Product_id]\" " . ($Product_N == $row["Product_N"]? " selected" : "") . ">$row[Products_N

 

but now it works but when i select the product it resets the form back to brand?

 

i need to keep it selected..

[awjudd]

Please expand ...

 

~juddster

[techker]

when i select brand it jumps to products but brand goes back to Brand instead of staying on the selected brand?

 

maybe the link would help

 

http://x81.tech-design.ca/script/test_m.php

 

[awjudd]

$Brand_N == $row["Brand_N"]? " selected" : ""

 

That should be looking at the Brand_Id because that is the value that gets submitted.

 

~juddster

[techker]

can i print in a input box the selected brand and product ?

 

cause i get what your saying but it still does not stay loaded on the selected brand and product..

[techker]

look at this one

 

http://x81.tech-design.ca/script/test_mp.php

 

i added

 

<? print_r ( $_POST["Product"]);?>

<? print_r ( $_POST["Brand"]);?>

[awjudd]

If you look at it, after you select a value from the first drop down list and it auto posts back, the drop down list is back to 'Brand' which isn't correct.  As I mentioned you need to change the condition where you are marking it as selected to be based on the ID rather than the name because that is the value being returned.

 

Because of this, you are seeing the issue you mentioned.

 

~juddster