drop down

[Kathy]

I need to populate a drop down box with a table named categories and a field called Category, how do I go about doing this?? ???

[ignace]

$query = 'SELECT id, category FROM categories';
$result = mysql_query($query, $db);
if ($result) {
    $htmlSelect = '<select name="category-select">';
    while (list($id, $category) = mysql_fetch_array($result, MYSQL_NUM)) {
        $htmlSelect .= "<option value=\"$id\">$category</option>";
    }
    $htmlSelect .= '</select>';
}

 

<form action="" method="post">
    ...
    <label>Category: <?php print $htmlSelect; ?></label>
    ...
</form>

[Kathy]

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource

 

I get this when I run it

 

The problem is over here :

$result = mysql_query($query, $db);

?

[trq]

Does $db store your connection?

[Kathy]

Is this right?

function select_db(){
mysql_connect("localhost","login","pass")or die(mysql_error());
mysql_select_db("dbname")or die(mysql_error());
}
$query = 'SELECT id, Category FROM categories';
$result = mysql_query($query, $db);
if ($result) {
    $htmlSelect = '<select name="category-select">';
    while (list($id, $category) = mysql_fetch_array($result, MYSQL_NUM)) {
        $htmlSelect .= "<option value=\"$id\">$category</option>";
    }
    $htmlSelect .= '</select>';
} 

 

[trq]

No, you never call you function to make the connection, nor do you pass your connection back out to where it can be useful.

 

Should always check your queries actually return records before trying to use them too.

 

function select_db(){
  $db = mysql_connect("localhost","login","pass")or die(mysql_error());
  mysql_select_db("dbname")or die(mysql_error());
  return $db;
}

$db = select_db();
$query = 'SELECT id, Category FROM categories';
$result = mysql_query($query, $db);
if ($result) {
  if (mysql_num_rows($result)) {
    $htmlSelect = '<select name="category-select">';
    while (list($id, $category) = mysql_fetch_array($result, MYSQL_NUM)) {
        $htmlSelect .= "<option value=\"$id\">$category</option>";
    }
    $htmlSelect .= '</select>';
  }
}

[Kathy]

I'm being a total newbie here, I've tried it like this, but it's obviously still wrong:

<?PHP
mysql_connect("dedi536.nur4.host-h.net","ivif09","ivifpa55")or die(mysql_error());
mysql_select_db("web_cat09")or die(mysql_error());
?>

getting the same mysql error

I appreciate any help you can give me

[Kathy]

Okay great, I've managed to populate my drop down:

<?PHP
$link = mysql_connect("host","login","pass");
mysql_select_db("dbname", $link);

$query = "select Category FROM categories"; 
$results = mysql_query($query, $link) or die("Error performing query"); 

if(mysql_num_rows($results) > 0){ 
echo("<select name=\"Category\">"); 
while($row = mysql_fetch_object($results)){ 
echo("<option value=\"$row->record_id\">$row->Category</option>"); 
} 
echo("</select>"); 
} 
else{ 
echo("<i>No values found</i>"); 
}?>

 

I now want to populate another drop down from a choice in this drop down using the WHERE query, can anyone offer me guidance?

[ignace]

Assuming you want to select a subcategory:

 

$link = mysql_connect("host","login","pass");
mysql_select_db("dbname", $link);

$query = "select Category FROM categories";
$results = mysql_query($query, $link) or die("Error performing query");

if(mysql_num_rows($results) > 0){
    echo("<select name=\"Category\">");
    while($row = mysql_fetch_object($results)){
        echo("<option value=\"$row->record_id\">$row->Category</option>");
    }
    echo("</select>");
}
else{
    echo("<i>No values found</i>");
}

if (!empty($_POST)) {
    if (empty($_POST['subcategory'])) {
        $query = 'SELECT id, subcategory FROM subcategories';
        $result = mysql_query($query, $link);
        if (mysql_num_rows($result)) {// no need for > 0; if mysql_num_rows() == 0 then the if body won't be executed as 0 == false
            echo('<select name="subcategory">');
            while (list($id, $subcategory) = mysql_fetch_array($result, MYSQL_NUM)) {
                echo("<option value=\"$id\">$subcategory</option>");
            }
            echo('</select>');
        }
    } else if (!empty($_POST['category'] && !empty($_POST['subcategory'])) {
        $cat = $_POST['category'];
        $subcat = $_POST['subcategory'];
        ..process..
    }
}

[Kathy]

Something is wrong with this piece of code here:

<?PHP
while (list($id, $coName) = mysql_fetch_array($result, MYSQL_NUM)) {
		echo("<option value=\"$id\">$coName</option>");
		}
		echo('</select>');
		}
	} else if (!empty($_POST['category'] && !empty($_POST['coName']))) {
		$cat = $_POST['category'];
		$subcat = $_POST['coName'];
		..process..
		}
	}
?>

I've added another ) after ['coName'], but I'm still receiving this error when I run the script?

Parse error: syntax error, unexpected T_BOOLEAN_AND, expecting ')'

[ignace]

Their is alot missing from this code:

 

..something here..
    ..something here..
       ..something here..
            while (list($id, $coName) = mysql_fetch_array($result, MYSQL_NUM)) {
                echo("<option value=\"$id\">$coName</option>");
            }
            echo('</select>');
        }
    } else if (!empty($_POST['category'] && !empty($_POST['coName'])) {
        $cat = $_POST['category'];
        $subcat = $_POST['coName'];
        ..process..
    }
}