DHTML and PHP Accessing MySQL

donald3

I want to be able to populate a couple drop-down menus using data from a database using PHP and MySQL. My problem begins at populating the menus.

I want users to be able to select from a couple categories in the first menu, and then move to a second menu, which has been generated by the database depending on the item choosen in the first menu... have I lost you?

If you can help, I would GREATLY appreciate it!!!

Thank You!
Donald
😃

toma42

STFI 🙂

http://www.mattkruse.com/javascript/dynamicoptionlist/source.html

saidbakr

I don't know a detailed solution for this problem.
However, you should notice that,
The solution may be found in a mechanism, that will recall the page again with the new data which users choosed it from the first menu, to add related data in the second menu.

You can make all the contents of the two menus extracted from the data base and sending it completely to the page on the first call.

Also you should keep clearly in yourmind the deference between the two sides in the web, Client Side and The Server Side.

mystrymaster

the solution is to poulate the output from mysql into javascript arrays and then you can access them with an onchange event.

donald3

Okay, this is what I am trying to do:

secondMenuArray = new Array()

function populateSecondMenu(mainItem)
{
// Create appropriate array full of Option
// objects. These are placed in the second menu
// later in this function
if (mainItem == "news")
{
secondMenuArray = new Array()
<?php
$location = localhost;
$database = ;
$passwd = ;

$connection = mysql_connect($location, $database, $passwd);
if ($connection == 1) {
$db = mysql_select_db("***", $connection);
}
else {
echo "Couldn't Connect to MySQL Server!";
exit;
}

$sql="Select * From noteCards";

$result = mysql_query($sql, $connection) or die ("<p>Error.");

while ($Myresult = mysql_fetch_array($result)) {
echo "populateArray(\"$Myresult[source]\",\"$Myresult[category]\")";
}
?> }
else if (mainItem == "research")
{
secondMenuArray = new Array()
populateArray("Downs Syndrome","research/downs.html")
populateArray("Cancer","research/cancer.html")
populateArray("Chromosome Capping","research/capping.html")
populateArray("AIDS","research/aids.html")
populateArray("Male Birth Control Pill","research/guy_pill.html")
}

        ... 
        ...
        ...

}

function populateArray(displayText, nextPage)
{
nextIndex = secondMenuArray.length
secondMenuArray[nextIndex] = new Option(displayText,nextPage)
}

...

</script>

So, this results in two forms with nothing being generated in the second drop-down menu... If I take the php stuff out, it works just fine...

I am not sure what the idea behind the page reload is? Is there a web site I could reference?

Thanks!!!
Donald

saidbakr

You should note that, you cannot embed or nest a php code or even an html tags inside JavaScript codes.

You should remeber that, your php code is executed only on the webserver, then the result is sent to the client browser as html tags and contents,
so you may revers the order of code you had written.
I.e nest the JavaScript Inside the PHP code.
this may be some thing like this:

<?php
for ($i = 0; $i < 10; ++$i)
{
?>
<script>
document.write("Hello Number<?=$i?>");
</script>
<?
}
?>

Try this simple example and you will catch the idea easily.
Don't forget to look at the page source of the resultant page.