Showing tables in dropdown menu

[phpbuild12]

Hello, I need to make so that a dropdown menu shows all tables from selected database, where tables starts with option blabla...

What would be the best way?

[Ashley_Sheridan]

Well, you could probably get the list of tables with

SHOW TABLES;

and then loop over that in PHP to filter out the ones you don't want before displaying them in a dropdown.

[phpbuild12]

Well could you please provide some more info I really dont know how would I do that as I have to set unique value to each one if I want to select 1, right?

[Ashley_Sheridan]

You didn't say what you wanted to do with them after, you only asked how to go about putting them into a dropdown.

If you want to identify them later, what's wrong with using the names of the tables as identifiers?

[phpbuild12]

Its too many of them, I need a dropdown list of tables inside specific database with the same perfix, and I need to be able to select one and do something with it...

[Ashley_Sheridan]

I'm not sure you're actually taking my advice on board. What code do you have already?

[phpbuild12]

Nothing, so far I just connected to database..

[Ashley_Sheridan]

OK. Well, I would say the next step is to run that query I originally posted. Then loop through the results that gives, and output those into the dropdown menu.

[Derokorian]

Note if you wanted to limit the tables returned to just those with the prefix, without filtering them in PHP you could use the INFORMATION_SCHEMA.TABLES like so:

SELECT TABLE_NAME FROM INFORMATION_SCHEMA 
   WHERE TABLE_NAME LIKE 'option%' 
     AND TABLE_SCHEMA = '<insert your database name here>' 
     AND TABLE_TYPE = 'BASE TABLE'

[nancy30]

where do you wish to place this dropdown window you can connect each menu item to a table data

[phpbuild12]

Ok, thanks. I got so far:

$query = mysql_query("SHOW TABLES FROM dbname123 LIKE 'account%'") or die(mysql_error());
$array = mysql_fetch_assoc($query);

Now I need to display tables inside a dropdown menu.

[Weedpacket]

phpbuild12 wrote:

Now I need to display tables inside a dropdown menu.

You'll need to answer yourself some questions, then:

1. How do I make a dropdown menu (what does the HTML look like)?

2. How do I turn the results of a database query into a list?

3. How do I turn a list into a dropdown menu?

Putting the answers to those questions together would provide a solution.

[phpbuild12]

Well I would need an answer fast so I can work further, but..

1. Inside while loop, I tried but I couldnt get it.
2. <select></select>?

Update...

Ok I can display them in a dropdown menu now, all I need now is to be able to select 1 and do something with it.

<?php
require("connect.php");

$sql = "SHOW TABLES FROM base123 LIKE 'acc%'";
$result = mysql_query($sql);

?>

<select>
<?php
while ($row = mysql_fetch_row($result)) {
echo "<option> $row[0] </option>";
}
?>
</select>

<?php

?>

[Ashley_Sheridan]

Well, what did you want to do? Is it something you need to happen as soon as someone selects an option from your select list (it's not a dropdown, that's something different), or do you need to submit it as part of a larger form?

[phpbuild12]

I am sorry for not beeing clear. I did that step now, now I have to do something else.

Right now I have a dropdown menu with the list of items inside. I need to make it so that when user select item something happens. If item name starts with a display something, if it starts with b display something else, could you please explain me how could I do that?

[Ashley_Sheridan]

I think you mean a select list, not a drop down? Assuming that, something like this in Javascript:

<select onChange="do_something(this);">
</select>

And in your Javascript file:

function do_something(option)
{
    alert("do something with " + option.value);
}