Take a look at my Website !!!

[gat]

Hello, Firstly can you take a look at my website..

www.uk-graffiti.co.uk

www.uk-graffiti.co.uk/characters.php (shows how images work)

ok, now as you can see i have nearly 1000 images on the site.

So, i need searches.....

I know that its all to do with querying the database with certain fields, but i have never done this before...

Each image has "pic_descrip and pic_area and pic_date"
these are the fields that contain the data to search on.

But i have no idea how or where to start.

I would like any search results to be displayed as the thumbnails are now on a results.php page but obviously it would only be the relevant results that are pulled into the tables.

Can any one help starting me off please...

---

An ideal way would be

Search by: (drop down box, "date" or "Writer" or "area")
then a text box for the user to type, then a go button or search button.

[superwormy]

$query = "SELECT * FROM table WHERE pic_descrip LIKE '%" . $pic_descrip . "%' AND pic_area LIKE '%" . $pic_area . "%' ";

www.php.net/mysql
www.mysql.com

[gat]

ok thats cool.........
that makes sence....

I presume the % is what the user types in...

BUT

where do i put that code?
if i simply want a dropdown box, a text box and a go button.

and all results to be displayed in results.php

Do i put the code behind the go button?
Do i simply add the drop down box, text box, and button in dreamweaver?

like i said, im not sure how or where to put the code?

Thanks, sorry for my ignorance....

[superwormy]

You should really go read the manual or some tutorials.... but...

Make a form in Dreamweaver with the options you want, for instance a text box for the person to type in the picture description.

Make that submit to your PHP page, stick the variable in the query BETWEEN the % signs, run the query, and display the result.

If your form looks like this:

<form method="post" action="/my_php_page.php">
<input type="text" name="pic_descrip" />
<input type="submit" value="Submit" />
</form>

Then your query should look like this:

<?php

$query = "SELECT * FROM table WHERE pic_descrip LIKE '%" . $_POST['pic_descrip'] . "%' ";

$result = mysql_query ($query) or die (mysql_error());
while ($array = mysql_fetch_array ($result)) {

print_r ($array);
print ("<br /><br /><br />");

}

?>

[gat]

many many thanks,
ill have a play, and get back if i have any problems...

Thanks

[gat]

many many thanks,
ill have a play, and get back if i have any problems...

Thanks