Just_Johnny Posted March 23, 2009 Share Posted March 23, 2009 I'm new to javascript. I've read MANY tutorials, but I'm just not getting it. What I'm trying to accomplish is very simple. (not for me) I have 3 drop-downs. 1st. Populates from simple mysql query. 2nd. Are just static values. 3rd. Needs to populate from database from a query that uses values from 1st and 2nd drop down. When I read tutorials on how to do this I keep getting lost. I've never done anything like this before. Can anybody help? Quote Link to comment Share on other sites More sharing options...
Maq Posted March 23, 2009 Share Posted March 23, 2009 Can anybody help? What are you having trouble with...? Quote Link to comment Share on other sites More sharing options...
Just_Johnny Posted March 23, 2009 Author Share Posted March 23, 2009 <html> <!-- Please see the full php-ajax tutorial at http://www.php-learn-it.com/tutorials/starting_with_php_and_ajax.html If you found this tutorial useful, i would apprciate a link back to this tutorial. Visit http://www.php-learn-it.com for more php and ajax tutrials --> <title>php-learn-it.com - php ajax form submit</title> <head> <script type="text/javascript" src="prototype.js"></script> <script> function sendRequest() { new Ajax.Request("test.php", { method: 'post', postBody: 'name='+ $F('name'), onComplete: showResponse }); } function showResponse(req){ $('show').innerHTML= req.responseText; } </script> </head> <body> <form id="test" onsubmit="return false;"> Name: <input type="text" name="name" id="name" > <input type="submit" value="submit" onClick="sendRequest()"> </form> <div id="show"></div> <br/><br/> <a href="../../starting_with_php_and_ajax.html"><< back to php ajax tutorial</a> </body> </html> This does this http://www.php-learn-it.com/tutorials/demos/starting_with_php_ajax/test.html I want to do the same thing but only with a drop down. Quote Link to comment Share on other sites More sharing options...
Just_Johnny Posted March 24, 2009 Author Share Posted March 24, 2009 Ok, please forget about the last post. Here is the code I'm working with. <html> <!-- Please see the full php-ajax tutorial at http://www.php-learn-it.com/tutorials/starting_with_php_and_ajax.html If you found this tutorial useful, i would apprciate a link back to this tutorial. Visit http://www.php-learn-it.com for more php and ajax tutrials --> <title>php-learn-it.com - php ajax form submit</title> <head> <script type="text/javascript" src="prototype.js"></script> <script> function sendRequest() { new Ajax.Request("test.php", { method: 'post', postBody: 'name='+ $F('name'), onComplete: showResponse }); } function showResponse(req){ $('show').innerHTML= req.responseText; } </script> </head> <body> <form id="test" onsubmit="return false;"> <select name="name" id="name"> <?php include("connect_database.php"); $query = mysql_query("SELECT * FROM cars"); while($rows=mysql_fetch_array($query)){ $make = $rows['make']; echo "<option value='".$make."'>".$make."</option>"; } ?> </select> <input type="submit" value="submit" onClick="sendRequest()"> </form> <div id="show"></div> <br/><br/> <a href="../../starting_with_php_and_ajax.html"><< back to php ajax tutorial</a> </body> </html> For some reason this isn't working. Quote Link to comment Share on other sites More sharing options...
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