onchange select做不同的mysql查询

Question

As title says I need help with onchange. I have select tag,and I need to do different mysql query when I choose something from select list. Example:

<select name="list">
<option>cars</option>
<option>busses</option>
<option>trucks</option>
</select>

and then when i select cars it do this

$query="select * from table where type='cars'";

and if I choose trucks it do

$query="select * from table where type='trucks'";

and so on...
then I need to display the result in div under the list example

<select name="list">
<option>cars</option>
<option>busses</option>
<option>trucks</option>
</select>
<div> this is where I need to display results from query</div> 

Please help!!!

Answer 1

If you want to change the result by ajax :

HTML :

 <select id="list" name="list">

    <option>cars</option>

    <option>busses</option>

    <option>trucks</option>
    </select>

in JS File:

$(document).ready(function() {
  $('#list').change(function() {
var selected=$(this).val();
  $.get("change_query.php?selected="+selected, function(data){
      $('.result').html(data);

    });
    });
});

and you should create change_query.php file and your query and code in it and return the result in it

$type = $_POST["selected"];

$query="select * from table where type='".$type."'";

print result here .... ;

• Tell me if you need any help ,I just guided you in Jquery not all code

Answer 2

You should add your SELECT code to a form with method attribute set to post , and also a submit button of course. Then , php will get the value of those INPUTS and do whatever you want.

   $term = $_POST['list'];

/*
In order to prevent unwanted queries or injection add an array with those terms and
check if the posted value is in this array:

*/
    $terms = array("cars" , "trucks");
    if(!array_key_exists($term , $terms))
     die(); //bad bad bad boy.

    $query = "select * from table where type='$term'";

Answer 3

you could use

onchange="this.form.submit();"

as well.