linking php to jquery

I am trying to build my first jquery web application, but I've hit a roadblock and can't seem to figure this out.

I have a PHP page and an HTML page. The HTML page has a form with a triple drop down list. The PHP page connects to the database but I am not sure how to pass the query result from the php page to populate the drop down list on the html/javascript page.

Here is my code thus far.

HTML:

<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">

$(document).ready(function() {
$("#selector").submit(function() {

    $.ajax({
       type: "GET",
       url: "DBConnect.php",
       success: function(msg){
         alert(msg);
       }
});

var select_car_make = $('#select_car_make').attr('value');
    var select_car_model = $('#select_car_model').attr('value');
    var select_car_year = $('#select_car_year').attr('value');
    alert("submitted");


    }); //end submit
});
</script>

<h1 style="alignment-adjust:center">Car information:</h1>
<hr  />
<div id="results">

<form action="get.php" id="selector" method="get" name="sizer">
<table width="451" height="70" border="0">

      <th width="145" height="66" scope="row"><label for="select_car_make"></label>
          <div align="center">
            <select name="select_car_make" id="select_car_make" onchange="">
            </select>
      </div></th>
    <td width="144"><label for="select_car_model"></label>
      <div align="center">
        <select name="select_car_model" id="select_car_mod开发者_运维知识库el">
        </select>
      </div></td>
    <td width="140"><label for="select_car_year"></label>
      <div align="center">
        <select name="select_car_year" id="select_car_year">
        </select>
      </div></td>
  </tr>
</table>

<input name="completed" type="submit" value="submit" />
</form>

Here is the PHP Page:

<?php

$DBConnect = mysqli_connect("localhost", "root", "password", "testing")
or die("<p>Unable to select the database.</p>" . "<p> Error code " . mysqli_errno($DBConnect) . ": " . mysqli_error($DBConnect)) . "<p>";

echo "<p>Successfully opened the database.</p>";


$SQLString1 = " SELECT car_make FROM cars";
$QueryResult = mysqli_query($DBConnect, $SQLString1)

---

Hey Justin if this is your first time dipping your toes, I would keep it ultra-simple. Put the select box inside something with an ID, such as a span or div. Then get your AJAX response to just rewrite the contents, this is an easy and clear way to start with AJAX, for example:

$.ajax({
    type: "POST",
    url: "/call.php",
    data: "var=" + myData,
    success: function(response){
        $("#someId").html(response);
    }
});

On your remote page just echo the whole select box:

echo "<select name='cars'>";
echo "<option value='".$value."'>".$name."</option>";
etc...

Again this isn't the best way, but its certainly not the worst.

---

You should be able to JSON encode your result from PHP into a variable which Javascript or the Jquery can read. I did it like this with an image string I received from PHP reading a directory of images:

var imageFiles = '<?=$images_js?>';
    imageFiles = $.parseJSON(imageFiles);

    var images = [];
    for(i = 0; i<imageFiles.length; i++){
        var image = document.createElement('img');
        image.src = imageFiles[i];
        images.push(image);

    }
    var count = imageFiles.length;
    var i = 0;

the is the variable which came from my php result. The $.parseJSON(imageFiles); does the interpretation for the variable.

I hope this helps, or puts you along the right path.

---

If you want the car make selection page to be populated by your database, you should give it a .php extension as well. I think you should reconsider if AJAX is the best option.

But I would put the database connection code inside a "conn.php" file to be included on your selection page then populate it with PHP.