Accessing a variable from inside a jQuery drop down list

Question

I have a jQuery dropdown list that has two dropdowns. The first one is a list of cars and the second one shows the car's models. This is all working fine. Now I would like to use the selection selected by the user to search a database (MySQL) and find a specific element within the database. This just gives me the else within my code and once it gave me a undefined variable error message.

How can I code this so that I can access the variables and display a search result? Here is my code:

This is just an example of the jQuery I'm using - I'm actually using cars but this shows cities:

<select class="select" id="province" onchange="filterCity();">
  <option value="1">RM</option>
  <option value="2">FI</option>
</select>

 <select class="select" id="city" disabled>
  <option data-province="RM" value="1">ROMA</option>
  <option data-province="RM" value="2">ANGUILLARA SABAZIA</option>
  <option data-province="FI" value="3">FIRENZE</option>
  <option data-province="FI" value="4">PONTASSIEVE</option>
</select>

<span id="option-container" style="visibility: hidden; position:absolute;"></span>

<script>

    function filterCity(){
      var province = $("#province").find('option:selected').text(); // stores province
      $("#option-container").children().appendTo("#city"); // moves <option> contained in #option-container back to their <select>
      var toMove = $("#city").children("[data-province!='"+province+"']"); // selects city elements to move out
      toMove.appendTo("#option-container"); // moves city elements in #option-container
      $("#city").removeAttr("disabled"); // enables select
 };
</script>

And here is the code where I'm trying to access the searched for variable (this code is echoing wrong (the else)): I used the name="car" in the second select:

if (isset($_POST['car'])) {
        $_SESSION['car'] = $_POST['car'];
    $car = $_SESSION['car'];
}
function find_tire(){
    if (isset($_POST['car'])) {
        $_SESSION['car'] = $_POST['car'];
    $car = $_SESSION['car'];
}

    global $db;

    if (isset($car)) {
    $query = $db->prepare("SELECT idtires FROM vehicle WHERE idcarmodel = $car");
 $query->execute();
    $tire = $query->fetchAll();
    var_dump($query);}
if (isset($tire)) {
            echo "<ul>";
        foreach ($tire as $name) {
                echo "<li id='tiresearch'>";
                echo "Tire Size is Available: " . $name    ['idtires'];
                echo "</li>";
            }echo "</ul>";
    } else {

            echo "Wrong";
        }
  }

Are you getting `$_POST['car']` because its seems you didn't specified `name="car"` within your second `select` — Narendrasingh Sisodia, Mar 21 '15 at 05:37
@NarendraSisodia - Yes, I used name="car" in the second select - I didn't post my actual car code because it was too long. — Regina Shepherd Riddle, Mar 21 '15 at 05:39
But please make sure to show it within your code else others over here will point out that error first — Narendrasingh Sisodia, Mar 21 '15 at 05:41
@NarendraSisodia, I wrote it just above the function so people would know. Is that what you meant? — Regina Shepherd Riddle, Mar 21 '15 at 05:42
Are you getting value within `$tire` and why do you have such redundant code does it make any difference `function find_tire()` and the one above it — Narendrasingh Sisodia, Mar 21 '15 at 05:50
@NarendraSisodia, I'm sorry if it seems redundant - I was just trying everything I could think of to get it to work. When I finish with it, I'll erase some of the redundancy. A var_dump just shows me the query without the $car - it says there is an undefined variable which would be $car. — Regina Shepherd Riddle, Mar 21 '15 at 05:52
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/73463/discussion-between-narendra-sisodia-and-regina-shepherd-riddle). — Narendrasingh Sisodia, Mar 21 '15 at 05:54

score 0 · Answer 1 · answered Mar 21 '15 at 05:35

0

you can use ajax for this to send your variable to another page and get response from there .something like this

var id = $('selector').val();
    var AjaxURL = 'http://example.com';
    var data = id;
    $.ajax({
        type: "POST",
        url: AjaxURL,
        data: {data: data},
        success: function(result) {
            window.console.log('Successful');
        }
    });

answered Mar 21 '15 at 05:35

Manoj Dhiman

5,096
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I'm not sure I understand this exactly. I'm just trying to access a variable. I don't think I would need a new page for that. – Regina Shepherd Riddle Mar 21 '15 at 05:41

score 0 · Answer 2 · edited Apr 18 '16 at 16:35

0

I had to put all the select inside form tags, use a different page for the search and add the function to the bottom of the search page instead of accessing it from a different page.

Thank you to anyone who posted answers and tried to help me with this. It's working (with a few bugs) but I'm so happy it's finally working.

edited Apr 18 '16 at 16:35

halfer

19,824
17
99
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answered Mar 22 '15 at 03:46

Regina Shepherd Riddle

123
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Accessing a variable from inside a jQuery drop down list

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