Display data when changing selected value using JQuery / PHP

Question

Hope your are fine!

I'm would like to execute a request (SQL) like this one:

SELECT Name FROM course WHERE IdSectionFK = '.$idSectionFK.';

I have the list of sections:

while($row = $selectAllSection->fetch(PDO::FETCH_OBJ)){
   echo "<option value=".$row->IdSection.">".$row->Name."</option>";
}

And I would like to display the data only for the selected value. I tried something like this to get the value of the list:

<script>

function displayVals() {

        var idSection = $("#sectionListe").val();

        $.post('index.php', { 'idSection': idSection },function (){
          alert("success");
        })

        .success(function() { alert("second success"); })
        .error(function() { alert("error"); })
        .complete(function() { alert("complete"); });
    }                                    

    $("select").change(displayVals);
    displayVals();

</script>

So, the variable "idSection" is equale to the PHP variable "idSectionFK" in my SQL request. But how can I execute the right SQL request ?

Thank you so much for your help!

Lapinou.

you can use jquery.post -> api.jquery.com/jQuery.post

hwsw
– hwsw

2013-07-20 18:48:49 +00:00
Commented Jul 20, 2013 at 18:48 — hwsw
– hwsw, Commented Jul 20, 2013 at 18:48

Hieu Nguyen · Accepted Answer · 2013-07-20 22:06:30Z

1

AJAX is the solution.

JS code:

function displayVals() {
    var idSection= $("#sectionListe").val();
    $("p").html(idSection);
    $.post('/url/to/php/file', { 'idSection': idSection }, function (response) {
        // do something with the response here
        // e.g: $('select').append(response);
        console.log(response);
    });
}                                    

$("select").change(displayVals);
displayVals();

PHP code to process the received data:

$idSectionFK = 0;

if (isset($_POST['idSection'])) {
    // get the ID from ajax, run SQL here
    $idSectionFK = intval($_POST['idSection']);

    $query = "SELECT Name FROM course WHERE IdSectionFK = '.$idSectionFK.';";

    // .... your code ...
}

Just to give you the basic idea.

edited Jul 20, 2013 at 22:06

answered Jul 20, 2013 at 19:34

Hieu Nguyen

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8 Comments

Lapinou Over a year ago

Thanks for your help ;) But it doesn't work. I have no POST response... I don't understand why :/

Hieu Nguyen Over a year ago

Do you really need POST response? I haven't put that in the code before. Answer updated, could you check now?

Lapinou Over a year ago

No I don't. I mean, when I check if(isset($_POST['idSection'])), it always return false... Like if $.post doesn't work...

Lapinou Over a year ago

I edited my script to check it. It seems ok but $_POST['idSection'] is not set :/

Hieu Nguyen Over a year ago

Are you sure that index.php is the file contains the SQL query? Also any error in the console (using Firebug / Chrome dev tool)?

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