Get MySQL Record using Php & JQuery

Question

Here's my scenario:

I have a drop down menu, with different values. When option 'A' is chosen, I want it to use JQuery to call a Php script. That Php script will make a database query which returns a record ID. When that record ID is returned to the javascript, javascript will direct the browser to a particular URL based on the returned ID (i.e. /products/$ID)

Right now, I have it so that the dropdown menu triggers a javascript function. I also have the Php script that does the database work. Im just not sure what should go in the javascript function, or how to make it connect with he Php script, and how I get the returned data (and how to deal it with the return type -- xml, html, etc)

Im using JQuery library, and Php 5.x.

yeah, i know I those are the steps, but im unsure of how the implementation works. Sorry, I always forget in the heat of things to click 'accept' when im finding my answer ! :) — djt
– djt, Commented Apr 19, 2011 at 20:05

shmeeps · Accepted Answer · 2011-04-19 20:13:52Z

A simple example would be to take a form like

Example.php

<form id="form" name="form">        
    <select name="dropdown" id="dropdown">
        <option value="A">A</option>
        <option value="B">B</option>
        <option value="C">C</option>
        <option value="D">D</option>
    </select>

    <input type="submit" id="submit" name="submit" value="Submit!" />
</form>

Then link it with some JQuery to intercept the form, and send it to a PHP file

JQuery.js

$(document).ready(function(){
    $("#dropdown").change(function(event) {
        $.ajax({
            type: "POST",
            url: "query.php",
            data: $('#form').serialize(),
                datatype: "json",
            success: function(data){
                var ret = jQuery.parseJSON(data);
                    // Redirect to ret.link
            }
    });
    event.preventDefault();
    });
});

Then, create the PHP file to interpret it

query.php

$stmt = $sql->dbh->prepare("SELECT `Link` FROM `Links` WHERE `ID` = :id");

$stmt->bindValue(':id', $_POST['dropdown']);

$stmt->execute();

$link = $stmt->fetch()['Link'];

echo json_encode(array('link' => "{$link}"));

die();

Dave Kiss · Accepted Answer · 2011-04-19 20:16:58Z

1

$("#menu-item").click(function() {

    jQuery.ajax({
        type: "POST",
        url: "yourScript.php",
        data: "data=foo",
        success: function(response){
            //redirect to id using response
            window.location.replace("http://yoursite.com/products/" + response);
        }
    });

});

Of course, this will need to be customized to your unique situation.

edited Apr 19, 2011 at 20:16

answered Apr 19, 2011 at 20:11

Dave Kiss

10.5k11 gold badges55 silver badges77 bronze badges

3 Comments

djt Over a year ago

but what format is 'response' in? and how would I access the data within it?

Dave Kiss Over a year ago

response is any data that is returned from your php script. this can be anything from JSON to HTML to a string. JSON is usually the best way to send the data back and forth, though, as jQuery has an excellent .parseJSON function

djt Over a year ago

thanks for that, im almost there. still having some issues with the return but ill figure it out

Collectives™ on Stack Overflow

Get MySQL Record using Php & JQuery

2 Answers 2

Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related