html - Implementing AJAX into a PHP page with multiple selections
very new coder here.
I have a page with four select inputs each with two possible options, and when all of the selections have a value and a submit button is clicked a certain song (1 of 16 based on the selections made) will echo out into an audioplayer.
In its current state I have been able to connect to my database which echos out the links and titles for the audio tracks.
My issue is that I want all of the selections to visually retain their option values once the submit button is clicked so that users can see what options they have selected for the current song that is playing.
I've found lots of examples online of implementing AJAX into a page with one selection that activates via an onchange event such as this one from W3 Schools http://www.w3schools.com/php/php_ajax_database.asp, but nothing with multiple selections and a submit button.
Someone from this community helped me out the other day so that the code for the W3 schools example could accommodate a submit button onclick event instead of the onchange on the select input, but with my lack of much fluency with PHP/Javascript I don't really have an idea of how to include multiple selections.
I was hoping someone could take a look at how far I got in my page with multiple selections but no AJAX implemented and explain to me in very simple terms how I could go about including AJAX so that the select options are visible once the submit button has been clicked. If one could even show me a version of my page with AJAX placed with comments to explain the process would be absolutely golden.
Here is my page...
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<link rel="stylesheet" href="styles.css">
<link rel="shortcut icon" href="favicon.ico"/>
<title>MoodTunes -- Tunes for your mood</title>
<script src="prefixfree.min.js"></script>
</head>
<body>
<?php
$myDatabase = mysql_connect("localhost","root","root");
if (!myDatabase) die ('Oh dear.' . mysql_error());
if(isset($_GET['submit'])){
mysql_select_db("tunes",$myDatabase);
$volume = $_GET['volume'];
$tempo = $_GET['tempo'];
$complexity = $_GET['complexity'];
$key = $_GET['key'];
$query = "SELECT * FROM music WHERE volumeOption='".$volume."' AND tempoOption='".$tempo."' AND complexityOption='".$complexity."' AND keySignatureOption='".$key."'";
$mydata = mysql_query($query, $myDatabase) or die(mysql_error());
while($row = mysql_fetch_array($mydata)){
echo "<div id='submitContent'>";
echo "<h3><span>Now Playing:</span> " . $row['title'] . "</h3>";
echo "<figure id='audioplayer' style='display:inline-block;'>";
echo "<audio id='audiotrack' controls loop>";
echo "<source src='" . $row['link'] . "'type='audio/mpeg'>";
echo "</audio>";
echo "</figure>";
}
mysql_close($myDatabase);
}
?>
</div>
<header>
<div>
<h1>
<img src="assets/images/logo.png" alt="">
</h1>
<h2>Tunes for <span>your</span> mood</h2>
</div>
</header>
<main>
<h4>Choose your tune criteria</h4>
<form>
<label for="volume"></label>
<select name="volume" id="volume">
<option>Select One</option>
<option value="0" id="loud">heavy</option>
<option value="1" id="quiet">soft</option>
</select>
</label>
<label for="tempo"></label>
<select name="tempo" id="tempo">
<option>Select One</option>
<option value="0" id="fast">fast</option>
<option value="1" id="slow">slow</option>
</select>
</label>
<label for="complexity"></label>
<select name="complexity" id="complexity">
<option>Select One</option>
<option value="0" id="complex">complex</option>
<option value="1" id="simple">simple</option>
</select>
</label>
<label for="key"></label>
<select name="key" id="key">
<option>Select One</option>
<option value="0" id="minor">minor key</option>
<option value="1" id="major">major key</option>
</select>
</label>
<div id="submitDiv">
<input type="submit" name="submit" id="submit" value="Get My Tune!">
</div>
</form>
</main>
</body>
</html>
Any help would be greatly appreciated. Like I said, I'm still very new to much coding so please answer simply if you can help me out. Thanks.
Answer
Solution:
I suggest splitting into 2 files: an HTML file and a PHP file. Keep the PHP separate and call it with an XHR object (Ajax).
music.html
music.php
And I'm using PHP MySQLi prepared statements for database access for security and the fact plain old PHP MySQL functions are deprecated.
Answer
Solution:
Disclaimer The
object will only be available while the
{-code-2}
. If you wish to retain the information regardless of the browser being closed, please change to the{-code-3}
object instead.Here's your working jsFiddle example
One of the easiest and quickest ways to retain form input information without even bothering with serverside language is using the
functions.
We can capture all of the change events on the form elements and push their values into the . We can then retrieve and apply them to each of the respective inputs on the page reload with very minimal code.
Answer
Answer
Basically, if the storage is available, we create an event handler that binds to the change event of all of our form elements. Now when they change, we'll have a bunch of keys we can retrieve from the element.
Now, we need to apply those values to our elements on page load.
Answer
Answer
Now in the above code we're simply looping over all of our form elements and applying the values that were stored in the sessionStorage to those elements.
Additionally
Let's get you away from the
{-code-7}
library as it's deprecated, and will be completely removed in future versions of PHP. Let's also focus on sanitizing user input, so you don't have SQL injection vulnerabilities which could have catastrophic side effects.Users can still bypass the
{-code-9}
conditional, so let's add some more sanity checking.Now we should check if all those are set at the same time.
Inside of this conditional, we can then apply our database logic using mysqli's object oriented approach. We'll also use prepared statements to avoid the SQL injection.
Answer
Answer
Answer
Answer