Difference between revisions of "PHPAJAX"
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Revision as of 16:33, 12 September 2023
Main Page >> Web Application Development >> Workbook >> Ajax with fetch()
Ajax with fetch
Let's implement this very simple example:
Client-side
First, let's have a look at the webpage itself:
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Ajax demo</title>
<link href="https://cdn.jsdelivr.net/npm/bootstrap@5.3.1/dist/css/bootstrap.min.css" rel="stylesheet">
</head>
<body>
<div class="container">
<h1>Game search</h1>
<form class="row g-3">
<div class="col-auto">
<input type="text" class="form-control" id="searchBox" placeholder="Keywords">
</div>
</form>
<p id="results"></p>
</div>
</body>
</html>
<script>
// We "listen" for key pressed in our search box
document.getElementById("searchBox").addEventListener("keyup", doSearch);
// Function called when searching
function doSearch() {
// Get keyword from search box
keywords = document.getElementById("searchBox").value;
// Call server script, passing our keyword
fetch('https://mi-linux.wlv.ac.uk/~in9352/ajax/ajax.php?search=' + keywords)
// Convert response string to json object
.then(response => response.json())
.then(response => {
// Clear result box
document.getElementById("results").innerHTML = '';
// Loop through data and add to result box
response.forEach(game => {
document.getElementById("results").append(game.game_name + ' ') ;
});
});
}
</script>
The HTML is fairly simple (I have used Bootstrap, but that's optional). Note the text box and blank paragraph to display the results later on.
Make sure you read the comments in the JS code.
Server-side
The server-side script is as follow:
<?php
// Connect to database and run SQL query
$mysqli = new mysqli("localhost","5cs023_user","5cs023_password","alix");
if ($mysqli -> connect_errno) {
echo "Failed to connect to MySQL: " . $mysqli -> connect_error;
exit();
}
$sql = "SELECT * FROM videogames WHERE game_name LIKE '%{$_GET['search']}%' ORDER BY released_date";
$results = $mysqli->query($sql)->fetch_all(MYSQLI_ASSOC);
print(json_encode($results));
?>
The code is very similar to our previous database example, but instead of looping and displaying each record, we fetch all records as a single object, convert it to JSON, and "dump" it with a print. This will return the JSON in the HTTP body.
More
Things to improve:
- Only trigger the search when at least 2 characters have been typed.
- Display the results in a nice list.
