问题
I'm having some troubles with my code, I want it to execute the php file whenever I enter something but it isn't working
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">
function getStates(value) {
$.post("search.php", {name:value},function(data)
$("#results").html(data);
});
}
</script>
</head>
<input type="text" onkeyup="getStates(this.value)"/>
<br>
<div id="results"></div>
<body>
</body>
</html>
php
<?php
mysqli_connect("localhost", "#", "#") or die(mysqli_connect_errno());
mysql_select_db("#") or die(mysql_error());
$search = $_POST["name"];
$players = mysql_query("SELECT firstname FROM players WHERE firstname LIKE '%search%'");
while($player = mysql_fetch_array($players)) {
echo "<div>" . $players["firstname"] . "</div>";
}
?>
回答1:
From what I can see,you should change this
'%search%'
to
'%{$search}%'
in
$players = mysql_query("SELECT firstname FROM players WHERE firstname LIKE '%search%'");
EDIT
@user3187651 Assuming you've done everything right on the server side. Change your javascript to:
function getStates(value) {
$.post("search.php", {name:value},function(data){
$("#results").html(data);
}
);
}
This should get rid of the error in the client side.
回答2:
You are missing {
. Just do:
function xyx(name) {
$.post("search.php", { name: value }, function(data) {
$("#results").html(data);
});
}
回答3:
There's something that is missing in your code:
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> //ur getting the jquery via online
<script>
$(document).ready(function(){
$("#textBoxId").change(function() //triggers when you change the value in your textbox
{
var value = $(this).val(); //gets the value of your textbox
$.post("search.php", {id:value},function(data)
$("#results").append(data);
});
}
});
</script>
</head>
<body>
<input type="text" id="textBoxId"/>
<br>
<div id="results"></div>
</body>
</html>
And in your php:
<?php
mysqli_connect("localhost", "#", "#") or die(mysqli_connect_errno());
mysql_select_db("#") or die(mysql_error());
$search = $_POST['id'];
$returnData = "";
$players = mysql_query("SELECT firstname FROM players WHERE firstname LIKE '%search%'");
while($player = mysql_fetch_array($players)) {
$returnData .= "<div>" . $players["firstname"] . "</div>";
}
echo $returnData;
回答4:
For more secure and creative back-end code, you can use this.
<?php
$host = 'localhost';
$user = 'root';
$password = '';
$database = 'dev_testing';
$mysqli = new mysqli($host, $user, $password, $database);
$username = $_GET['username'];
$username = trim(htmlspecialchars($username));
$like = '%' . strtolower($username) . '%';
$statement = $mysqli -> prepare('
SELECT name, picture, description
FROM users
WHERE lower(name) LIKE ?
ORDER BY INSTR(title, ?), title
LIMIT 20'
);
if (
$statement &&
$statement -> bind_param('ss', $like, $username) &&
$statement -> execute() &&
$statement -> store_result() &&
$statement -> bind_result($name, $picture, $description)
) {
$array = [];
while ($statement -> fetch()) {
$array[] = [
'name' => $name,
'picture' => $picture,
'description' => $description
];
}
echo json_encode($array);
exit();
}
Advantages of the code
- Prevents SQL Injection
- Orders results from the best match
- Sends a JSON response (JSON is light-weight)
Full Tutorial: Live Search with AJAX, PHP, and MYSQL
来源:https://stackoverflow.com/questions/21144023/livesearch-php-and-ajax