I have a PHP/AJAX search script working just fine. The problem is when I type something in the text field the suggestions appear below, but they cannot be selected. In short, I want to select one of the suggestion and show the value in the search text field upon selecting it.
I have tried a solution in JavaScript, but it doesn't work. It shows an error in the console like "the select value is null".
ajax.php
<head>
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
</script>
</head>
<body>
<input id="sear" autocomplete="off" type="text" name="search"
placeholder="Seaarch here" onkeyup="search(this.value)">
<div onclick="ali()" id="list" ></div>
<script>
function search(str){
if(str.length ==0){
document.getElementById("list").innerHTML="Please Enter Something";
return;}
xml=new XMLHttpRequest();
xml.onreadystatechange=function(){
if(xml.readyState==4 && xml.status==200){
document.getElementById("list").innerHTML=xml.responseText;}
}
xml.open("GET","search.php?char="+str,true);
xml.send();
}
</script>
<script>
document.addEventListener("DOMContentLoaded",function(){
function ali(){
var acs;
var x=document.getElementById("#list").select();
acs=x.value;
document.getElementById("#sear").innerHtml=acs;
}
});
</script>
</body>
search.php
<?php
$con=mysqli_connect("localhost","root","","userdiary");
$str=$_GET['char'];
$sql="SELECT * FROM `login` WHERE `user_id` LIKE '$str%' OR `user_name` LIKE
'$str%'";
$query=mysqli_query($con,$sql);
if(mysqli_num_rows($query)>0){
while($res=mysqli_fetch_assoc($query)){
echo "<option>" .$res['user_name']."</option>"."<hr>";
}
}
else{
echo "No match Found..";
}
?>
If the #sear element looks something like this once suggestions populate into it:
<div id="list">
<option>test</option>
<option>test a</option>
<option>test b</option>
</div>
then you can try this for your jQuery script:
$('#list option').on('click', function() {
$('#sear').val($(this).text());
});
You can check this code I used it yesterday. You just have to change the name of SQL part according to your goal
Also what I give you is jQuery ui you have to add
<script type="text/javascript" src="http://code.jquery.com/ui/1.10.1/jquery-ui.min.js"></script>
<link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.10.1/themes/base/minified/jquery-ui.min.css" type="text/css" />
and your input must have auto class
search.php:
<?php
define('DB_SERVER', 'localhost');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'some');
if (isset($_GET['term'])){
$return_arr = array();
try {
$conn = new PDO("mysql:host=".DB_SERVER.";port=8889;dbname=".DB_NAME, DB_USER, DB_PASSWORD);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $conn->prepare('SELECT Tname FROM Topic WHERE Tname LIKE :term');
$stmt->execute(array('term' => '%'.$_GET['term'].'%'));
while($row = $stmt->fetch()) {
$return_arr[] = $row['Tname'];
}
} catch(PDOException $e) {
echo 'ERROR: ' . $e->getMessage();
}
/* Toss back results as json encoded array. */
echo json_encode($return_arr);
}
?>
and jQuery UI
$(function() {
$(".auto").autocomplete({
source: "search.php",
minLength: 1,
});
});
and css
.ui-menu .ui-menu-item a {
text-decoration: none;
display: block;
padding: 5px .4em;
line-height: 1.5;
min-height: 0;
font-weight: normal;
z-index:11;
}
ul#ui-id-1{
z-index:11;
display: none;
top: 132px;
left: 354.5px;
width: 283px;
}
You typed select in worng way. It should be something as follows:
<select onchange="ali()" id="list" ></select>