I used an autosuggest script (using jquery, ajax, php and mysql) in one of my pages and that worked perfectly.
but when i tried to add the autosuggest for another Form (Text Input) on the same page i got a lot of errors, it displays the correct information but when i click them that fills the first input, i tried modifying some variable but now its not working at all for the second input. and im really new to JS, Jquery, Ajax ( never used that before ).
Here's is the code of the page containing the inputs (Test.html).
<html>
<head>
<title>Test</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js">
</script>
<script>
function suggest(inputString){
if(inputString.length == 0) {
$('#suggestions').fadeOut();
} else {
$('#country').addClass('load');
$.post("autosuggest.php", {queryString: ""+inputString+""}, function(data){
if(data.length >0) {
$('#suggestions').fadeIn();
$('#suggestionsList').html(data);
$('#country').removeClass('load');
}
});
}
}
function fill(thisValue) {
$('#country').val(thisValue);
setTimeout("$('#suggestions').fadeOut();", 600);
}
#result {
height:20px;
font-size:16px;
font-family:Arial, Helvetica, sans-serif;
color:#333;
padding:5px;
margin-bottom:10px;
background-color:#FFFF99;
}
#country{
padding:3px;
border:1px #CCC solid;
font-size:17px;
}
.suggestionsBox {
position: absolute;
left: 0px;
top:40px;
margin: 26px 0px 0px 0px;
width: 200px;
padding:0px;
background-color: #000;
border-top: 3px solid #000;
color: #fff;
}
.suggestionList {
margin: 0px;
padding: 0px;
}
.suggestionList ul li {
list-style:none;
margin: 0px;
padding: 6px;
border-bottom:1px dotted #666;
cursor: pointer;
}
.suggestionList ul li:hover {
background-color: #FC3;
color:#000;
}
ul {
font-family:Arial, Helvetica, sans-serif;
font-size:11px;
color:#FFF;
padding:0;
margin:0;
}
.load{
background-image:url(loader.gif);
background-position:right;
background-repeat:no-repeat;
}
#suggest {
position:relative;
}
</style>
</head>
<body>
<form id="form" action="#">
<div id="suggest">type a country: <br />
<input type="text" size="25" value="" id="country" onkeyup="suggest(this.value);" onblur="fill();" class="" />
<div class="suggestionsBox" id="suggestions" style="display: none;"> <img src="arrow.png" style="position: relative; top: -12px; left: 30px;" alt="upArrow" />
<div class="suggestionList" id="suggestionsList"> </div>
</div>
</div>
<div id="suggestcity">type a city: <br />
<input type="text" size="25" value="" id="city" class="" />
</div>
</form>
</body>
</html>
And the code of my the autosuggest.php is
<?php
$db_host = '';
$db_username = '';
$db_password = '';
$db_name = '';
$db = new mysqli($db_host, $db_username ,$db_password, $db_name);
if(!$db) {
$db->set_charset("utf8");
echo 'Could not connect to the database.';
} else {
if(isset($_POST['queryString'])) {
$queryString = $db->real_escape_string($_POST['queryString']);
if(strlen($queryString) >0) {
$db->set_charset("utf8");
$query = $db->query("SELECT Ville FROM Villes WHERE Ville LIKE '$queryString%' LIMIT 10");
if($query) {
echo '<ul>';
while ($result = $query ->fetch_object()) {
echo '<li onClick="fill(\''.addslashes($result->Ville).'\');">'.$result->Ville.'</li>';
}
echo '</ul>';
} else {
echo 'OOPS we had a problem :(';
}
} else {
// do nothing
}
} else {
echo 'There should be no direct access to this script!';
}
}
?>
So My question is, what should i change ( variables and functions etc... ) in the two pages so i can use the autosuggest on the second input with the id="city" from another table on the same database.
if you can help that would be wonderful, i have been trying to modify whatever on the code for the last couple days and nothing worked. Thank you verry much
