我正在尝试为我的网站制作一个类似Google的搜索栏，它会自动更新搜索结果
我正在使用php和AJAX来做这件事，这是AJAX的一部分，所以我可以进一步解释
edited below
所以希望你们都知道AJAX，我在html表单中的一个事件属性中调用search()函数onkeyup
它正在运行AJAX，我用一点alert('hello!');来测试它
现在我的问题是我在以下代码中得到一个未定义的索引错误：
<?php
require 'connection.php';
require '../search.php';
$term = $_POST['searchword'];
$type = $_POST['searchtype'];
echo $term . $type;
$term = mysqli_real_escape_string($con,$term);
$type = mysqli_real_escape_string($con,$type);
$query = "SELECT * FROM `users` WHERE `username` LIKE '$term%' AND `type` = '$type'";
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_assoc($result)){
$name = $row['username'];
$ntype = $row['type'];
$id = $row['user_id'];
echo "<a href = \"user.php?id=$id\"><div class=\"col-lg-12 result\">$name</br>$ntype</div></a>";
}if($rowcnt = mysqli_num_rows($result) == 0){
echo"<div class=\"col-lg-12 result center\">No results found</div>";
}
?>
未定义索引错误适用于$term和$type，即$_POST['searchword'];和$_POST['searchtype'];
我仔细检查了代码，并意识到search.php页面不包括在内......所以＆＃34; searchword＆＃34;和＆＃34; searchtype＆＃34;没有定义，所以我尝试包括它，现在每次我点击搜索栏中的按钮我得到第二个搜索表单，因为代码也运行search.php onkeyup，所以我想知道是否有任何方法可以避免这种情况从发生中
顺便说一下，我仍然得到未定义的索引错误，即使它被包括在内
这些是我正在使用的所有文件：
search.php：
<?php
session_start();
require 'res/connection.php';
include 'res/loggedinmenu.php';
if(empty($_SESSION['id'])){
echo"<script>
setTimeout(function () {
window.location.href = 'index.php';},2000);
</script>";
echo "
<div class=\"alert alert-info\">
<strong>You are not logged in!</strong> in 2 seconds you will be redirected to the login page
</div>
";
}
?>
<!DOCTYPE>
<html>
<head>
<title> search for other developers or owners</title>
<link rel="stylesheet" type="text/css" href="css/profile.css">
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/js/bootstrap.min.js"></script>
<style>
body{
background-color:#dcdcdc;
}
.all{
margin-top:110px;
}
.search{
border:1px solid #b0b0b0;
background-color:#e4e4e4;
display:inline;
box-shadow: 2px 2px 3px #ccc;
margin-bottom:35px;
}
.header{
width:100%;
background-color:#3928f2;
color:white;
text-align:center;
padding-top:15px;
padding-bottom:15px;
font-weight:bold;
font-size:20px;
}
.form{
padding-top:20px;
border-bottom:1px solid #ccc;
}
.result{
font-size:20px;
margin-top:15px;
padding-bottom:5px;
border-bottom:1px solid #ccc;
text-decoration:none;
}
.result :hover{
text-decoration:none;
}
.center{
text-align:center;
}
</style>
</head>
<body>
<?php menu(); ?>
<div class="container all">
<div class="row">
<div class="col-lg-2">
</div>
<div class="col-lg-8 search">
<div class="row">
<div class="col-lg-12 header">search for a developer or an owner</div>
<div class="col-lg-12 form">
<form action="" method="post" style="text-align:center;">
<div class="row">
<div class="col-lg-6">
<label for="term">Search term :</label>
<input type="text" onkeyup="search()" class="form-control" name="searchword" id="searchbar" value = "<?php if (isset($_GET['term'])){ echo $_GET['term']; }?>"/>
</div>
<div class="col-lg-6">
<label for="type">looking for a :</label>
<select name="searchtype" class="form-control" id="stype">
<option>owner</option>
<option>developer</option>
</select></br>
</div>
</div>
</div>
</form>
<hr>
</div>
<div class="row">
<div id="results" class="col-lg-12">
<?php
if(isset($_GET['term']) && isset($_GET['type'])){
$term = $_GET['term'];
$type = $_GET['type'];
}else{
$term = $_POST['searchword'];
$type = $_POST['searchtype'];
}
if($term == ""){
echo '<div class="col-lg-12 result center">Type in the box above to start searching</div>';
}else{
$term = mysqli_real_escape_string($con,$term);
$type = mysqli_real_escape_string($con,$type);
$query = "SELECT * FROM `users` WHERE `username` LIKE '$term%' OR `username` LIKE '%$term' AND `type` = '$type'";
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_assoc($result)){
$name = $row['username'];
$ntype = $row['type'];
$id = $row['user_id'];
echo "<a href = \"user.php?id=$id\"><div class=\"col-lg-12 result\">$name</br>$ntype</div></a>";
}if($rowcnt = mysqli_num_rows($result) == 0){
echo"<div class=\"col-lg-12 result center\">No results found</div>";
}
}
?>
</div>
</div>
</div>
</div>
<div class="col-lg-2">
<!--leave empty -->
</div>
</div>
</div>
</body>
</html>
<script>
function search(){
var xhttp = new XMLHttpRequest();
var url = "res/searchsyntax.php";
var trm = document.getElementById('searchbar').value;
var tpe = document.getElementById('stype').value;
var params = "term=" + term + "&type=" + tpe;
http.open("POST", url, true);
//Send the proper header information along with the request
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
document.getElementById("results").innerHTML = http.responseText;
}
}
http.send(params);
}
</script>
res / searchsyntax.php：
<?php
require 'connection.php';
$term = $_GET['term'];
$type = $_GET['type'];
if($term == ""){
echo '<div class="col-lg-12 result center">Type in the box above to start searching</div>';
}else{
$term = mysqli_real_escape_string($con,$term);
$type = mysqli_real_escape_string($con,$type);
$query = "SELECT * FROM `users` WHERE `username` LIKE '$term%' AND `type` = '$type'";
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_assoc($result)){
$name = $row['username'];
$ntype = $row['type'];
$id = $row['user_id'];
echo "<a href = \"user.php?id=$id\"><div class=\"col-lg-12 result\">$name</br>$ntype</div></a>";
}if($rowcnt = mysqli_num_rows($result) == 0){
echo"<div class=\"col-lg-12 result center\">No results found</div>";
}
}
?>
答案 0 :(得分：1)
您收到此错误是因为您没有通过AJAX将帖子参数（即“searchword”和“searchtype”）发送到PHP代码。
This code演示了如何执行此操作。
var url = "get_data.php";
var params = "lorem=ipsum&name=binny";
http.open("POST", url, true);
//Send the proper header information along with the request
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);