我有一个搜索栏，用于显示MySQL，PHP和JS的AJAX实时搜索结果。
问题是当查询与MySQL数据库中的任何“名称”都不匹配时，我不知道如何使搜索结果显示“未找到匹配项”或完全隐藏结果div
当前，当用户在搜索栏中输入与数据库中任何“名称”都不匹配的内容时，AJAX实时搜索结果下方会弹出空白结果。相反，我希望消息“未找到匹配项”取代该空白结果。
我已经尝试了许多其他/ if /回声代码以及不同顺序的组合，但到目前为止没有任何效果。我还尝试根据结果显示/隐藏显示“未找到匹配项”的div的另一种方法。
如何一劳永逸地修复此代码，以便当用户在MySQL数据库中搜索与任何名称都不匹配的任何名称时，它会显示“找不到匹配项”？
以下是我当前正在使用的文件和代码：
index.php
<form>
<input type="text" id="search" class="search" data-js="form-text"
placeholder="Search Over 100+ Resources..." autocomplete="off">
<button type="submit" class="Button" value="Submit"><i class="fa fa-
search"></i></button>
<div id="display"></div>
<div id="no-results" style="display:none"><ul><li id='hover'>No matches
found</li></ul></div>
</form>
ajax.php
<?php
//Including Database configuration file.
include "db.php";
//Getting value of "search" variable from "script.js".
if (isset($_GET['search'])) {
//Search box value assigning to $Name variable.
$Name = $_GET['search'];
//Search query.
$Query = "SELECT Name FROM search WHERE Name LIKE '$Name%' LIMIT 5";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
echo '<ul>';
//Fetching result from database.
while ($Result = MySQLi_fetch_array($ExecQuery)) {
?>
<!-- Creating unordered list items.
Calling javascript function named as "fill" found in "script.js" file.
By passing fetched result as parameter. -->
<li onclick='fill("<?php echo $Result['Name']; ?>")'>
<a>
<!-- Assigning searched result in "Search box" in "index.php" file. -->
<?php
if ($ExecQuery > "0") {
echo $Result['Name'];
}
else {
echo "<li id='hover'>No matches found</li>";
}
?>
</li></a>
<!-- Below php code is just for closing parenthesis. Don't be confused. -->
<?php
}}
?>
</ul>
script.js
//Getting value from "ajax.php".
function fill(Value) {
//Assigning value to "search" div in "index.php" file.
$('#search').val(Value);
//Hiding "display" div in "index.php" file.
$('#display').hide();
}
$(document).ready(function() {
//On pressing a key on "Search box" in "index.php" file. This function will
be called.
$('#no-results').hide();
$("#search").keyup(function() {
//Assigning search box value to javascript variable named as "name".
$('#display').hide();
$('#no-results').css("display", "none");
var name = $('#search').val();
//Validating, if "name" is empty.
if (name == "") {
//Assigning empty value to "display" div in "index.php" file.
$('#no-results').css("display", "none");
}
//If name is not empty.
else {
//AJAX is called.
$.ajax({
//AJAX type is "Post".
type: "GET",
//Data will be sent to "ajax.php".
url: "ajax.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "name" into "search" variable.
search: name
},
//If result found, this funtion will be called.
success: function(html) {
//Assigning result to "display" div in "index.php" file.
$("#display").html(html).show();
}
});
}
});
});
答案 0 :(得分：5)
已更新
您应检查数据是否有效，是否从数据库查询中得到任何结果，如果没有记录，则可以打印未找到数据消息。
您应该检查$ExecQuery的输出，并据此设置if条件。
现在让我输入您的输出和结果，希望对您有所帮助。
更新ajax.php 上次更新的部分
echo "<li onclick='fill(`".$Result['Name']."`)'>".$Result['Name']."</li>";
完成ajax.php
<?php
//Including Database configuration file.
include "db.php";
//Getting value of "search" variable from "script.js".
if (isset($_GET['search'])) {
//Search box value assigning to $Name variable.
$Name = $_GET['search'];
//Search query.
$Query = "SELECT Name FROM search WHERE Name LIKE '$Name%' LIMIT 5";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
if ($ExecQuery->num_rows > 0) {
echo "<ul>";
while ($Result = MySQLi_fetch_array($ExecQuery)) {
// use the onclick function that defined in js file. you can use the ` sign in js instead of ' if you needed.
echo "<li onclick='fill(`".$Result['Name']."`)'>".$Result['Name']."</li>";
}
echo "</ul>";
}else{
echo "<ul><li>No Result Found!</li></ul>";
}
}
die();
?>
和您的Ajax代码。
function fill(value) {
console.log(value);
$('#search').val(value);
$('#display').hide();
}
$(document).ready(function() {
//On pressing a key on "Search box" in "index.php" file. This function will be called.
$("#search").keyup(function() {
//Assigning search box value to javascript variable named as "name".
$('#display').hide();
$('#no-results').css("display", "none");
var name = $('#search').val();
//Validating, if "name" is empty.
if (name == "") {
//Assigning empty value to "display" div in "index.php" file.
$('#no-results').css("display", "block");
}
//If name is not empty.
else {
//AJAX is called.
$.ajax({
//AJAX type is "Post".
type: "GET",
//Data will be sent to "ajax.php".
url: "ajax.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "name" into "search" variable.
search: name
},
//If result found, this funtion will be called.
success: function(html) {
if (html == '<ul><li>No Result Found!</li></ul>') {
$('#no-results').css("display", "block");
}else{
//Assigning result to "display" div in "index.php" file.
$("#display").html(html).show();
}
}
});
}
});
});
根据需要更改其他部分。
答案 1 :(得分：0)
AJAX是异步Javascript和XML。为什么要发回HTML？
指针
如果您通过Ajax进行此操作，我强烈建议您不要发送纯HTML。您应该发送回一些JSON数据并相应地在客户端进行处理。
使用PDO代替MySQLi
解决方案PHP
<?php
include "db.php";
if (isset($_POST['search'])) {
$Name = $_POST['search'];
$Query = "SELECT Name FROM search WHERE Name LIKE '$Name%' LIMIT 5";
$ExecQuery = MySQLi_query($con, $Query);
$res = [];
while ($Result = MySQLi_fetch_array($ExecQuery)) {
$res[] = $Result['Name'];
}
echo json_encode($res);
}
解决方案Javascript：
$.ajax({
//AJAX type is "Post".
type: "POST",
//Data will be sent to "ajax.php".
url: "ajax.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "name" into "search" variable.
search: name
},
//If result found, this funtion will be called.
success: function(data) {
//Assigning result to "display" div in "search.php" file.
const list = JSON.parse(data);
const html = list.length > 0 ?
list.map(name=>{
`<li onclick="${name}">
<a>${name}</a>
</li>`
}).join("") :
`<li>No matches found</li>`
$("#display").html(`<ul>${html}</ul>`).show();
}
});