顯示“未找到匹配項”或隱藏DIV結果（AJAX和MySQL）

Question

我有一個搜索欄，用於顯示MySQL，PHP和JS的AJAX實時搜索結果。

問題是當查詢與MySQL數據庫中的任何“名稱”都不匹配時，我不知道如何獲取搜索結果以顯示“未找到匹配項”或完全隱藏結果div 。

當前，當用戶在搜索欄中鍵入與數據庫中任何“名稱”都不匹配的內容時，AJAX實時搜索結果下方將彈出空白結果。 相反，我希望消息“未找到匹配項”接管該空白結果。

我已經嘗試了許多其他/如果/回聲代碼和組合以不同的順序，到目前為止沒有任何工作。 我還嘗試根據結果顯示/隱藏顯示“未找到匹配項”的div的另一種方法。

我該如何徹底修復此代碼，以便當用戶搜索與MySQL數據庫中的任何名稱都不匹配的任何名稱時，它會顯示“找不到匹配項”？

以下是我當前正在使用的文件和代碼：

的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>

的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();
           }
       });
   }
 });
 });

Answer 1

更新

您應該檢查數據是否有效，是否從數據庫查詢中得到任何結果，如果沒有記錄，則可以打印未找到數據消息。 你應該檢查的輸出$ExecQuery並設置if根據這一條件。 現在讓我提供您的輸出結果，希望對您有所幫助。

更新ajax.php最近更新的部分

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();
?>

和你的ajax代碼。

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();
             }

           }
       });
   }
 });
 });

根據需要更改其他零件。

Answer 2

AJAX是異步Javascript和XML。 為什么要發回HTML？

指針

• 如果您通過Ajax進行此操作，那么我強烈建議您不要發送純HTML。 您應該發送回一些JSON數據，並相應地在客戶端進行處理。

• 使用PDO代替MySQLi

解決方案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);
}

解決方案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();
  }
});