显示“未找到匹配项”或隐藏DIV结果（AJAX和MySQL）

时间：2019-03-18 18:30:51

标签： javascript php mysql ajax

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

2 个答案:

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();
  }
});
相关问题
最新问题