我尝试使用php创建一个搜索框。然而，当我搜索代码时，它不工作

Question

<div class="row">
                <?php
                $connection = mysqli_connect('localhost','root','','product-store');
                if(isset($_POST['search'])) {
                    $searchKey = $_POST['search'];
                     $sql = "SELECT * FROM products WHERE code_no LIKE '%$searchKey%'";


                } else {


                    $sql = "SELECT * FROM products order by code_no desc";
                    $searchKey = "";


                }

            ?>
            <form action="tabledata.php" method="POST"> 
                <div class="col-md-12 col-sm-12 col-xs-12">
                    <input type="text" name="search" class="form-control" placeholder="Search By Code" value="<?php echo $searchKey; ?>" > 
                </div>
                <br>
                <div class="input-group">
                    <button class="btn btn-success">Search</button>
                </div>
                <br>
            </form>

            <br>
            <br>

            </div>



        </div>
    </div>
</div>

​

       $(document).ready(function(){        $("#search").keyup(function() {           var query = $(this).val();           if (query != "") {       $.ajax({        url:"tabledata.php",        method:"POST",        data:{query:query},        success:function(data)        {         $('#result').html(data);        }       });      }      $('#searchKey').keyup(function(){       var search = $(this).val();       if(search != '')       {        load\_data(search);       }       else       {        load\_data();       }      });     });

Answer 1

您在输入字段中缺少id

<input type="text" id="search"  name="search" class="form-control" placeholder="Search By Code" value="<?php echo $searchKey; ?>" >

Answer 2

您不仅在输入字段中缺少id，而且也从未在php代码中执行查询并获取结果。和sql注入是可能的，清理您的$searchKey或使用准备好的语句。

也许您也可以将搜索查询放在一个不同的php文件中，并在那里创建您希望返回给ajax函数的html。

也许这会有所帮助：

为您的ajax搜索功能创建一个新的php文件，如searchData.php

    <?php //searchData.php

if ($_SERVER['REQUEST_METHOD'] === 'POST' && isset($_POST['query'])) {
  $connection = mysqli_connect('localhost','root','','product-store');

  $searchKey  = $_POST['query']; // Needs to be sanitized to prevent sql injections
  $sql   = mysqli_query($connection, "SELECT * FROM products WHERE code_no LIKE '%$searchKey%'");

  if (mysqli_num_rows($sql) > 0) {
    // Fetch the rows
    $rows = mysqli_fetch_array($sql, MYSQLI_ASSOC);

    $html = '';
    // Loop through all the rows
    foreach ($rows as $row) {
      // Create here your html you want to return to your ajax function
    }

    echo $html;

  } else {
    // No results, echo html/ text to the ajax function to inform the user
  }
}

将ajax函数中的url更改为searchData.php，然后ajax函数将获得您在此文件中回显的所有文本/html。