<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>
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 $searchKey; ?>" >
您不仅在输入字段中丢失了id，而且您也从未在php代码中执行查询并获取结果。sql注入是可能的，请清理您的$search chKey或使用准备好的语句。
也许您还可以将搜索查询放在不同的php文件中，并在那里创建html，返回ajax函数。
也许这会有所帮助：
为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search chData.phpajax函数将获得该文件中所有的text/html
