Question

好的，所以基本上我想为某些网站产品建立一个简单的搜索系统。我的目标是确保用户每次搜索内容时，他/她只会看到标题与搜索字词匹配的项目。

这是我的代码产品页面代码：

<form action="search.php" id="search_form" name="search_form" method="post">
<div class="form-field">
<label for="search-field">Search</label>
<input type="search" name="search_form" />
<input type="submit" value="Search">
</div>
</form>

这是我的search.php代码：

<?php
  include 'config.php';
  $x = $_POST['search_form'];
  $stmt = $conn->prepare("SELECT * FROM  WHERE product_name='".$x."'");
  $rowName = $row['name'];
  $stmt->execute();
  $result = $stmt->get_result();
  while($row = $result->fetch_assoc()): 
?>

Answer 1

SQL应该看起来像这样：

SELECT * FROM  WHERE product_name LIKE '%socks%'

这将符合以下条件：

'socks for cats'
'green socks'
'knitted socks for giraffes'

更改PHP代码以使用LIKE实现以上目的。

假设您正在使用PDO：

include 'config.php';
$x = '%' . $_POST['search_form'] . '%';
$stmt = $conn->prepare("SELECT * FROM  WHERE product_name LIKE ?");
$stmt->execute([$x]);
while($row = $stmt->fetch()){
   // do something with product data
}

mysqli

include 'config.php';
$x = '%' . $_POST['search_form'] . '%';
$stmt = $conn->prepare("SELECT * FROM  WHERE product_name LIKE ?");
$stmt->bind_param("s", $x);
$stmt->execute();
$res = $stmt->get_result();
while($row = $res->fetch_assoc()){
   // do something with product data
}

SELECT * FROM表WHERE name LIKE“'.search_variable。' “不起作用

1 个答案: