php - 带有 LIKE '% ? %' PHP 的 SQL 参数化查询

Question

我在 php 中有一个搜索功能，并使用参数化查询创建它以确保其安全。

$words = $_POST['words']//words is the form that has the words submitted by the user 
$array = explode(',', $words);
$con = mysqli_connect("localhost","user","pass","database");

$stmt = $con->prepare(" SELECT column_name FROM table WHERE column_name LIKE ?")
foreach($array as $key) { //searches each word and displays results   
  $stmt->bind_param('s', $key)
  $stmt->execute();
  $result = $stmt->get-result();

  while($row = $result->fetch_assoc(){
    echo $row["column_name"]
  }
}

但是我希望$stmt声明是

  $stmt = $con->prepare(" SELECT column_name FROM table WHERE column_name LIKE '%?%' ")

否则人们必须输入 column_name 的整个值才能找到它。

Answer 1

您可以使用 CONCAT() ，像这样:

LIKE CONCAT ('%', ?, '%')