Question

一种Stack Overflow成员在将不安全的查询转换为使用参数的查询方面给了我一些帮助，但我很难让它工作。我本可以一直打扰他，但最初的问题是无关紧要的。所以我发布了一个新的！

<?php


$mysqli = new mysqli("host", "user", "password", "database");// connection
$key = $_POST['key'];
$query = $mysqli->prepare("SELECT * FROM handover  WHERE hs LIKE ? OR WHERE nv    

LIKE ? OR WHERE vsa LIKE ?"); 
$stmt->bind_param("sss", $key, $key, $key); 
$stmt->execute();
$result = mysqli_query($mysqli,$query);

$count=mysqli_num_rows($result);

if($count==0 ){

echo "</br></br></br></br></br></br></br><h2>Handover Details</h2><p>  No results    

found</p>";
}
else{
while($row = mysqli_fetch_array($result)) {

These are my errors:

我收到此错误：

注意：未定义的变量：第112行的E：\ NEW1 \ xampp \ htdocs \ FINTAN \ keyword.php中的stmt

致命错误：在第112行的E：\ NEW1 \ xampp \ htdocs \ FINTAN \ keyword.php中的非对象上调用成员函数bind_param（）

<?php


$mysqli = new mysqli("localhost", "root", "", "test3");// connection
$key = $_POST['key'];
$query = $mysqli->prepare("SELECT * FROM handover WHERE hs LIKE ? OR nv LIKE ? OR vsa 

LIKE    ?"); 
$stmt->bind_param("sss", $key, $key, $key); 
$stmt->execute();
$stmt->store_result();
$rows = $stmt->num_rows;

if($rows==0 ){

echo "</br></br></br></br></br></br></br><h2>Handover Details</h2>

<p> WORKING none    

found</p>";
}
else{
while($row = mysqli_fetch_array($result))

Answer 1

SELECT * FROM handover  WHERE hs LIKE ? OR nv    
LIKE ? OR vsa LIKE ?"

删除额外的WHERE，你也需要

$query->bind_param("sss", $key, $key, $key); 
$query->execute();

$query->store_result();
$rows = $query->num_rows;

if($rows==0 )

如何解决PHP程序中的“未定义变量”？

1 个答案: