Question

页面设置为运行简单的数据库查询。表单和提交按钮呈现，但是当我键入搜索项时，它会返回错误。当我在下拉菜单中使用它来调用查询时，数据库连接脚本运行良好，所以我不确定为什么它在这里不起作用。

我编辑了我的原始帖子以删除MySQL并将其替换为如下：我现在遇到的问题是错误是说我在第28行有一个未定义的变量if if（$ result-＆gt; num_rows＆gt ; 0）{

    <?php

$servername = "localhost";
$username = "xxx";
$password = "xxx";
$dbname = "oldga740_SeniorProject";


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
}


$output= '';
//collect 
if(isset($_POST['search'])) {
  $searchq = $_POST['search'];
  $searchq = preg_replace("#[^0-9a-z]#i","",$searchq);

  $query = "SELECT * FROM Customer WHERE Client LIKE '%$searchq%'";
  $result = $conn->query($query);
}

//Display results
if ($result->num_rows > 0) {
     echo $result;

}

else {
     echo "0 results";
}


?>

<html>

<head>

</head>

<body>

<form action="Index.php" method="post">
  <input type="text" name="search" placeholder="Search...." />
  <input type="submit" value=">>" />
</form>



</body>

</html>

Answer 1

请替换您的查询

 $query = mysql_query("SELECT * FROM Customer WHERE Client LIKE '%$searchq%'") or die("could not search");

通过

$query = mysql_query("SELECT * FROM Customer WHERE Client LIKE '%".$searchq."%'") or die("could not search");

我收到一个语法错误访问被拒绝用户'root'@'localhost'（使用密码：NO）

1 个答案: