我收到一个语法错误 Access denied for user 'root'@'localhost' (using password: NO)

Question

该页面设置为运行简单的数据库查询。 表单和提交按钮呈现，但当我输入搜索项时，它返回错误。 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，即 if ($result->num_rows > 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");

经过

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

我收到一个语法错误 Access denied for user 'root'@'localhost' (using password: NO)

问题描述

1 个解决方案

解决方案1
-1 2015-11-21 13:08:44

我收到一个语法错误 Access denied for user &#39;root&#39;@&#39;localhost&#39; (using password: NO)

问题描述

1 个解决方案

解决方案1 -1 2015-11-21 13:08:44

我收到一个语法错误 Access denied for user 'root'@'localhost' (using password: NO)

解决方案1
-1 2015-11-21 13:08:44