[英]Problems with mysqli_fetch_array() expects parameter 1 to be mysqli_result
[英]mysqli_fetch_array & row expects parameter 1 to be mysqli_result
我是SQL的新手,正在嘗試建立一個登錄系統。 我已按照指南進行操作,但是嘗試登錄時會收到以下2條消息:
警告:mysqli_fetch_array()期望參數1為mysqli_result,在第13行的C:\\ xampp \\ htdocs \\ loginguide / login.php中給出布爾值
警告:mysqli_num_rows()期望參數1為mysqli_result,在第16行的C:\\ xampp \\ htdocs \\ loginguide \\ login.php中給出布爾值
我在MyPHPAdmin上創建了一個數據庫,但是我也不知道該怎么做,我是要向該數據庫添加表嗎?
這是login.php代碼:
<?php
include("config.php");
session_start();
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$myusername = mysqli_real_escape_string($db,$_POST['username']);
$mypassword = mysqli_real_escape_string($db,$_POST['password']);
$sql = "SELECT id FROM admin WHERE username = '$myusername' and passcode = '$mypassword'";
$result = mysqli_query($db,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$active = $row['active'];
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
session_register("myusername");
$_SESSION['login_user'] = $myusername;
header("location: welcome.php");
}else {
$error = "Your Login Name or Password is invalid";
}
}
?>
<html>
<head>
<title>Login Page</title>
<style type = "text/css">
body {
font-family:Arial, Helvetica, sans-serif;
font-size:14px;
}
label {
font-weight:bold;
width:100px;
font-size:14px;
}
.box {
border:#666666 solid 1px;
}
</style>
</head>
<body bgcolor = "#FFFFFF">
<div align = "center">
<div style = "width:300px; border: solid 1px #333333; " align = "left">
<div style = "background-color:#333333; color:#FFFFFF; padding:3px;"><b>Login</b></div>
<div style = "margin:30px">
<form action = "" method = "post">
<label>UserName :</label><input type = "text" name = "username" class = "box"/><br /><br />
<label>Password :</label><input type = "password" name = "password" class = "box" /><br/><br />
<input type = "submit" value = " Submit "/><br />
</form>
<div style="font-size:11px;<?php echo $error; ?></div>
</div>
</div>
</div>
</body>
</html>
1)您需要在SQL查詢中包括“活動”列,或僅將*替換為*
select * from admin where ....
(or)
select id, active from admin where ....
2)檢查給定示例中包含的config.php文件,如果DB_USERNAME,DB_PASSWORD,DB_DATABASE名稱輸入錯誤(語法錯誤),則會收到該警告。
<?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', 'root');
define('DB_DATABASE', 'testDB');
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
?>
我認為您的查詢失敗並返回了錯誤的值。 檢查您的查詢,如果它運行正常,請將其放在您的代碼中:
$result = mysqli_query($db,$sql);
if (!$result) {
printf("Error: %s\n", mysqli_error($db));
exit();
}
欲獲得更多信息。
您只需選擇ID,然后使用“活動”
因此,您應該首先解決該問題:
SELECT * FROM admin WHERE username = '$myusername' and passcode = '$mypassword'
現在應該可以工作了
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