Question

在登录时遇到此错误，我安装了新版本的xampp / php仍然面临着很多mysqli的问题......

下面是login.php的代码，然后转到lock.php，代码在下面......

<?php

include("config.php");
session_start();

if($_SERVER["REQUEST_METHOD"] == "POST")
{
// username and password sent from form 

$myusername=addslashes($_POST['username']); 
$mypassword=addslashes($_POST['password']); 


$sql="SELECT id FROM mc_admin WHERE username='$myusername' and passcode='$mypassword'";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
$active=$row['active'];

$count=mysql_num_rows($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: home.php");
}
else 
{
$error="Your Login Name or Password is invalid";
}
}
?>

下面的lock.php代码：

<?php
include('config.php');
session_start();
$user_check=$_SESSION['login_user'];

$ses_sql=mysql_query("select username from mc_admin where username='$user_check' ");

$row=mysql_fetch_array($ses_sql);

$login_session=$row['username'];

if(!isset($login_session))
{
header("Location: login.php");
}
?>

Answer 1

MYSQL查询在出错时返回FALSE，因此您的一个查询在执行时出错。

根据PHP文档mysql_query() returns a resource on success, or FALSE on error所以在尝试调用mysql_fetch_array检查之前，如下所示：

您可以打印错误

$result = mysql_query('your query');
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

此外，您必须注意，自PHP5.5起，现在不推荐使用mysql_ *函数。您可以使用mysqli或PDO

来代替mysql_ *

Answer 2

转到mysqli！检查一下我是如何解决我的。当查询是INSERT时，我收到相同的错误，因为它返回空，所以我只在需要时调用fetch_assoc()。

<强> mysqlcon.php

<?php

function mysqlquery($query,$dbcon,$returnValue = false){
  $result = $dbcon->query($query);

  if(!$result){
    die('There was an error running the query [' . $dbcon->error . ']');
  }

  //Do not use `fetch_assoc()` in INSERT query to prevent
  //`PHP Fatal error:  Call to a member function fetch_assoc() on a non-object` error.
  if($returnValue){
    return $result->fetch_assoc();
  }
}


function dbcon(){

  $db_address = 'localhost';
  $db_user = 'root';
  $passwd = 'your_password';
  $db = 'your_db';

  $dbcon = new mysqli($db_address, $db_user, $passwd, $db);
  if($dbcon->connect_errno > 0){
    die('Unable to connect to database [' . $dbcon->connect_error . ']');
  }

  return $dbcon;
}

使用示例：

<强> exaple.php

<?php
require_once('mysqlcon.php');

$dbcon = dbcon();

$result = mysqlquery("SELECT * FROM TABLE WHERE COLUMN='hello'",$dbcon,true);
echo $res = $result['db_column'];

错误：mysqli_fetch_array（）期望参数1为mysqli_result，给出布尔值

2 个答案: