PHP警告:mysqli_fetch_array()希望参数1是mysqli_结果,布尔值在
我得到警告: mysqli_fetch_数组期望参数1是mysqli_结果,布尔值在 这是我的密码:PHP警告:mysqli_fetch_array()希望参数1是mysqli_结果,布尔值在,php,mysqli,Php,Mysqli,我得到警告: mysqli_fetch_数组期望参数1是mysqli_结果,布尔值在 这是我的密码: <?php session_start(); $db=mysqli_connect("localhost","aaron","","demo"); $id=$_GET["id"]; $sql=mysqli_query($db,"SELECT * FROM usres"); $check=mysqli_fetch_array($db,$sql);
<?php
session_start();
$db=mysqli_connect("localhost","aaron","","demo");
$id=$_GET["id"];
$sql=mysqli_query($db,"SELECT * FROM usres");
$check=mysqli_fetch_array($db,$sql);
if(isset($_POST['update'])){
$id=$_POST['id'];
$name=$_POST['name'];
$email=$_POST['email'];
$password=$_POST['password'];
$bankbookno=$_POST['bankbookno'];
$adharno=$_POST['adharno'];
$pancard=$_POST['pancard'];
$result = mysqli_query($db, "UPDATE users SET name='$name',email='$email',password='$password',bankbookno='$bankbookno' ,adharno='$adharno',pancard='$pancard'WHERE id=$id");
header("location:view.php");
}
?>
作为第一个参数。您可以在mysqli_查询返回时获得mysqli_结果
例如:
$db = mysqli_connect("localhost","aaron","","demo");
$sql = mysqli_query($db,"SELECT * FROM usres");
$check = mysqli_fetch_array($sql);
还请注意,在SQL语句中直接通过$id使用$_POST['id'],将启用针对应用程序的SQL注入攻击。有很多不同的方法,其中之一是。mysqli\u fetch\u数组只需要一个参数,由mysqli\u查询、mysqli\u存储或mysqli\u使用结果返回
$sql=mysqli_query($db,"SELECT * FROM usres");
$check=mysqli_fetch_array($sql);//removed $db, which is not needed here
mysqli_fetch_array函数将结果行作为关联数组、数字数组或两者都提取
mysqli_fetch_数组参数应为:
由mysqli_查询、mysqli_存储结果或mysqli_使用结果返回
$sql=mysqli_query($db,"SELECT * FROM usres");
$check=mysqli_fetch_array($sql);//removed $db, which is not needed here
编辑代码并提出问题。您的帖子可能会被删除。请添加解释以改进您的问题警告:您对参数化预处理语句持开放态度,应使用参数化预处理语句,而不是手动生成查询。它们由或提供。永远不要相信任何形式的输入!即使您的查询仅由受信任的用户执行。同样的错误再次出现