PHP错误: mysqli_query()期望参数1为mysqli，布尔给定

Question

这不是问题，我需要你的帮助。我读过类似的线程，但无法调试代码中的问题。你能给出正确的答案吗？

<?php
/*variable declaration*/
$host="localhost";
$user="root";
$pass="";
$dbname='mydatabase';

/*connection to mysql server*/
$connect = mysqli_connect($host,$user,$pass);

/*selecting database*/
$selectdb=mysqli_select_db($connect,$dbname);

if(!$selectdb){
    echo 'Failed to connect. Wrong username or database.';
}else{
    echo 'Connection successful.';
}

/*creating task*/
$query = "SELECT 'Name', 'Password' FROM 'db' ORDER BY 'id'";

if(mysqli_query($selectdb,$query)){
echo 'Success';
}else{
echo '<br>Failed';
}
?>

在这里输入图像描述

Answer 1

修改

$query = "SELECT 'Name', 'Password' FROM 'db' ORDER BY 'id'";

至

$query = "SELECT `Name`, `Password` FROM db ORDER BY 'id'";

您不能在查询中使用列名上的引号，也不能使用backticks或根本不使用。

Answer 2

<?php
/*variable declaration*/
$host="localhost";
$user="root";
$pass="";
$dbname='mydatabase';
$connect = mysqli_connect($host,$user,$pass,$dbname);

/*I avoided this line & used $connect inside if() statement which worked    perfectly*/
//$selectdb = mysqli_select_db($connect,$dbname);

if(!$connect){
    echo 'Failed to connect. Wrong username or database.';
}else{
    echo 'Connection successful.';
}

/*creating task*/
$query = "SELECT `Name`, `Password` FROM `db` ORDER BY `id`";

if($query_run = mysqli_query($connect,$query)){
echo 'Success';
}else{
echo '<br> Failed';
}

Answer 3

if(mysqli_query($connect,$query)){...

在这种情况下，您需要使用$connect而不是$selectdb。