警告：mysql_select_db（）期望参数2是资源，

Question

<?php
$servername = "localhost";
$username = "root";
$password = "Rachel";
$db = "hairdressingapointments";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $db);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 
echo "Connected Sussessfully";
mysql_select_db('Hairdressingapointments', $conn) or die(mysql_error());
$sql = "SELECT `ApointmentDate`, `ApointmentTime` FROM `apointments` WHERE          `staff_id`=1 && `quantity`>0";
if(!mysql_query($sql)){
die('Error: ' . mysql_error());
}
echo $sql;

mysql_close();

?>

花了好几个小时试图解决这个问题并且我猜测它的东西如此简单。找回以下错误：   警告：mysql_select_db（）期望参数2是资源，第15行的C：\ wamp2 \ www \ hairdressingapointments \ TeresaApointments.php中给出的对象，

mysql_select_db('Hairdressingapointments', $conn) or die(mysql_error());

Answer 1

1. 您已使用

   连接到数据库

   mysqli_connect（...）;

   所以，你不需要

   mysql_select_db（....）;

2. 同时将查询更改为此

   $ sql =“SELECT ApointmentDate，ApointmentTime FROM apointsments WHERE staff_id = 1 AND quantity＆gt; 0”;

   如果您使用SQLWorkbench或SQLYog或其他工具，您可以输入SQL并确保它在将其添加到脚本之前有效。

3. 另外，请确保表名真的是

  

apointments

而不是

  

约会

我从php.net - mysqli_connect

获得了这些信息