So here is my php file that connects to my database "pizza"
所以这是我的PHP文件,连接到我的数据库“披萨”
<?php
//Connect to database
$link = new mysqli('127.0.0.1', 'root', '', 'pizza');
if (!$link) {
//output this message if connection is unsuccessful
$output = 'Unable to connect to the database server.';
exit();
}
?>
Next is my attempt at trying to submit the data
接下来是我尝试提交数据的尝试
<?php
//Include our file that connects us to Pizza database
include("connect.php");
$userAddress = $_POST["address"];
$userPhone = $_POST["phoneNo"];
//Insert new data into database
$sql = "INSERT INTO orders (address, email, phone)
VALUES ('$userAddress', '$userEmail', '$userPhone')";
if (mysqli_query($sql)) {
echo "New record created successfully";
} else {
echo "Error ";
}
mysql_query($sql);
?>
Can anyone see what is wrong here?
有谁能看到这里有什么问题?
2 个解决方案
#1
Replace this line:
替换此行:
if (mysqli_query($sql)) {
echo "New record created successfully";
}
With this:
if (mysqli_query($link, $sql)) {
echo "New record created successfully";
}
#2
Considering you have instantiated a mysqli
object, you should use its methods to query your database.
考虑到你已经实例化了一个mysqli对象,你应该使用它的方法来查询你的数据库。
$link = new mysqli('127.0.0.1','root','','pizza');
$query = "INSERT INTO orders (address) VALUES ('$userAddress')";
if ($link->query($query)) {
//do stuff
}
#1
Replace this line:
替换此行:
if (mysqli_query($sql)) {
echo "New record created successfully";
}
With this:
if (mysqli_query($link, $sql)) {
echo "New record created successfully";
}
#2
Considering you have instantiated a mysqli
object, you should use its methods to query your database.
考虑到你已经实例化了一个mysqli对象,你应该使用它的方法来查询你的数据库。
$link = new mysqli('127.0.0.1','root','','pizza');
$query = "INSERT INTO orders (address) VALUES ('$userAddress')";
if ($link->query($query)) {
//do stuff
}