Question

我收到此错误：

警告：mysql_select_db（）期望参数1为string，object 在第11行的C：\ xampp \ htdocs \ apply \ insert.php中给出未选择数据库未插入

<?php

$con = mysqli_connect('localhost','root','');

if (!$con)
{
    echo 'Not connected to server';
}

if (!mysql_select_db($con,'staff'))
{
    echo "Database not selected";
}

$Name = $_POST['username'];
$Email = $_POST['email'];

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')";

if(!mysqli_query($con,$sql))
{
    echo 'Not inserted';
}
else {
    echo 'Inserted';
}

header("refresh:2; url=index.html");

?>

Answer 1

您正在将mysql_与mysqli_个功能混合使用。

将if (!mysql_select_db($con,'staff'))更改为if (!mysqli_select_db($con,'staff'))

Answer 2

编写如下代码：

<?php


$con = mysqli_connect('localhost','root','');

if (!$con)
{
    echo 'Not connected to server';
}

if (!mysqli_select_db($con,'staff'))
{
    echo "Database not selected";
}

$Name = $_POST['username'];
$Email = $_POST['email'];

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')";

if(!mysqli_query($con,$sql))
{
    echo 'Not inserted';
}
else {
    echo 'Inserted';
}

Answer 3

像这样编写你的代码

<?php
$con = mysqli_connect('localhost','root','','staff');
if (!$con)
{
echo 'connection not established';
}

$Name = $_POST['username'];
$Email = $_POST['email'];

$sql = "INSERT INTO person (Name, Email) VALUES ('$Name','$Email')";

if(!mysqli_query($con,$sql))

{
echo 'Not inserted';
}

else {
echo 'Inserted';
}

警告：mysql_select_db（）期望参数1为字符串，

3 个答案: