Question

我正在尝试通过制作一个html表单然后通过$ _GET方法添加接收数据来接收数据。每当我尝试提交数据时，我都会收到一条错误消息：“找不到您的文件”＃39;在我的Chrome浏览器中，所以我尝试通过输入＆＃34; localhost /....＆＃34来打开我的Chrome页面。在我的地址栏中，它显示了“未找到数据库”

这是我的PHP代码： -

<html>
<?php


$user_name = "root";
$password = "";
$database = "mysql"; //mysql is name of my database
$server = "localhost";

$db_handle = mysql_connect($server , $user_name, $password,"addresbook");      //adressbook is where all the tabels are
$db_found = mysql_select_db($database, $db_handle);


if (!$db_found) {

print "Database Found ";
$x=$_GET['fname'];
$y=$_GET['sname'];
$z=$_GET['address'];

$sql="INSERT INTO addresbook(First_Name,Surname,Address)     VALUES('".$x."','".$y."','".$z."')";
mysql_query($sql,$db_handle);

}
else {

print "Database NOT Found ";

}

?>
</html>

这里是mt html代码： -

<form action="practic.php"method="get">
Firstname:<input type="text" name="fname"><br>
Lastname:<input type="text" name="sname"><br>
Address:<input type="text" name="address"><br>
<input type="submit">
</form>

顺便说一句，我正在使用wamp server.Thanks提前。

Answer 1

$db_found成功时会成立，所以你的条件应该是

if ($db_found) { // make DB changes etc.

并切换到MySQLi或PDO并使用已经提到的预准备语句，请参阅手册：

http://php.net/manual/en/mysqli.prepare.php

Answer 2

使用mysqli_connect因为mysql_connect现已弃用。

$db_handle = mysqli_connect($server , $user_name, $password,$database);

请参阅此处的连接

http://www.w3schools.com/php/func_mysqli_connect.asp

Answer 3

您可以更改代码以使用if(video.readyState == 4){ $(window).scrollTop($(window).scrollTop()+1); $(window).scrollTop($(window).scrollTop()-1); }或使用MysqLi

PDO

您的<?php $user_name = "root"; $password = ""; $database = "mysql"; //mysql is name of my database $server = "localhost"; $con = mysqli_connect($server,$user_name,$password,'adresbook');//adresbook is where all the tabels are // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); exit(); } // Change database to "adresbook" $db_found = mysqli_select_db($con,'adresbook'); if ($db_found) { print "Database Found "; $x=$_GET['fname']; $y=$_GET['sname']; $z=$_GET['address']; $sql="INSERT INTO addresbook(First_Name,Surname,Address) VALUES('".$x."','".$y."','".$z."')"; mysqli_query($con, $sql); } else { print "Database NOT Found "; } ?>应该与包含phpMyadmin的数据库addresbook一样。

在数据库中接收数据时遇到问题

3 个答案: