php實現簡單的登入
阿新 • • 發佈：2019-01-02
我用了bootstrap框架，不知為什麼一直表單不能正常排序顯示，不過主要學習php邏輯，先不糾結。
登入頁：test.php
<!DOCTYPE html>
<html lang="zh-CN">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1"> 處理頁：
produce_action.php
<?php
header("Content-Type: text/html;charset=utf-8");
$name = $_POST["name"];
$pwd = $_POST["pwd"];
if(is_null($name) || strlen($name) ==0 ){
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('請輸入使用者名稱！');</script>";
}else if(is_null($pwd)||strlen($pwd) == 0){
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('請輸入密碼！');</script>";
}else{
echo "name= ".$name;
echo "<br/>pwd=".$pwd;
if(!is_null($name))
$_SESSION['userId']=$name;
echo "<br/> session=".$_SESSION['userId'];
}
?>進階版，增加資料庫判斷：
<?php
header("Content-Type: text/html;charset=utf-8");
$name = $_POST["name"];
$pwd = $_POST["pwd"];
if(is_null($name) || strlen($name) ==0 ){
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('請輸入使用者名稱！');</script>";
}else if(is_null($pwd)||strlen($pwd) == 0){
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('請輸入密碼！');</script>";
}else{
$conn=mysql_connect("localhost", "root","root");
if(!$conn){
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('連線資料庫失敗！');</script>";
return;
}
mysql_select_db("test",$conn);
$sql = "select * from users where userName='".$name."'";
$info1=mysql_query($sql,$conn);
if($row = mysql_fetch_object($info1)){
if($row->UserPwd==$pwd){
echo "登入成功！";
echo '<br>';
break;
}else{
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('密碼錯誤！');</script>";
break;
}
}else{
echo "<script LANGUAGE='Javascript'> location.href='test.php'; alert('不存在該使用者！');</script>";
}
}
?>注意，這裡的test要替換成資料庫名，資料庫裡面要新增資料方便測試