Question

我使用以下ajax脚本将用户登录到我的网站。提交时的ajax脚本将表单数据传递给我的页面validate_login.php：

的index.php：

<script src="assets/jquery/jquery.min.js"></script>
<script type="text/javascript" src="assets/jquery/flip/jquery.flip.js"></script>
<script type="text/javascript" src="assets/jquery/flip/jquery.flip.min.js"></script>
<script src="assets/jquery/jquery-ui.js"></script>

<script type="text/javascript"> 
$(document).ready(function() {
$(document).on('click', '#submit', function() {
var myusername = $("#myusername").val();
var mypassword = $("#mypassword").val();
if (myusername == null || myusername == "" || mypassword == null || mypassword == "") {
if (myusername == null || myusername == "") { document.forms["form"]["myusername"].style.border = "2px solid #963634";}
if (mypassword == null || mypassword == "") { document.forms["form"]["mypassword"].style.border = "2px solid #963634";}
$( ".home_column" ).effect( "shake" ); 
} else {
// Returns successful data submission message when the entered information is stored in database.
$.post("include/validate_login.php", {
username1: myusername,
password1: mypassword
}, function(data) {
if(data == 'login_wrong') {
$(".home_column").flip({
    direction:'lr',
        color: 'rgba(138, 138, 138, 0.2)', 
            content:'<h21>Incorrect Login Details</h21>'
})  
setTimeout(
  function() 
  {
   $(".home_column").revertFlip()
  }, 2500);  }else{
  if(data == 'login_success') {
  $(".home_column").flip({
    direction:'lr',
        color: 'rgba(138, 138, 138, 0.2)', 
            content:'<h21>Hello <? echo $_SESSION['username']; ?></h21><br/><br/><h21>Please Wait...</h21>'
})  
setTimeout(
  function() 
  {
   window.location = 'dashboard.php'; 
  }, 2500);

  } }
$('#form')[0].reset(); // To reset form fields
});
}
});
});

</script>

validate_login.php，在此页面中我运行MySQL查询以检查用户名和密码是否匹配。成功查询后，我回复了“login_success”＃39;或者一旦失败，我就会回复＃login_wrong＆＃39;。

validate_login.php：

session_start();
$query = "SELECT * FROM $tbl_name WHERE user_name = '$username' UNION
SELECT * FROM $tbl_name2 where user_name = '$username'"; 
$result = mysql_query($query) or die( mysql_error() );
$row = mysql_fetch_assoc($result);

$hashed_pass = crypt($password, $Blowfish_Pre . $row['salt'] . $Blowfish_End);

if(mysql_num_rows($result) > 0) {   // Check Username Exists

if($hashed_pass === $row['user_password_hash']) {   // Check Password is valid
$_SESSION['username']=$row['user_first_name'];      //Start User Session  
$_SESSION['id']=$row['user_id'];      
$_SESSION['user']=$username;
$_SESSION['username2']=$row['user_last_name'];
$_SESSION['allowance']=$row['allowance'];
$_SESSION['privelleges']=$row['privelleges'];
$sql2 = mysql_query("UPDATE $tbl_name2 SET user_failed_logins = '0', user_status = 'online' WHERE user_name = '".$username."'");
$result2=mysql_query($sql2); 
$sql3 = mysql_query("UPDATE $tbl_name SET user_failed_logins = '0', user_status = 'online' WHERE user_name = '".$username."'");
$result3=mysql_query($sql2); //Start User Session

echo "login_success";
}else{
echo "login_wrong";

然后我使用我的ajax代码收集这些回显的值作为数据，我在我的ajax中执行检查以查看数据是否等于＆＃39; login_success＆＃39;或者login_wrong＆＃39;

if(data == 'login_success') {

如果登录成功，我会执行一个效果，使我的div翻转并显示欢迎信息：

if(data == 'login_success') {
$(".home_column").flip({
        direction:'lr',
            color: 'rgba(138, 138, 138, 0.2)', 
                content:'<h21>Hello <? echo $_SESSION['username']; ?></h21><br/><br/><h21>Please Wait...</h21>'
    })

我放<?php echo $_SESSION['username']; ?>这是假设显示我们之前在validate_login.php中设置此变量的用户名。但是，在会话有时间设置/启动之前，ajax似乎正在运行。这意味着会话为空，而不是Hello Michael，我得到Hello。

所以我试图在我的validate_login.php中回复<?php echo $_SESSION['username']; ?>和我的echo "login_success";而不是像我这样的ajax：

echo "login_success";
echo $_SESSION['username']; 
    }else{
    echo "login_wrong";

理论上，数据（即用户的名字）应该与其他数据一起收集＆＃39; login_success＆＃39;。

但是，我可以在ajax中收集多个数据吗？因为如果我这样做，我就不能再执行if语句来查看数据=＆＃39; login_success＆＃39;它只是阻止整个脚本完全工作，提前谢谢

Answer 1

您可以通过以下方式将数据作为JSON发回：

echo json_encode( array(
        "item1" => $value1,
        "item2" => $value2
    ));

然后在JS：

$.post("include/validate_login.php", {
    username1: myusername,
    password1: mypassword
}, function(json) {

    console.log(json.item1);
    console.log(json.item2);

}, 'json'); // Don't forget this

从PHP页面收集多个数据？

1 个答案: