我的代码运行良好，但在成功调用时失败。在 ajax 部分“if (result.success){”上，它在浏览器源选项卡中显示错误，结果为空。
登录页面：
<form id="login_form_header">
<input type="hidden" name="action" value="sc_ajax_callback" />
<input type="hidden" name="func" value="sc_login" />
Email:
<input type="text" name="username" />
Password:
<input type="password" name="password"/>
<input type="submit" value="Log in" />
</form>
<script>
$("#login_form_header").submit(function(event){
event.preventDefault();
$.ajax({
type: 'post',
url: '<?php echo get_stylesheet_directory_uri(); ?>/connect.php',
data: $('#login_form_header').serialize(),
dataType: 'json',
success: function(result){
if (result.success){
window.location = "my-dashboard/";
return false;
};
},
error: function(e){console.log("Could not retrieve login information")}
});
return false;
});
</script>
连接页面：
<?php
mysql_connect("localhost", "%user%", "%pass%") or die(mysql_error()); // Connect to database server(localhost) with username and password.
mysql_select_db("%db%") or die(mysql_error()); // Select registration database.
# Make sure form data was passed to the script
if(isset($_POST['username']) && !empty($_POST['username']) AND isset($_POST['password']) && !empty($_POST['password'])){
# Define Variables
$active_var = "1";
$given_username = mysql_escape_string($_POST['username']);
$given_password = dosomethingtopass;
$matched_username = "";
$matched_password = "";
# See if there is matching info in the database
$sql = 'SELECT username, password, active FROM %table% WHERE username="'.$given_username.'" AND password="'.$given_password.'" AND active="'.$active_var.'"';
$result = mysql_query($sql);
while($row = mysql_fetch_assoc($result)){
if($given_password == $row['password']){
$matched_username = $row['username'];
$matched_password = $row['password'];
}
};
# If there was a match
if($matched_username != "" && $matched_password != ""){
#If there is only one result returned
echo json_encode(array("$matched_password"));
$session_sql = 'SELECT id, username, password, last_login FROM %table% WHERE username="'.$matched_username.'" AND password="'.$matched_password.'"';
$session_result = mysql_query($session_sql);
while($returned_row = mysql_fetch_assoc($session_result)){
if($matched_password == $returned_row['password']){
$_SESSION['id'] = $returned_row['id'];
//$_SESSION['last_login'] = $returned_row['last_login'];
$_SESSION['username'] = $returned_row['username'];
}
}
$date = date('Y-m-d H:i:s');
$update_sql = "UPDATE %table% SET last_login='".$date."'";
mysql_query($update_sql);
echo json_encode(array("success"=>"user logged in", "session"=>$_SESSION));
}
}
?>
此外，update_sql 正在更新我表中的每一行，这有点奇怪。不知道我哪里出错了有什么帮助吗？
编辑：
更新
$.ajax({
type: 'post',
url: '<?php echo get_stylesheet_directory_uri(); ?>/connect.php',
data: $('#login_form_header').serialize(),
dataType: 'json',
success: function(result){
window.location = "my-dashboard/";
};
},
error: function(e){console.log("Could not retrieve login information")}
});