Javascript 关于提交问题的PHP Ajax Html

因此，我有下面的代码，所有的工作，但成功/错误，我可以找到它

这是settings.php

<li><button type="button" onclick="ApimanagementFunctioncall()" id="button1" class="btn btn-default btn-lg"><a href="#">Api management</button></a></li>
<script>
    function ApimanagementFunctioncall(){
    var xhttp = new XMLHttpRequest();
        xhttp.onreadystatechange = function() {
    if (xhttp.readyState == 4 && xhttp.status == 200) {
    document.getElementById("display").innerHTML = xhttp.responseText;
    }
    }
    xhttp.open("GET", "Apimanagement.php", true);
    xhttp.send();   
    }

    </script>

函数ApimanagementFunctioncall（）{
var xhttp=newXMLHttpRequest（）；
xhttp.onreadystatechange=函数（）{
如果（xhttp.readyState==4&&xhttp.status==200）{
document.getElementById（“display”）.innerHTML=xhttp.responseText；
}
}
open（“GET”，“Apimanagement.php”，true）；
xhttp.send（）；
}

apimagement.php

这里是MySQLI表

<div class="container">
    <div class="row text-left">
     <button type="button" class="btn btn-default btn-lg"  data-toggle="modal" data-target="#AddApi"><span class="glyphicon glyphicon-plus"></span>Add Api</button>
    </div>
</div>

<div class="modal fade" id="AddApi" role="dialog">
<?php
include ("ApiAdd.php");
?>
</div>

添加Api

这是Addapi.php

<?php
    $errorMessage = "";
    $UserId = "11";

     if(isset($_POST['submit'])){
       $VarKeyID = $_POST["KeyID"];
       $VarVCode = $_POST["VerificationCode"];
       $errorMessage1 = ""; 
       $errorMessage2 = "";
       $VkeyAdd= ""; $VCodeAdd = "";
        if(empty($VarKeyID) == false){
            if(is_numeric($VarKeyID) == true){
                if(strlen($VarKeyID) !== 8){
                    echo '<script type="text/javascript">alert("Key Id is not the correct length of 8 numbers!\n");</script>';}
                    else{ $VKeyAdd = "true";}
            }else{$errorMessage1 = '<script type="text/javascript">alert("Key Id should only be numbers!");</script>';}
        }else{$errorMessage1 ='<script type="text/javascript">alert("Please Enter a Key ID!");</script>';}

        if(empty($VarVCode) == false){
            if(ctype_alnum($VarVCode) == true){
                if(strlen($VarVCode) !== 64){
                    $errorMessage2 = "Verification Code should be 64 characters!";
                    }else{$VCodeAdd = "true";}              
            }else{$errorMessage2 = '<script type="text/javascript">alert("Verification Code should only be letters and numbers!");</script>';}
        }else{$errorMessage2 ='<script type="text/javascript">alert("Please Enter a Verification Code!");</script>';}

        If($VKeyAdd == "true"){
            if($VCodeAdd == "true"){
            echo "Next";}else{echo $errorMessage2;}
        }else{echo $errorMessage1;}
     }
?>

<div  id="AddApi">
    <div class="modal-dialog">
        <div class="modal-content">
            <div class="modal-header">
            <a href="Settings.php" title="Close" class="close"><span class="glyphicon glyphicon-remove"></span></a>
            <h2> Add Api Key </h2> 
            <a href="https://community.eveonline.com/support/api-key/CreatePredefined?accessMask=1073741823" target="_blank" class="btn btn-link" role="button">Api create key here</a>
            <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="POST">
            Key ID</br> <input type="text" name="KeyID" maxlength="8" value=""></br>
            Verification Code</br> <input type="text" name="VerificationCode" maxlength="64" min="64" value=""></br>
            <input type="submit" name="submit" value="Submit" formtarget="_Self">
            </form>
        </div>
    </div>
<div>

---

永远不要调用表单中的任何内容
name=“submit”
-这可能不是您的问题，但是如果您想使用脚本提交，请更改名称。还要在php中设置错误消息，并测试它是否为空，如果不是echo。使查看错误isI不希望脚本的位置变得更加容易，同时显示错误消息，并且在代码运行时不会出现任何错误。只是我不想这样。我真的不确定你说的是错的。你期待什么样的行为？这仅仅是因为当一切都正确时，它不会运行
Settings.php
？您应该可以使用
window.location.href=“Settings.php”来实现这一点
或
标题（“位置：Settings.php”）。您的代码太复杂了。写一张什么叫什么的图表。现在，当您使用jQuery$.get和$.post处理与服务器的所有通信时，您正在将普通表单发布与普通AJAX混合在一起。下面是我希望实现的流程图