好吧,我现在有1个表单和1个文本输入字段:
echo '<form class="changePassForm" action="" method="post" enctype="multipart/form-data">';
echo '<input class = "passwordText" type="password" placeholder="Change Password" name="passwordText">';
echo '<input class = "oldPass" type="password" placeholder="Enter Old Password" name="oldPass">';
echo '</form>';
当用户在passwordText输入上按Enter键时,输入的文本将在变量中传递给php文件:
$(".passwordText").keydown(function(event){
if(event.keyCode == 13){
var pass = $(this).val();
//var pass = document.getElementById("p2");
$.ajax({
url: "../php/passwordchange.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: 'passwordText=' + pass, // data sent to php file
success: function(data) // A function to be called if request succeeds
{
console.log(data);
}});
console.log("WORKS!!");
}
});
当发生这种情况时,我需要.passwordText输入字段滑过并隐藏(无法与之交互),这样我就可以从第二个文本输入中获取值。我不知道如何做到这一点,因为我是一个javascript的初学者。
以下是我的CSS供参考:
.changePassForm input {
-webkit-appearance: none;
-moz-appearance: none;
appearance: none;
outline: 0;
border: 1px solid rgba(255, 255, 255, 0.4);
background-color: rgba(255, 255, 255, 0.2);
width: 250px;
border-radius: 3px;
padding: 10px 15px;
margin: 0 auto 10px auto;
display: block;
text-align: center;
font-size: 18px;
color: white;
-webkit-transition-duration: 0.25s;
transition-duration: 0.25s;
font-weight: 300;
}
.changePassForm input:hover {
background-color: rgba(255, 255, 255, 0.4);
}
.changePassForm input:focus {
background-color: white;
width: 300px;
color: #53e3a6;
}
.changePassForm button {
-webkit-appearance: none;
-moz-appearance: none;
appearance: none;
outline: 0;
background-color: white;
border: 0;
padding: 10px 15px;
color: #53e3a6;
border-radius: 3px;
width: 250px;
cursor: pointer;
font-size: 18px;
-webkit-transition-duration: 0.25s;
transition-duration: 0.25s;
}
.changePassForm button:hover {
background-color: #f5f7f9;
}
如何移动第一个输入以从第二个文本输入中获取值?
答案 0 :(得分:0)
您需要在AJAX成功中做到这一点。
KUBERNETES_PROVIDER=ubuntu ./kube-up.sh,
上面的代码是&#34;告诉&#34; $.ajax({
url: "../php/passwordchange.php",
type: "POST",
data: 'passwordText=' + pass, // data sent to php file
success: function(data){
console.log(data);
//HERE'S THE PLACE TO DO IT.
$('.passwordText').slideUp(500)
}
});
在500毫秒(或半秒)的时间范围内隐藏自己。
我还假设当你的php执行时,它会返回一个成功的响应。 如果脚本返回失败的响应,您可能还想执行其他操作。
您可能希望了解如何构建JSON响应的JSEND。