成功提交后让对话框开始工作

发布于 2025-01-01 00:44:54 字数 880 浏览 0 评论 0原文

我试图让我的表单在成功提交表单后显示一个对话框（可能通过 Javascript）。我该怎么做？

<script type="text/javascript">
$.validator.setDefaults({
});

$().ready(function() {
    // validate alphaRegister form on keyup and submit
    $("#alphaRegister").validate({
        rules: {
            email: {
                required: true,
                email: true
            }
        },
        messages: {
            email: "*"
        }
    });
});
</script>

HTML

    <form id="alphaRegister" action="src/php/newSubscriber.php" method="post">
        <input type="email" name="email" id="cemail" value="" class="required" />
        <div style="float:right; margin:0 5px 2px 0;"><input type="submit" id="submit" name="submit" value="" /></div>
    </form>

原文

I am trying to make my form display a dialog box (possibly via Javascript) upon successful submission of my form. How do I do this?

<script type="text/javascript">
$.validator.setDefaults({
});

$().ready(function() {
    // validate alphaRegister form on keyup and submit
    $("#alphaRegister").validate({
        rules: {
            email: {
                required: true,
                email: true
            }
        },
        messages: {
            email: "*"
        }
    });
});
</script>

HTML

    <form id="alphaRegister" action="src/php/newSubscriber.php" method="post">
        <input type="email" name="email" id="cemail" value="" class="required" />
        <div style="float:right; margin:0 5px 2px 0;"><input type="submit" id="submit" name="submit" value="" /></div>
    </form>

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心欲静而疯不止 2025-01-08 00:44:54

提交/处理表单后，使用服务器端脚本将 JS 变量添加到页面，然后在 $().ready 函数中检查该变量是否已设置。如果是，则显示您的对话框。

更新：

为此，您需要将 PHP 变量 formSubmitResult 设置为 true/false 或 1/0（如果它通过/未通过 PHP 验证例程）。

然后将其放在您提交的页面的头部：

<script type="text/javascript">
var submitIsValid=<?php echo formSubmitResult;?>;
$().ready(function() {
 if(submitIsValid)
 {
  alert("Submission was valid!"); // or some code to open your custom dialog box
 }
});
</script>

Use your server side script to add a JS variable to your page once the form has been submited/processed and then in your $().ready function, check to see if this variable is set. If it is then display your dialog box.

Update:

To do this you will need to set a PHP variable formSubmitResult to true/false or 1/0 if it passes/fails your PHP validation routine.

Then put this in the head of the page that you submit to:

<script type="text/javascript">
var submitIsValid=<?php echo formSubmitResult;?>;
$().ready(function() {
 if(submitIsValid)
 {
  alert("Submission was valid!"); // or some code to open your custom dialog box
 }
});
</script>

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