【问题标题】:Show a loading image after clicking the submit button when the form is valid表单有效时点击提交按钮后显示加载图片
【发布时间】:2026-02-02 10:55:01
【问题描述】:

我有一个非常基本的表单,如果表单有效,我想在单击提交按钮后显示加载图像,然后显示操作页面。这是验证的javascript代码:

<script type="text/javascript">
function validateForm()
{
    valid = true;

    if(document.grades.fullName.value == "")
    {
        alert ("Please fill out your Full Name.");
        valid = false;
    }

    if(document.grades.idNo.value == "")
    {
        alert ("Please fill out your ID Number.");
        valid = false;
    }

    if(document.grades.pCS.value == "")
    {
        alert ("Please fill out your grade for Prelim Class Standing.");
        valid = false;
    }

    if(document.grades.pExam.value == "")
    {
        alert ("Please fill out your grade for Prelim Examination.");
        valid = false;
    }

    if(document.grades.mCS.value == "")
    {
        alert ("Please fill out your grade for Midterm Class Standing.");
        valid = false;
    }

    if ( document.grades.mExam.value == "" )
    {
        alert ("Please fill out your grade for Midterm Examination.");
        valid = false;
    }

    if(document.grades.fCS.value == "")
    {
        alert ("Please fill out your grade for Finals Class Standing.");
        valid = false;
    }

    if(document.grades.fExam.value == "")
    {
        alert ("Please fill out your grade for Finals Examination.");
        valid = false;
    }

    return valid;
}
</script>

请有人告诉我如何完整。

【问题讨论】:

  • "请有人告诉我有多完整。"抱歉,这不是出租编码器
  • 也许这个逻辑可能对你有所帮助 - loading image once file submit clicked
  • 请告诉我们您尝试过的代码。

标签: php javascript loader


【解决方案1】:

请更换,

return valid;

由,

if(valid)
{
  document.getElementById('display').style.display="block";
  return true;
}

在 html 中,

<div id="display" style="display:none;"><img src="{image_url}" /></div>

【讨论】:

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