这是我的HTML代码:
<form name="frmMain" action ="<?php echo(htmlentities($_SERVER['PHP_SELF'])); ?>"
method="post" onsubmit="return validateform(this);" enctype="multipart/form-data">
...
...
...
<button type="submit" name="btnSubmit" class="btn btn-default"
style="position: relative; float:right; margin-right: 50px;">Submit Evaluation
<i class="glyphicon glyphicon-chevron-right"></i></button>
</form>
单击提交评估按钮时,javascript中的此功能会触发:
function validateform(frm)
{
var html=""; // to store the html content
// Build the content of the create vendor popup form
html+="<div class='col-md-12 divPreview' id='divPreview'>\n\
<div class='col-md-12 divBackground' id='divBackground'>\n\
<button type='submit' name='btnSubmit' class='btn btn-default' onclick='return SubmitForm(frm);'>Submit</button>\n\
</div></div>";
$("#divPreview").html(html); // display processing popup form
return(false);
}
它实际上是一个预览div,其中包含一个提交以下方法的按钮:
function SubmitForm(frm)
{
.. do some processing here
frm.submit(); // submit the form
return(false); // to prevent the calling form from executing
}
如何将frm form参数从validateform()传递给submitform()?
我这样做了,现在它正在运作:
function validateform(frm)
{
var html=""; // to store the html content
// Build the content of the create vendor popup form
html+="<div class='col-md-12 divPreview' id='divPreview'>\n\
<div class='col-md-12 divBackground' id='divBackground'>\n\
<button type='submit' name='btnSubmit' class='btn btn-default'>Submit</button></div></div>";
$("#divPreview").html(html); // display processing popup form
$("#btnSubmit").on("click", function(){ SubmitForm(frm); });
return(false);
}