javascript javascript动态更改表单提交
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javascript change form onsubmit dynamically
提问by user1507558
I have a form with some action and onsubmit values, which is submitted through a submit input tag. The problem is that it should be submittable by two buttons, so I wrote a function for the second button to change the action and onsubmit values of the form:
我有一个带有一些 action 和 onsubmit 值的表单，它是通过提交输入标签提交的。问题是它应该可以通过两个按钮提交，所以我为第二个按钮编写了一个函数来更改表单的 action 和 onsubmit 值：
<a href="javascript:submitCompare()" class="submit">Compare</a>
function submitCompare()
{
document.myForm.action = "anotherAction.php";
document.myForm.onsubmit = function() {return countChecked()};
document.myForm.submit();
}
function countChecked()
{
var n = $(".reports input:checked").length;
if (n >= 3 ) {
alert ('You must select less than 3 reports.');
return false;
}
else return true;
}
After clicking on the Compare link it sends me to the anotherAction.php page correctly, but even when I have more than 2 selected checkboxes (which is the validation rule). Can somebody help me make the onsubmit function work correctly?
单击“比较”链接后，它会将我正确发送到 anotherAction.php 页面，但即使我选择了 2 个以上的复选框（这是验证规则）。有人可以帮助我使 onsubmit 功能正常工作吗？
采纳答案by Otto Allmendinger
In submitCompare(), you explicitly and unconditionally call
在 中submitCompare()，您明确且无条件地调用
document.myForm.submit();
What you probably want instead there is
你可能想要的是
if (countChecked()) {
document.myForm.submit();
}
回答by Trevor
document.myForm.onsubmit = function() {return countChecked()};
should be
应该
document.myForm.onsubmit = function( e ) {
e = e || window.event;
if ( !countChecked() ) {
e.preventDefault();
e.returnValue = false;
}
};
Returning false on a submit will just end any further function execution. You want to preventDefault submission behavior if you don't want it to submit.
在提交时返回 false 只会结束任何进一步的函数执行。如果您不希望它提交，您希望阻止默认提交行为。
回答by Vlad
It is a late reply, but if someone else is looking at this...
这是一个迟到的回复，但如果其他人正在看这个......
instead of:
代替：
document.myForm.onsubmit = function() {return countChecked()};
I think you wanted:
我想你想要：
document.myForm.setAttribute("onsubmit", "return countChecked()");
