当我克隆表单并想要提交克隆表单时,它会提交父表单。 我更新了这个问题。
var mainPart = wDoc.MainDocumentPart;
var res = from bm in mainPart.Document.Body.Descendants<BookmarkStart>()
where bm.Name == "vendos_tekstin"
select bm;
var bookmark = res.SingleOrDefault();
if (bookmark != null)
{
var parent = bookmark.Parent; // bookmark's parent element
// build paragraph piece by piece
Text text = new Text(DateTime.Now.ToString() + " , ");
Text text1 = new Text(gjenerimi + " , ");
Text text2 = new Text(merreshifren());
var run = new Run();
run.Append(text,text1,text2);
Paragraph newParagraph = new Paragraph(run);
run.PrependChild<RunProperties>(runProp);
// insert after bookmark parent
parent.InsertAfterSelf(newParagraph);
答案 0 :(得分:0)
您可以尝试下面的内容,
<div class="cl">
<form id="myform_1">
<input type="text" name="field1" />
<br/>
<input type="text" name="field2" />
<br/>
<input type="submit" class="submit"/>
</form>
</div>
我刚刚为您的提交按钮分配了一个课程
export const signIn = ( action$, state ) => {
return action$
.ofType(types.SIGN_IN_REQUEST)
.switchMap(( { params, } ) => (services.signInRequest( // it is ajax observable
mappers.mapSignInRequest(params)
))
.map(response => actions.signInJWTSuccess( // dispatch success
mappers.mapUser(response)
))
.catch(error => of(actions.signInError( // dispatch error
mappers.mapSignInError(error)
)))
};
答案 1 :(得分:0)
function onValidate(val){
$('#'+val).each(function(key, form) {
$(form).validate({ //intit plugin
rules: {
field1: {
required: true
},
field2: {
required: true
}
},
submitHandler: function (form) {
alert('valid form submitted');
return false;
}
});
});
}
$(document).ready(function () {
var i=1;
$('body').on('click','input[type="submit"]',function(e) {
e.preventDefault();
$(this).closest('form').submit();
});
$('#clone').click(function(e) {
e.preventDefault();
i++;
$('#myform_1').clone(false).prop('id','myform_'+i ).appendTo('.clone');
onValidate( 'myform_'+i);
});
onValidate("myform_"+i);
});