我有一个脚本将表单提交到弹出窗口但不显示表单的操作(process.php),它不显示任何内容(空白窗口)。继承人我的剧本:
function redirectOutput() {
var myForm = document.getElementById('formID');
var w = window.open('about:blank','Popup_Window','toolbar=0,scrollbars=0,location=0,statusb
ar=0,menubar=0,resizable=0,width=400,height=300,left = 312,top = 234');
myForm.target = 'Popup_Window';
return true;
}
答案 0 :(得分:28)
虽然有效,但您的window.open
突然出现换行符。
这对我来说很合适:http://jsfiddle.net/pimvdb/N3YSG/。
var myForm = document.getElementById('formID');
myForm.onsubmit = function() {
var w = window.open('about:blank','Popup_Window','toolbar=0,scrollbars=0,location=0,statusbar=0,menubar=0,resizable=0,width=400,height=300,left = 312,top = 234');
this.target = 'Popup_Window';
};
答案 1 :(得分:13)
在弹出窗口中提交表单的简便方法:
HTML:
<form action="..." method="post" onsubmit="target_popup(this)">
<!-- form fields etc here -->
</form>
使用Javascript:
function target_popup(form) {
window.open('', 'formpopup', 'width=400,height=400,resizeable,scrollbars');
form.target = 'formpopup';
}
来源: http://www.electrictoolbox.com/post-form-popup-window-javascript-jquery/
答案 2 :(得分:2)
为什么不在<form>
?
<form target="_blank">...</form>