如何使用Javascript/jQuery发布到弹出窗口?
var eids = selected_Ids.join();
var url = 'http://www.zyx.com';
$.post (url, {
eId : eids
},
function (data) {
// newpage = result;
var new_window = window.open(url, "abc");
});弹出窗口不会出现。我做错什么了。
发布于 2014-01-23 09:54:31
function formPost(actionUrl, a) {
var sHTML = "<form id='form1' action='" + actionUrl + "' method='post'>";
for (var i = 0; i < a.length; i = i + 2) {
sHTML += "<input name='" + a[i] + "' type='hidden' value='" + a[i + 1] + "' />";
}
sHTML += "</form>";
var ooo = window.open("", "test");
ooo.document.body.innerHTML = sHTML;
ooo.form1.submit();
}只需调用此函数
formPost('http://www.zyx.com', ["key1", "111", "key2", "222"])它在IE和Chrome中运行得很好
享受吧~ :)
https://stackoverflow.com/questions/21296832
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