Jquery post,在新窗口中响应

使用弹出窗口文档的write()-方法将标记放在那里:

$.post(url, function (data) {
    var w = window.open('about:blank');
    w.document.open();
    w.document.write(data);
    w.document.close();
});

Accepted answer不适用于“use strict”,因为“with”语句会抛出错误。因此:

$.post(url, function (data) {
    var w = window.open('about:blank', 'windowname');
    w.document.write(data);
    w.document.close();
});

另外,请确保“windowname”中没有任何空格,因为这将在IE中失败:)

如果您不需要有关请求数据的反馈,也不需要在打开程序和弹出窗口之间进行任何交互,则可以在弹出窗口中发布隐藏表单:

例子:

<form method="post" target="popup" id="formID" style="display:none" action="https://example.com/barcode/generate" >
  <input type="hidden" name="packing_slip" value="35592" />
  <input type="hidden" name="reference" value="0018439" />
  <input type="hidden" name="total_boxes" value="1" />
</form>
<script type="text/javascript">
window.open('about:blank','popup','width=300,height=200')
document.getElementById('formID').submit();
</script>

我用一个ajax帖子完成了它,然后用一个数据url返回:

$(document).ready(function () {
    var exportClick = function () {
        $.ajax({
           url: "/api/test.php",
           type: "POST",
           dataType: "text",
           data: {
              action: "getCSV",
              filter: "name = 'smith'",
           },
           success: function(data) {
              var w = window.open('data:text/csv;charset=utf-8,' + encodeURIComponent(data));
              w.focus();
           },
           error: function () {
              alert('Problem getting data');
           },
        });
    }
});