按钮 onclick 应连同参数一起重定向到另一个 xhtml 页面

Question

我有一个 table 从后端获取的值呈现为 json，

在其中一个单元格中，我有一个按钮，单击此按钮我想呈现另一个 xhtml 页面，但我还希望将参数与超链接一起传递如何实现，在此先感谢我正在发布我的代码以供参考，

 $.get('../Paid_deep_dive', {type: "paid_performance", campaign_id: $campaign, s_date: start_date, e_date: end_date}, function (response) {

        var data = response.paid_performance;
        response_data = response;

        $('table#tb2 TBODY').append('<tr><td width ="12%"><form><input type="button" class="btn btn-block btn-lg btn-inverse" onclick="loadtable2(' + elem.site_id + ')" value="' + elem.site_name + '" style="width:100%"  ></form></td>' + elem.site_name + '</center></td><td align="center" width="33%"><center>' + elem.impressions.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",") + '</center></td><td align="center" width="25%"><center>' + elem.ontarper + '%</center></td>;<td align="center"><center>' + elem.viewable_imp.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",") + '</center></td></tr>');
});

在上面的代码中，我想在

处有一个超链接

onclick = "http://localhost:8080/Digiview_v2/faces/xhtml_files/paid_drill_down.xhtml?variable=" + parameter; 

以及传递参数。有什么办法可以正确做到吗

Answer 1

您通过分配给 window.location.href:

在 JS 中重定向

$('table#tb2 TBODY').append('<tr><td width ="12%"><form><input type="button" class="btn btn-block btn-lg btn-inverse" onclick="window.location.href=\'http://localhost:8080/Digiview_v2/faces/xhtml_files/paid_drill_down.xhtml?variable=' + encodeURIComponent(parameter) + '\'" value="' + elem.site_name + '" style="width:100%"  ></form></td>' + elem.site_name + '</center></td><td align="center" width="33%"><center>' + elem.impressions.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",") + '</center></td><td align="center" width="25%"><center>' + elem.ontarper + '%</center></td>;<td align="center"><center>' + elem.viewable_imp.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",") + '</center></td></tr>');