<script> //@创建表单方法 function post(URL, PARAMS) { var temp = document.createElement("form"); temp.action = URL; temp.method = "post"; temp.style.display = "none"; for (var x in PARAMS) { var opt=document.createElement(&qu…
今天遇到这么一个需求,携带一个编号一个名字跳转到另一个JSP页面,直接页面跳转(get携带数据)的话不太安全,于是想到到后台转发一下. 第一种:直接以表单提交方式的进行 JS代码: var form = $("<form action='"+contextPath+"/trainacontentType_forwardToAddTraincontent.action"+"' method='post'></form>")…
document.body.appendChild(jForm) won't work because jForm is not a dom element, it is a jQuery object so add the below script before jForm.submit(); jForm.appendTo('body') function loadPage(url, projectName) { var jForm = $('<form></form>', {…
JS下载 function downloadFile(id) { var url = "<%=request.getContextPath()%>/cer/downloadFile"; var form = $("<form></form>").attr("action", url).attr("method", "post"); form.ap…