【GET请求转POST】JS Form提交实现页面跳转

 1   //path:请求地址;key、value:参数键值
 2   var postForm = document.createElement("form");
 3         postForm.method = "post";
 4         postForm.action = path;
 5         postForm.style = "display:none";
 6         
 7   var parameterInput = document.createElement("input");
 8         parameterInput.setAttribute("name", key); 
 9         parameterInput.setAttribute("value",value); 
10 
11   postForm.appendChild(parameterInput);    
12              
13   document.body.appendChild(postForm);
14 
15   postForm.submit();  

 

posted @ 2020-09-16 09:15  爪哇不洼  阅读(562)  评论(0编辑  收藏  举报