Possible Duplicate:
07000
我有Apache2安装和Python工作.
我遇到了问题.我有两页.
一个是Python页面,另一个是带有JQuery的Html页面
我可以将Src改为google jquery链接.
有人可以告诉我如何让我的ajax帖子正常工作.
- $(function()
- {
- alert('Im going to start processing');
- $.ajax({
- url: "saveList.py",type: "post",data: {'param':{"hello":"world"}},dataType: "application/json",success : function(response)
- {
- alert(response);
- }
- });
- });
和Python代码
- import sys
- import json
- def index(req):
- result = {'success':'true','message':'The Command Completed Successfully'};
- data = sys.stdin.read();
- myjson = json.loads(data);
- return str(myjson);
解决方法
这是一个示例html文件和附带的python CGI脚本,可以帮助您:
使用这个HTML:
- <html>
- <head>
- <Meta http-equiv="content-type" content="text/html; charset=utf-8">
- <title>test</title>
- <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
- <script>
- $(function()
- {
- $('#clickme').click(function(){
- alert('Im going to start processing');
- $.ajax({
- url: "/scripts/ajaxpost.py",datatype:"json",data: {'key':'value','key2':'value2'},success: function(response){
- alert(response.message);
- alert(response.keys);
- }
- });
- });
- });
- </script>
- </head>
- <body>
- <button id="clickme"> click me </button>
- </body>
- </html>
这个脚本:
- #!/usr/bin/env python
- import sys
- import json
- import cgi
- fs = cgi.FieldStorage()
- sys.stdout.write("Content-Type: application/json")
- sys.stdout.write("\n")
- sys.stdout.write("\n")
- result = {}
- result['success'] = True
- result['message'] = "The command Completed Successfully"
- result['keys'] = ",".join(fs.keys())
- d = {}
- for k in fs.keys():
- d[k] = fs.getvalue(k)
- result['data'] = d
- sys.stdout.write(json.dumps(result,indent=1))
- sys.stdout.write("\n")
- sys.stdout.close()
单击按钮后,您可以看到cgi脚本返回:
- {
- "keys": "key2,key","message": "The command Completed Successfully","data": {
- "key2": "value2","key": "value"
- },"success": true
- }