2017-09-19 20 views
-1

这里是我的代码:HTML表单 - 摆脱问号和方程的

{% if request.path == '/employees' %} 
    <form action="{{ url_for('employees_name', employee_id=x) }}" /> 
    EmployeeId: <input type="text" name=x><br> 
    <input type="submit" value="Submit"> 
    </form> 
{% endif %} 

例如当我输入 “1” 作为输出我得到:

http://127.0.0.1:5002/employees/?x=1 

但我需要这样的:

http://127.0.0.1:5002/employees/1 

这里是我的Python代码:

app = Flask(__name__) 
api = Api(app) 

class Employees_Name(Resource): 
    def get(self, employee_id): 
     conn = db_connect.connect() 
     query = conn.execute("select * from employees where EmployeeId =%d " %int(employee_id)) 
     result = {'data': [dict(zip(tuple (query.keys()) ,i)) for i in query.cursor]} 
     return Response(render_template('test.html', result=result, mimetype='text/html')) 

api.add_resource(Employees, '/employees') 

有没有办法做到这一点?谢谢

+0

我试图做类似的搜索引擎.. – pingwin850

+0

你在做什么?请解释你的代码 – Nabin

+0

html代码是用于在浏览器中显示来自数据库的数据.. Python代码从数据库中获取数据..我需要这种模式'http://127.0.0.1:5002/employees/1'来显示它在浏览器 – pingwin850

回答

1

使用'GET'处理表单时,表单数据将被格式化为附加到URL后面的查询字符串。正常的方法是使用端点来处理表单并验证表单数据,然后重定向到像您的那样的端点GET

因此,代码可能是这样的:

from flask import request, url_for, redirect 
... 

class Employees(Resource): 
    def post(self): 
     employee_id = request.form.get('employee_id') 
     # validate... 
     return redirect(url_for('employee', employee_id=employhee_id)) 

class Employee(Resource): 
    def get(self, employee_id): 
     conn = db_connect.connect() 
     ... 

api.add_resource(Employees, '/employees') 
api.add_resource(Employee, '/employees/<int:employee_id>') # this rule is waht you want 

在模板:

<form action="{{ url_for('employees') }}"> 
    EmployeeId: <input type="text" name="employee_id"><br> 
    <input type="submit" value="Submit"> 
</form>