authenticate user using form in the dialog window












0














I have implemented using OAuth.
Now I have a popup window with a form that supposed to allow a user to login by providing his credentials. The form is expected to submit its data to /login url, where I should validate user parameters and authenticate him.



The thing is that I supposed to see a default page with limited functionality by default.



Originally I thought I should use formLogin of Spring security. In my configure method of WebSecurityConfigurerAdapter class I defined configuration



    http
.antMatcher("/**").authorizeRequests()
.antMatchers(
"/",
"/csrf-token",
"/oauth**",
"/error**",
"/swagger-resources/**",
"/swagger-ui.html",
"/v2/api-docs",
"/webjars/**",
"/login")
.permitAll()
.anyRequest().authenticated()
.and().formLogin().permitAll().loginProcessingUrl("/login").defaultSuccessUrl("/")
...;


I am able to see a form on /login page when I directly access it in the browser.



But how to implement the case when the form is not on a separate page and just a part of other functionally? Should I just implement a separate POST handler for some url like /login?










share|improve this question





























    0














    I have implemented using OAuth.
    Now I have a popup window with a form that supposed to allow a user to login by providing his credentials. The form is expected to submit its data to /login url, where I should validate user parameters and authenticate him.



    The thing is that I supposed to see a default page with limited functionality by default.



    Originally I thought I should use formLogin of Spring security. In my configure method of WebSecurityConfigurerAdapter class I defined configuration



        http
    .antMatcher("/**").authorizeRequests()
    .antMatchers(
    "/",
    "/csrf-token",
    "/oauth**",
    "/error**",
    "/swagger-resources/**",
    "/swagger-ui.html",
    "/v2/api-docs",
    "/webjars/**",
    "/login")
    .permitAll()
    .anyRequest().authenticated()
    .and().formLogin().permitAll().loginProcessingUrl("/login").defaultSuccessUrl("/")
    ...;


    I am able to see a form on /login page when I directly access it in the browser.



    But how to implement the case when the form is not on a separate page and just a part of other functionally? Should I just implement a separate POST handler for some url like /login?










    share|improve this question



























      0












      0








      0


      1





      I have implemented using OAuth.
      Now I have a popup window with a form that supposed to allow a user to login by providing his credentials. The form is expected to submit its data to /login url, where I should validate user parameters and authenticate him.



      The thing is that I supposed to see a default page with limited functionality by default.



      Originally I thought I should use formLogin of Spring security. In my configure method of WebSecurityConfigurerAdapter class I defined configuration



          http
      .antMatcher("/**").authorizeRequests()
      .antMatchers(
      "/",
      "/csrf-token",
      "/oauth**",
      "/error**",
      "/swagger-resources/**",
      "/swagger-ui.html",
      "/v2/api-docs",
      "/webjars/**",
      "/login")
      .permitAll()
      .anyRequest().authenticated()
      .and().formLogin().permitAll().loginProcessingUrl("/login").defaultSuccessUrl("/")
      ...;


      I am able to see a form on /login page when I directly access it in the browser.



      But how to implement the case when the form is not on a separate page and just a part of other functionally? Should I just implement a separate POST handler for some url like /login?










      share|improve this question















      I have implemented using OAuth.
      Now I have a popup window with a form that supposed to allow a user to login by providing his credentials. The form is expected to submit its data to /login url, where I should validate user parameters and authenticate him.



      The thing is that I supposed to see a default page with limited functionality by default.



      Originally I thought I should use formLogin of Spring security. In my configure method of WebSecurityConfigurerAdapter class I defined configuration



          http
      .antMatcher("/**").authorizeRequests()
      .antMatchers(
      "/",
      "/csrf-token",
      "/oauth**",
      "/error**",
      "/swagger-resources/**",
      "/swagger-ui.html",
      "/v2/api-docs",
      "/webjars/**",
      "/login")
      .permitAll()
      .anyRequest().authenticated()
      .and().formLogin().permitAll().loginProcessingUrl("/login").defaultSuccessUrl("/")
      ...;


      I am able to see a form on /login page when I directly access it in the browser.



      But how to implement the case when the form is not on a separate page and just a part of other functionally? Should I just implement a separate POST handler for some url like /login?







      java spring-boot spring-security






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 12 at 19:04

























      asked Nov 12 at 18:54









      lapots

      3,6251260127




      3,6251260127





























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