I want to place the Login Form of the generated Login View from the Grails Spring Security Plugin at the main layout in layouts/main.gsp. I just copied it over, but it doesn't work, because of cause, now no controller is available to pass the form data to. The form-code looks like this:
<form action='${postUrl}' method='POST' id='loginForm' class='cssform' autocomplete='off'>
<p>
<label for='username'><g:message code="springSecurity.login.username.label"/>:</label>
<input type='text' class='text_' name='j_username' id='username'/>
</p>
<p>
<label for='password'><g:message code="springSecurity.login.password.label"/>:</label>
<input type='password' class='text_' name='j_password' id='password'/>
</p>
<p id="remember_me_holder">
<input type='checkbox' class='chk' name='${rememberMeParameter}' id='remember_me' <g:if test='${hasCookie}'>checked='checked'</g:if>/>
<label for='remember_me'><g:message code="springSecurity.login.remember.me.label"/></label>
</p>
<p><input type='submit' id="submit" value='${message(code: "springSecurity.login.button")}'/></p>
</form>
How should I change this code, so that I send the form code to the Login auth action?
Thanks.
As I udenrstand, your question is what to use for form action? postUrl should be same as grails.plugins.springsecurity.apf.filterProcessesUrl property in config, and by default it's /j_spring_security_check.
If you didn't change this value, you can use /j_spring_security_check instead of ${postUrl}.
Spring Securty Plugin have special filter that handles this URL, and authorizes user. Next steps are depends on configuration, but if I remember correctly, by default it redirects user to /.
See configuration options at http://grails-plugins.github.io/grails-spring-security-core/guide/urlProperties.html
Related
I am programming in Spring and using Thymeleaf as my view, and am trying to create a form where users can update their profile. I have a profile page which lists the user's information (first name, last name, address, etc), and there is a link which says "edit profile". When that link is clicked it takes them to a form where they can edit their profile. The form consists of text fields that they can input, just like your standard registration form.
Everything works fine, but my question is, when that link is clicked, how do I add the user's information to the input fields so that it is already present, and that they only modify what they want to change instead of having to re-enter all the fields.
This should behave just like a standard "edit profile" page.
Here is a segment of my edit_profile.html page:
First Name:
Here is the view controller method that returns edit_profile.html page:
#RequestMapping(value = "/edit", method = RequestMethod.GET)
public String getEditProfilePage(Model model) {
model.addAttribute("currentUser", currentUser);
System.out.println("current user firstname: " + currentUser.getFirstname());
model.addAttribute("user", new User());
return "edit_profile";
}
currentUser.getFirstname() prints out the expected value, but I'm getting blank input values in the form.
Thanks.
Solved the problem by removing th:field altogether and instead using th:value to store the default value, and html name and id for the model's field. So name and id is acting like th:field.
I'm slightly confused, you're adding currentUser and a new'd user object to the model map.
But, if currentUser is the target object, you'd just do:
<input type="text" name="firstname" value="James" th:value="${currentUser.firstname}" />
From the documentation:
http://www.thymeleaf.org/doc/tutorials/2.1/usingthymeleaf.html
I did not have a form with input elements but only a button that should call a specific Spring Controller method and submit an ID of an animal in a list (so I had a list of anmials already showing on my page). I struggled some time to figure out how to submit this id in the form. Here is my solution:
So I started having a form with just one input field (that I would change to a hidden field in the end). In this case of course the id would be empty after submitting the form.
<form action="#" th:action="#{/greeting}" th:object="${animal}" method="post">
<p>Id: <input type="text" th:field="*{id}" /></p>
<p><input type="submit" value="Submit" /> </p>
</form>
The following did not throw an error but neither did it submit the animalIAlreadyShownOnPage's ID.
<form action="#" th:action="#{/greeting}" th:object="${animal}" method="post">
<p>Id: <input type="text" th:value="${animalIAlreadyShownOnPage.id}" /></p>
<p><input type="submit" value="Submit" /> </p>
</form>
In another post user's recommended the "th:attr" attribute, but it didn't work either.
This finally worked - I simply added the name element ("id" is a String attribute in the Animal POJO).
<form action="#" th:action="#{/greeting}" th:object="${animal}" method="post">
<p>Id: <input type="text" th:value="${animalIAlreadyShownOnPage.id}" name="id" /></p>
<p><input type="submit" value="Submit" /> </p>
</form>
I want to end a param in URL in OFBiz, but I get this error
from security
org.apache.ofbiz.webapp.event.EventHandlerException: Found URL parameter [twCompagneDeRecrutementId]
passed to secure (https) request-map with uri [showCompagne] with an event that calls service [showCompagne];
this is not allowed for security reasons! The data should be encrypted by making it part of the request body (a form field) instead of the request URL.
Moreover it would be kind if you could create a Jira sub-task of https://issues.apache.org/jira/browse/OFBIZ-2330 (check before if a sub-task for this error does not exist). If you are not sure how to create a Jira issue please have a look before at http://cwiki.apache.org/confluence/x/JIB2
This is my ftl
<#if listCompagne??>
<#list listCompagne as newCompagne>
<div style="display:flex; width:100%;">
<div style="padding:6px; width:24%;"><label>${newCompagne.nom}</label></div>
<div style="padding:6px; width:24%;"><label>${newCompagne.dateDebut ?date}</label></div>
<div style="padding:6px; width:24%;"><label>${newCompagne.dateFin ?date}</label></div>
<input type="hidden" value="${newCompagne.twCompagneDeRecrutementId}"/>
<div style="padding:6px; width:20%;"> <a class="btn btn-outline-danger btn-block " href="<#ofbizUrl>showCompagne?twCompagneDeRecrutementId=${newCompagne.twCompagneDeRecrutementId}</#ofbizUrl>">afficher</a>
</div>
</div>
</#list>
I don't use ofbiz , but you should send parameter in body instead of in query.
In your case I would add a form and submit it using hidden form input, form should be unique per iteration using index
<form action="<#ofbizUrl>showCompagne?</#ofbizUrl>" id="afficher${newCompagne?index}" method="post" style="display: none;">
<input type="hidden" name="twCompagneDeRecrutementId" value="${newCompagne.twCompagneDeRecrutementId}" />
</form>
<a class="btn btn-outline-danger btn-block "
href="javascript:;" onclick="javascript:
document.getElementById('afficher${newCompagne?index}').submit()">afficher</a>
the solution is to go to url.properties and change parameters.url.encrypt to no
I'm building the website based on Joomla 3. I need to disable the validation of the fields: name, username, password1 (the second, because the first is password2) and email2 and use email1 as username (I installed the plugin Email for user authorization).
I have tried to remove these fields in file components/com_users/models/forms/registration.xml but the validation is still remaining. If I don't remove them but only change the rows required="true" to false for these fields the registration doesn't work at all and any user stored in the DB. How can I disable these fields?
It's not an easy workaround, and you will need some basic knowledge of Joomla and PHP, but I'll try to explain it to you as simple as i can.
>>> Creating view template override
First of all you will need to create your Registration view template override (to keep it Joomla update proof). To do so, create folder /templates/YOUT_TEMPLATE/html/com_users/registration and copy /components/com_users/views/registration/tmpl/default.php file there.
From now on you can modify registration output in your template folder.
>>> Modifying registration form output
By default Joomla takes all fields from form file /components/com_users/models/forms/registration.xml, where they are defined, and outputs them in a view. But if we don't want to use ALL the fields, we need to output fields manually.
My example code only output E-mail and Password fields for registration. Here's a sample code to do so: (default.php file)
<?php
defined('_JEXEC') or die;
JHtml::_('behavior.keepalive');
?>
<div class="grid_8" id="register_block">
<div class="content_block">
<h1>Registracija</h1>
<div class="login<?php echo $this->pageclass_sfx?>">
<form id="member-registration" action="<?php echo JRoute::_('index.php?option=com_users&task=registration2.register'); ?>" method="post" enctype="multipart/form-data">
<div>
<div class="login-fields">
<label id="jform_email1-lbl" for="jform_email1">E-mail:</label>
<input type="text" name="jform[email1]" id="jform_email1" value="" size="30">
</div>
<div class="login-fields">
<label id="jform_password1-lbl" for="jform_password1">Password:</label>
<input type="password" name="jform[password1]" id="jform_password1" value="" autocomplete="off" size="30">
</div>
<button type="submit" class="button"><?php echo JText::_('JREGISTER');?></button>
<input type="hidden" name="option" value="com_users" />
<input type="hidden" name="task" value="registration2.register" />
<?php echo JHtml::_('form.token');?>
</div>
</form>
</div>
</div>
</div>
Please note, that I've also replaced task value from registration.register to registration2.register, I did this to bypass some of validation rules using my own controller.
>>> Creating controller override
Locate file /components/com_users/controllers/registration.php and create a copy of it called registration2.php in same folder.
Open file registration2.php and change It's class name from UsersControllerRegistration to UsersControllerRegistration2
From now on Joomla registration form will use this class to create a new user.
Find a method called register and find this line:
$requestData = JRequest::getVar('jform', array(), 'post', 'array');
This is where Joomla get's registration form data. Add the following lines:
$requestData['name'] = $requestData['email1'];
$requestData['username'] = $requestData['email1'];
$requestData['email2'] = $requestData['email1'];
$requestData['password2'] = $requestData['password1'];
It will add missing registration info, and help you pass validation.
NOTE: This is a sample code, to show the main logic. If anyone has a better solution, I'd be more than happy to hear it.
Following same idea, a simpler solution might be just including hidden inputs with values in com_users\views\registration\tmpl\default.php above
<button type="submit" class="btn btn-primary validate"><?php echo JText::_('JREGISTER');?></button>
add
<input type="hidden" id="jform[username]" name="jform[username]" value="username" />
<input type="hidden" id="jform_name" name="jform[name]" value="name" />
It would pass the validation and you would no longer need to override controllers etc.
I need some help with the following issue. I am pretty new to JavaScript so this one is pretty tough for me. Any help would be greatly appreciated!
We have a form for registration that is already in use and I need to make some amends to it to enhance its usability. The form itself already contains several validation scripts and I need to add in a new one - an error message to appear next to the mobile field. I am unable to amend the form HTML itself so I have to do everything by javascript – I need to bind the new function to the submit button so that it fires AFTER all the other validations (field highlight, pop ups), but before the page turns over.
Here is the form (I have just left in the feild i need to amend) :
<form method="post" action="#" id="register_form" class="validate">
<input type="hidden" name="page" value="" />
<input type="hidden" name="unregistered" value="false" />
<fieldset>
<legend>I am a new online customer</legend>
<div class="element">
<label for="mobile_number">Mobile:**</label>
<input name="mobileNumber" type="text" id="mobile_number" title="Please enter at least one number" class="text validate {validate:{minLength:10, eitherOr:'#telephone_number', messages:{minLength:'Please enter a valid phone number (at least #NUM digits long)', eitherOr:'Please enter either your Telephone or Mobile number'}}}" />
</div>
<div class="element continue_shopping_cont submit">
<button type="submit" class="submit button small_button"><span>Register</span></button>
</div>
<input type="hidden" name="secure_from" value="" />
</fieldset>
</form>
Could someone help show me how i am supposed to do this and how is best to target this field?
I am trying to invoke a form submit using javascript (jquery) to invoke a webflow transition. It works and the submit invokes the desired transition. But, the updated radio button values is not reflected on the model object which is posted.
Here is the code:
<form:form method="post" action="#" commandName="infoModel" name="pageForm">
<form:input type="input" path="testMsg" id="success" />
<input type="button" id="clearSelections" value="Clear Selections">
<div class="question">
<h4><c:out value="${infoModel.questionInfo.description}"/> </h4>
<form:radiobuttons path="infoModel.answerId"
itemValue="answerId" itemLabel="answerDescription" items="${infoModel.answers}" delimiter="<br/>" />
</div>
<input type="submit" name="_eventId_saveQualitativeInput" value="Save" id="save" />
$(document).ready(function() {
$('#tabs').tabs();
//Clear selections (copy is server-side)
$('#clearSelections').click(function() {
//$('input[type="radio"]').prop('checked', false);
$('input[type="radio"]').removeAttr('checked');
$('#save').trigger('click');
});
});
</form:form>
The form:radiobutton, generates the below html:
<div class="question">
<h4>Is this a general obligation of the entity representing a full faith and credit pledge? </h4>
<span>
<input type="radio" checked="checked" value="273" name="infoModel.answerId" id="infoModel.answerId1">
<label for="infoModel.answerId1">Yes</label>
</span>
<span><br>
<input type="radio" value="274" name="infoModel.answerId" id="infoModel.answerId2">
<label for="infoModel.answerId2">No</label>
</span>
<br>
<span class="error"></span>
</div>
The input id= "success" value is registered and when the control goes to the server, the value of input id= "success" is updated in the "infoModel" object. But the value of answerId is not updated on the "infoModel" object.
Thoughts if i am missing something in the form:radiobutton element or if there is something else wrong?
Thanks in advance!
EDIT:::::::
Thanks mico! that makes sense. I stripped of some of the code first time to make it precise, but i have a list which is being used for building the radio-buttons, below is the code:
<c:forEach items="${infoModel.list["index"]}" var="qa" varStatus="rowCount">
<div class="question">
<h4><c:out value="${question.questionInfo.description}"/> </h4>
<form:radiobuttons path="list["index"][${rowCount.index}].answerId" itemValue="answerId" itemLabel="answerDescription" items="${question.answers}" delimiter="<br/>" />
<br>
</div>
</c:forEach>
Could you please suggest how i could try this one out?
NOTE: The same code works on a regular form submit on click of a button of type submit. Its the javascript form submit which is not working. I also tried to do whatever i want to do in javascript and then invoke the button.trigger('click'); form got submitted but the changes made on form in my javascript didnt reflect.
With commandName inside a form:form tag you set "Name of the model attribute under which the form object is exposed" (see Spring Documentation). Then in path you should tell the continuation of the path inside the model attribute.
With this said I would only drop the extra word infoModel from path="infoModel.answerId" and have it rewritten as path="answerId" there under the form:radiobutton.