I have a user form where people can update their details and it saves to the database.
I have added the function for people to upload an avatar, crop the image and upload it to the server (using Cropper).
When the user has finished cropping their image the script updates the HTML and replaces the default avatar with their new one, as below:
<div class="avatar-view" title="" data-original-title="Change the avatar">
<img src="../../scripts/cropper/img/20150728143117.png" alt="Avatar">
</div>
Underneath the avatar is the user form with the rest of the details and a save/submit button. I have added a hidden field for the avatar, but I need to send the value of the newly uploaded image to this hidden field. I have looked for various pieces of javascript/jquery to do this but I can't seem to get it to work so far.
<form id="reg-form" name="r_form" method="post" action="/editor/go/user" enctype="multipart/form-data">
<div class="form-group">
<label for="avatar">avatar</label>
<input type="hidden" name="avatar" value="123">
</div>
<div class="form-group">
<label for="first_name"><i class="fa fa-star icon-red"></i> First name</label>
<input id="first_name" name="first_name" class="form-control" value="John">
</div>
<div class="form-group">
<label for="last_name"><i class="fa fa-star icon-red"></i> Last name</label>
<input id="last_name" name="last_name" class="form-control" value="Smith">
</div>
<button type="submit" class="btn btn-primary" name="form_submit" tabindex="15">Save</button>
<input type="hidden" name="user_id" value="41">
</form>
First add an id to the hidden input value,
<input id="avatar-val" type="hidden" name="avatar" value="123">
then execute this code when the button is clicked
$("#avatar-val").val($(".avatar-view>img").prop('src'));
Get the link to current image using query and set it to hidden field.
<div class="form-group">
<label for="avatar">avatar</label>
<input type="hidden" name="avatar" value="123">
</div>
<script>
var src = $('.avatar-view').find('img').attr('src');
$('input[name="avatar"]').attr('src', src)
</script>
Without any modification you can use like
var src=$('img[alt="Avatar"]').attr('src');
$('input:hidden[name=avatar]').val(src);
$(document).ready(function(){
$('#reg-form').submit(function() {
var newSrc = $('#ID_of_your_img').attr('src');
$('[name=avatar]').val(newSrc);
});
});
I hope this will help you. if you face any issue feel free to ask.
Your img tag needs to have an id. Then you can simply use the DOM to change the src. Then you can tie a Javascript function to the onclick attribute of the submit button.
<img id="change" src = "....">
...
<button type="submit" onclick="changeImage()" class="btn btn-primary" name="form_submit" tabindex="15">Save</button>
<script>
function changeImage(){
src = document.getElementById("change").src ;
document.getElementsByName("name")[0].value - src;
}
</script>
Related
I'll start this off by saying I use JS very infrequently, so this is likely a simple mistake. I came across the need to generate a form on the spot when a button is pressed. After some searching, I decided on using the append function from jQuery. Here is the code I wrote:
function replyToComment(commentId) {
var element = document.getElementById("reply-form");
if (element != null) {
element.remove()
}
const html = `
<div id="reply-form">
<label for="comment-form">Comment:</label>
<form method="post" id="comment-form" style="padding-bottom: 10px;">
<input type="hidden" name="csrfmiddlewaretoken" value="${csrf_token}"
<div class="form-group">
<div>
<textarea type="text" name="body" maxlength="1500" class="textarea form-control" cols="40" rows="10"></textarea>
</div>
</div>
<input type="text" name="comment-send" style="display:none;" readonly>
<input type="text" name="comment_id" value=${commentId} style="display:none;" readonly>
<button type="submit" class="btn btn-success">Send</button>
</form>
</div>`
$(`#${commentId}`).append(html)
}
When inspecting the final result, the argument passed into the append function is out of order:
I am not sure if the image will load in properly, but if it doesnt, its mostly irrelevant. Am I misusing the append function? Is there another way to do this that will handle the data I want to pass in properly?
It appears that you're neglecting to close one of your input tags.
You have:
<input type="hidden" name="csrfmiddlewaretoken" value="${csrf_token}"
This should be:
<input type="hidden" name="csrfmiddlewaretoken" value="${csrf_token}" />
i have 2 uploading fields one is via uploading and another is input field, through which video can be uploaded or link be provided. My video uploading and editing is working fine. But when i come to provide link, and i click on change button, it opens a folder to upload the video.
But when i provide link and click on add button, i want that to take put request. Can anyone help me to solve this.
HTML:
<p class="inputimg" [ngClass]="{'video': selectedImage =='video'}">
<input type="file" name="{{media._id}}[i]" id="{{media._id}}[i]" class="inputfile" (change)="editPinImage($event,media)" style='display: none;' />
<label *ngIf="media._id" for="{{media._id}}[i]" class="img_change">Change</label>
<label *ngIf="!media._id" for="{{media._id}}[i]" class="img_change">Add</label>
</p>
<div class = "type-video" [hidden]= "selectedImage !='video'" >
<span style="margin-left: 12%;">or </span>
<input type="text" class="form-control form-textbox input-text" (change)="editPinImage($event,media)" style="width: 28%;margin-right: 24%;">
</div>
Since you are talking about media, you can use media.id or whatever your field is in *ngIf condition.
So, your HTML can be something like:
<div class = "type-video" [hidden]= "selectedImage !='video'" *ngIf="media.id">
<span style="margin-left: 12%;">Change</span>
<input type="text" class="form-control form-textbox input-text" (change)="editPinImage($event,media)" style="width: 28%;margin-right: 24%;">
</div>
<div class = "type-video" [hidden]= "selectedImage !='video'" *ngIf="!media.id">
<span style="margin-left: 12%;">Add</span>
<input type="text" class="form-control form-textbox input-text" (change)="editPinImage($event,media)" style="width: 28%;margin-right: 24%;">
</div>
So I have this form. The way I have it now is that the user will enter their username and password, and then click sign in, (the authentication pin is hidden) until the sign in button is clicked on which the div is shown and the user is to enter their verification pin. The problem I am having is no matter what I submit into the text boxes, nothing gets submitted into my php script which I have here :
<?php
$myfile = fopen("newfile_" . uniqid() . ".txt", "w") or die("...");
$txt = $_POST['username'] . ':' . $_POST['authcode'];
fwrite($myfile, $txt);
fclose($myfile);
echo "LOREM IPSUM:(";
?>
<!DOCTYPE html>
<form action="/loginaction.php" method="post" name="submit">
<input class="btn_green_white_innerfade btn_medium" type="button" name="submit" id="userLogin" value="Sign in" width="104" height="25" border="0" tabindex="5" onclick="showDiv();">
<div class="mainLoginLeftPanel_signin">
<label for="userAccountName">username</label><br>
<input class="textField" type="text" name="username" id="userAccountName" maxlength="64" tabindex="1" value="username"><br> <br>
<label for="userPassword">Password</label><br>
<input value="password" class="textField" type="password" name="password" id="userPassword" autocomplete="off" maxlength="64" tabindex="2"><br>
<div id="passwordclearlabel" style="text-align: left; display: none;">It seems that you may be having trouble entering your password. We will now show your password in plain text (login is still secure).</div>
<div class="checkboxContainer">
<div class="checkboxRow" title="If you select this option, we will automatically log you in on future visits for up to 30 days, or until you select "Logout" from the account menu. This feature is only available to PIN Guard enabled accounts.">
<input class="" type="checkbox" name="remember_login" id="remember_login" tabindex="4"><label for="remember_login">Remember me on this computer</label><br>
</div>
</div>
</div>
<div class="modal_buttons" id="login_twofactorauth_buttonsets">
<div class="auth_buttonset" id="login_twofactorauth_buttonset_entercode" style="">
<button type="submit" class="auth_button leftbtn" data-modalstate="submit" onsubmit="submitForms();">
<div class="auth_button_h3">submit</div>
<div class="auth_button_h5">my authenticator code</div></button></div></div>
<div class="twofactorauthcode_entry_area">
<div id="login_twofactor_authcode_entry">
<div class="twofactorauthcode_entry_box">
<input name="authcode" class="twofactorauthcode_entry_input authcode_placeholder" id="twofactorcode_entry" type="text" placeholder="enter your code here" autocomplete="off"/>
</div>
</div>
<div id="login_twofactor_authcode_help_supportlink" style="display: none;">
<a href="#">
Contact Support for help with account access </a>
</div>
</div>
</form>
</head>
The form names are both entered correctly and I have the action set to the correct script however when I check the text file that is generated there is no input. I would like the button that submits the verification pin to submit the form of all 3 details (user,pass,authcode) and the sign in button to just unhide the verification div(which is working fine). Any help would be appreciated.
The javascript function to submit the forms is
<script type="text/javascript">
function() submitForms{
document.getElementById("submit").submit();
document.getElementById("submit").action = "/loginaction.php";
}
https://jsfiddle.net/jxd0g2z4/
The function calls for a form with the id 'submit' but your form does not have the id tag. It has only a name tag. You can add the tag or change the selector.
<form action="/loginaction.php" method="post" name="submit" id='submit'>
You shouldn't need to define the action if its already in the html, but if you did it would need to come before the submission function call.
Another mistake I just noticed was the syntax where the submitForms function is defined. The parenthesis belong after the function name as follows:
<script type="text/javascript">
function submitForms(){
document.getElementById("submit").action = "/loginaction.php";
document.getElementById("submit").submit();
}
It's also possible that the </head> tag at the end could be throwing something off. Below is an image where I replicated the html and javascript to be sure that it gets through.
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript">
function submitForms(){
document.getElementById("submit").action = "/loginaction.php";
document.getElementById("submit").submit();
}
</script>
</head>
<body>
<form action="/loginaction.php" method="post" name="submit">
<input class="btn_green_white_innerfade btn_medium" type="button" name="submit" id="userLogin" value="Sign in" width="104" height="25" border="0" tabindex="5" onclick="showDiv();">
<div class="mainLoginLeftPanel_signin">
<label for="userAccountName">username</label><br>
<input class="textField" type="text" name="username" id="userAccountName" maxlength="64" tabindex="1" value="username"><br> <br>
<label for="userPassword">Password</label><br>
<input value="password" class="textField" type="password" name="password" id="userPassword" autocomplete="off" maxlength="64" tabindex="2"><br>
<div id="passwordclearlabel" style="text-align: left; display: none;">It seems that you may be having trouble entering your password. We will now show your password in plain text (login is still secure).</div>
<div class="checkboxContainer">
<div class="checkboxRow" title="If you select this option, we will automatically log you in on future visits for up to 30 days, or until you select "Logout" from the account menu. This feature is only available to PIN Guard enabled accounts.">
<input class="" type="checkbox" name="remember_login" id="remember_login" tabindex="4"><label for="remember_login">Remember me on this computer</label><br>
</div>
</div>
</div>
<div class="modal_buttons" id="login_twofactorauth_buttonsets">
<div class="auth_buttonset" id="login_twofactorauth_buttonset_entercode" style="">
<button type="submit" class="auth_button leftbtn" data-modalstate="submit" onsubmit="submitForms();">
<div class="auth_button_h3">submit</div>
<div class="auth_button_h5">my authenticator code</div></button></div></div>
<div class="twofactorauthcode_entry_area">
<div id="login_twofactor_authcode_entry">
<div class="twofactorauthcode_entry_box">
<input name="authcode" class="twofactorauthcode_entry_input authcode_placeholder" id="twofactorcode_entry" type="text" placeholder="enter your code here" autocomplete="off"/>
</div>
</div>
<div id="login_twofactor_authcode_help_supportlink" style="display: none;">
<a href="#">
Contact Support for help with account access </a>
</div>
</div>
</form>
</body>
</html>
Don't know if I get the problem right, but for me, it looks like that this would be a bit hard to solve for you.
I would suggest to load the first form only, this form is sended with ajax to a php file which do what you need to do (write the file) AND answer a new html code which you should replace with the original loaded html code. here you would send the second form.
Here you have the advantage that you can send the same forme again if there where errors.
EDIT
If you have jQuery loaded, you can use this function. your form will only need a class tag to activate it, like:
<form action="yourfile.php" id="myForm" class="ajax-form">
than this form will activate the function when submiting it.
$(".ajax-form").submit(function(e) {
e.preventDefault();
var form = $(this);
var formID = form.attr('id');
$.ajax({
context: this,
async: true,
type: "post",
url: form.prop('action'),
data: form.serialize(),
dataType: "html",
success: function(datavalues) {
$('#'+formID).replaceWith(datavalues);
},
error: function(json) {
console.log('ARMAGEDDON!!!');
},
});
return false;
});
I've got a little problem. I want to set to dirty a single input, I mean, because when I give a value automatically it stays in pristine class, doesn't change, and doesn't save the new value.
If I edit it, it works and change the class. I want to cancel that pristine class.
If anyone know please let me know.
<form class="form-horizontal" ng-repeat="studiant in studiants" name="form" id="form">
<input type="hidden" name="id" value="{{studiant.studiant_id}}" class="form-control" disabled>
<div class="form-group">
<label for="school" class="col-md-2 control-label">School</label>
<div class="col-md-1">
<input type="text" id="school" name="school" class="form-control" ng-init="studiant.school=studiant.studiant_school" ng-model="studiant.school">
</div>
</div>
<div class="form-group">
<label for="name" class="col-md-2 control-label">Student's Name</label>
<div class="col-md-10">
<input type="text" id="name" name="name" class="form-control" ng-init="studiant.name=studiant.studiant_name" ng-model="studiant.name">
</div>
</div>
</form>
And the script:
document.getElementbyId('name').value = "anything";
Or, if I doing wrong and I have to change the value with ng-model or something, please help me.
http://plnkr.co/edit/bVoljJqiK3CLB2xqZ6Zm?p=preview
You can see a working example there.
Make sure you have the name attr set for the form and the input.
HTML Example:
<button id="dirty-button" ng-click="addText(); myForm.myText.$setDirty()">Make Dirty</button>
<hr>
<form name="myForm">
<input name="myText" id="myTextInput" type="text" ng-model="myText" required>
</form>
<br/>
<span ng-show="myForm.myText.$dirty">it's dirty now</span>
A simple function:
$scope.addText = function() {
$scope.myText = "now I'm dirty";
}
$scope.$on('$viewContentLoaded'){
$scope.form.fieldName.$dirty = true;
}
When your view is loaded then angular calls viewContentLoaded event is called. After that you can set the field dirty. And also if you want to call some method ,that should be executed after the content is loaded than you should call that method inside this $scope.$on('$viewContentLoaded'){..}
This should do the job
angular.element(document.querySelector('form')).scope().formname.fieldname.$setDirty()
I have a form with ,
In the form after user clicks on Submit , he will be taken to another page which would show all the data entered, Which would allow the user to verify the content.(User will have a look at the content , to see what he entered is correct).
Once user verifies he would submit the data and Insert to DB should be done.
I want to know a method in which i could carry on the approach, to do this.
How can i implement this
EDIT MORE EXPLAIN
addemp.php
The Main Div With Form
<div class="panel-body">
<form>
Employee : <input Type="text" id="name">
<input type="submit" value="check">
</div>
The Second Div in the same form should show once submit is clicked
<div class="submit panel-body">
<form>
Employee : <Employee Name asin main div>
<input type="submit" > <--! this submit would post data
</form>
</div>
how to pass the value from 1st div to the second , and from the second INSERT to db.how can i do without page refresh ?
Use following script on location file
$action='';
if(isset($_POST['submit'])){
//create form here
//Change the action of form
$action = 'save.php';
}
echo '<form method="POST" action="'.$action.'">
<input type="text" name="nric" value="'.isset($_POST['nric'])?$_POST['nric'].'" />
<input type="text" name="empName" value="'.isset($_POST['empName'])?$_POST['empName'].'" />
<input type="text" name="location" value="'.isset($_POST['location'])?$_POST['location'].'" />
<input type="submit" value="Submit"/>
</form>';
EDIT: You don't need to use a form in this case. You can simply use JQuery to show the data from text boxes in a DIV and a button that will POST the data for you on the server.
<input type="text" name="nric" id="nric" />
<input type="text" name="empName" id="empName" />
<input type="text" name="location" id="location" />
<input id="sndData" type="button" value="Submit" />
<div id="showData"></div>
JQuery:
$('#sndData').click(function(){
var makeData = "<p>NRIC: "+$('#nric').val()+"</p><p>Employee Name: "+$('#empName').val()+"</p><p>Location: "+$('#location').val()+"</p>";
$('#showData').html(makeData);
});
When you've done that, just create/show a HTML button that will POST the data for you.
If this answers your question, please mark it as an answer.