View image of form here before reading the question.
Using jQuery when the Complete Safety Walk button is pressed I want to make sure that if any radio buttons are set to Fail that the Action Required, Assigned To and ECD fields for that row are not empty and the Complete Safety Walk button gets disabled.
The table has an id safety_walk_items and each row has a unique id item_1, item_2, etc.
Note: There are NO form open/close elements as each row is saved using AJAX (when the icon at the end of the row is clicked).
I'm not using any fancy input highlighting, just showing a hidden div with an error message.
<div class="validation_errors hidden"></div>
Related
I tried lot of posts but none seem to work for me.
I have 2 drop downs inside same form element and they should work in toggle manner, with following properties:
selecting any drop down should clear the other one and auto submit the form.
Also on form submit I am retaining the value of selecting drop down to display user what value had been selected.
For auto submit I have used .change and fformid.submit.
But issue is if i selected dropdown 1, form get submitted and page is displayed with dropdown1_value 1 selected and now when i make selection of 2nd drop down, both fdropdown 1 and dropdown 2 value gets submitted.
I tried using dropdown2_id.remove(), .emtpy(). .val("").
All these seem to visually remove elements, but form gets submitted with the value previously selected.
And i dont want to use Ajax, since I am using Django, doing Ajax will return result to js which is difficult for me to render than using html itself.
I have a form that saves user entered values on submit with php.
Some of the fields are in div's that are display:hidden until an onclick or onchange function changes that div to show.
How can I show only the divs containing fields with saved values after the form has been submitted? I have saved the values in the always visible fields but cannot trigger their functions.
I am using very little jquery because I am new to the syntax and would prefer to implement solutions I can understand and adapt. Simple jquery is acceptable if it is a better/quicker/easier solution.
Thanks
Code Example:
<input type="radio" id="customer" name="jobtype" value="customer" onclick="getJobType()" autofocus>Customer
<input type="radio" id="store" name="jobtype" value="store" onclick="getJobType()">Store
<span id="customerjobs" style="display:none">
<select id="customer" name="customer" onchange="createJobsList(this.value)">
*various options*
</select>
<span id="jobslist"><br></span>
</span>
The first span (id=customerjobs) is initially hidden. Upon selection from the radio's, all but the corresponding span is set to display:none and the selected is set to display:block. On submit, the selected radio is saved, but the onclick isn't called to show the span.
The second span (id=jobslist) content is populated by innerHTML using the results of an ajax call to PHP when a selection is made. On submit, the selected option is saved, but the onchange isn't called to fill the span.
So I need to trigger the onclick of a saved radio value to show my content and trigger the onchange of a saved select to populate additional content.
Note: I used onblur with javascript to set the focus initially so any action would trigger the content but it caused an unnecessary pause in filling the form that I didn't want.
Page loads with only a radio selection.
User clicks radio button.
Onclick function changes style of span id=customerjobs to display:block.
The select input inside the span is now visible. The user selects an option.
Onchange function makes an ajax call to request information from the server which is placed in span id=jobslist.
User submits form to same page.
Form saves entered values so they are still selected when page reloads.
Onclick and onchange functions are not triggered by PHP saved values so steps 3 and 5 never occur. Page is left with only the radio buttons unless it is clicked again.
Well, I have a jsfiddle to illustrate my problem using default selections because I cannot use PHP to save entered values.
Imagine the form has just been submitted and the values saved are the checked radio button(customer) and the selection from the drop down(1) which adds the word "customer".
Ideally, the entire form would still be visible (The selected radio, the selected option and the content added to the last span "customer").
Instead, only the selected radio is visible unless it is clicked again to unhide the select drop down. The drop down retains its' value as well, but the content in the last span will only appear onchange.
http://jsfiddle.net/L5H2u/31/
Try it out and advance thanks for any suggestions.
Can you hook a function to onload that checks the radio buttons and simulates the click by calling getJobtype()? That will get the initial case where the radio button is already the way the user wants it. Further clicks will work as you planned.
Edited to add: If I understand you right, all is well the first time the page is loaded because the user has to click something and that runs your getJobType() function. However, when the page is reloaded, the correct radio button is already checked, the user doesn't change anything, and your function doesn't run. If that's correct, running getJobType() from onload should fix it.
You may need something like <input type="hidden" id="firstrun" value="true"> The PHP would set that to false on subsequent loads of the page, and the onload function would only make things happen if "firstrun" was false. Edit: You don't need this because getJobType() has no default action; keep reading.
Edited still more: You have checked="checked' on the Customer radio button, so if the user is a customer, even the initial run doesn't reveal the hidden material.
Remove checked="checked" when the page is initially loaded so that on the initial presentation, neither button is checked.* Then add window.onload=getJobType; to the end of your JavaScript.
On the initial load, getJobType() will do nothing since it checks both buttons and has no default action. When a button is clicked, getJobType() will run and act based on the button that was clicked.
When the page is loaded a subsequent time, one of the buttons will be checked and when onload runs getJobType() it will perform the corresponding action.
The radio buttons, SELECT values, and any other form elements that are preserved and "reflected" by the server-side PHP will be correct when the form is loaded the second (and subsequent) times. Where you need an onload JavaScript function is when one of those values also changes something else, such as making a hidden DIV visible. If there are functions other than getJobType() that manipulate the DOM, it will likely be cleaner to write an init function that sets up the DOM based on the values of the form elements as preserved by the PHP process.
* I normally advocate having some button checked by default so that the user can always get back to the initial state. This case seems to be an exception.
I have one HTML page. It has 3 tag contains below details.
(when Form is shown other 2 are hidden & likewise.)
1. One Form
2. Calculated Values from Form
3. Summary
After form filling, Whenever I click on "proceed" button, a JS function is called & result is calculated. I want to fill this result in summary table fields. & every time "proceed" button is clicked new values should be added to old values & set new values to that field.
One time, I have put values in summary, but whenever I click on "show form" & summary gets hidden, the table gets reset. What should be the problem? I want to use only js & html.
What should I do ?
Help..
I have a login page with 2 radio buttons "new account" and "existing user". when "new account" is selected the "User" field auto populates with the text "New Account" and the "Password" field remains blank. I need to grey out the fields so that they are uneditable when the "new account" radio button is selected, but still pass along the information in the fields because it is used to gain access to the database. I can disable the fields to get the desired uneditable greyed out fields, but the information does not get passed along to the database.
I have tried to fix this by creating two hidden fields (which auto populate with the needed information for database access) to take the place of the "user" and "password" field which allows me to disable the visible fields while "new account" radio button is clicked and which still passes along the new user login info that never changes. This works fine until I try to login as an existing user, in which case my two hidden fields do not auto populate with the users input for their existing account information.
There may be a much simpler approach to fixing this problem, but all of my research and trials have not been successful yet. I have been reluctant to ask this question as it seems so simple and frequently used approach for a login page, but all of my searching has not yielded any thing that has worked yet. I appreciate any input or navigation in the right direction.
I'm pretty sure the correct and painless solution for your problem is to use two different form tags (take care to not nest them) and show/hide the form depending on the selected radio button.
And for the convenience of your user you should copy the username from one form to the other if he has already filled the user field and switches to the other version later.
EDIT
The complete solution:
HTML
<label><input class="formSwitcher" type="radio" name="formSwitch" data-form="#divForm1"> Form 1</label>
<label><input class="formSwitcher" type="radio" name="formSwitch" data-form="#divForm2"> Form 2</label>
<hr>
<div class="hiddenForm" id="divForm1">Put form 1 in here</div>
<div class="hiddenForm" id="divForm2">Put form 2 in here</div>
JS
// if someone clicks on one radio button
$('.formSwitcher').change(function(){
// get the form id we want to show (from the "data-form" attribute)
var formIdToChange = $(this).data('form');
// hide all forms first
$('.hiddenForm').hide();
// show the specific form
$(formIdToChange).show();
});
// initially hide all forms
$('.hiddenForm').hide();
// initially call the change method if one radio is already selected on page load
$('.formSwitcher:checked').change();
Are you saying that you need to change the existing user fields because they will grant access to the database? I think what you want to do is to change the properties of fields depending on what is selected. For example if you need the auto populate fields to be set when the user clicks new user then write something like
document.form.AutoPopUser.value="required value"
and when the person clicks on the existing user you can do
document.form.AutoPopUser.value=""
or if even having that part exist will mess up the existing user log in then you could delete that entire section by putting it inside a div and creating or destroying it depending on the selected option. I feel like I'm not being very clear but I think what you need is a clever set of onClick functions to get rid of things you dont need and add in the ones you do need. If you could post some code to go with this that would be awesome.
I have a grid. In a each cell in a column in the grid, there is a text box. Next to each text box is a button. Attached to the button is a click event handler (jQuery .click()) What I want to do is when the user clicks the button, I want to fetch the value of the text box immediately next to it. There are multiple rows in the grid. Each row how a text box and a button.
I know how to use jQuery to fetch an individual, or group of items (using selectors). I also know how to attach the click event handler to all the buttons.
But what are some good ways to reference the text box next to the button -- aside from "walking the DOM" (i.e. using .parent() or .next())0
An easy way to do it would be to get the TR then find the input from there:
$(this).closest('tr').find('input');
You want .siblings('input') - this will find those input elements "next to the button" that was clicked.