I need help making this timed quiz so when you hit submit it goes into a mysql database. right now it just tells you if you answered correctly or not Here is the code:
<input type = "text" id = "answer">
<input type = "button" id = "time" onclick="answer()">
<script type = "text/javascript">
[code]....
Few days ago since I started using JQuery amazing world, I got some basics to get started, I created a dialog in which I included a form that contains two text fields and a button to submit, what I want (in which I stuck now ^^) is the data interred within the form to be sent to the database, I used to do the following:
$("#sendRequest").click(function() {
$.post("Request.php", {
iduC:$("#iduC").val(),
idu:$("#idu").val(), idp:$("#idp").val()
[Code]....
I could send the data by lot of ways but using dialog is pretty cool!
I'm trying to convert a html form to use jQuery. The submit asp page works just fine via standard html post. Yet I can't figure out how to use jQuery
//dataString = data from the form
$.ajax({
type:'POST',
dataType: "application/x-www-form-urlencoded",
[Code]....
On my form I have 3 submit buttons which handle different things.I am looking for a way to stop or continue form execution with a confirm box on the third submit button and the third only.I can't use onsubmit because that will trigger on all three buttons.
View 2 Replies View RelatedIn the following .submit function, I am attempting to grab the value of the selected option in the facilityCodes dropdown list when I click the submit button and then during the submit function,select the facilityCode again in the dropdown list when the page reloads and then disable the list so that the user cannot change it.However,the situation is when I reload the page after the submit button is clicked the dropdown defaults to the first option and the list is enabled.I apparently am not understanding how.submit works so that I'm selecting the option I'm defining in my code below and then disabling the list AFTER the page reloads due to an error on the page. My question is how can I accomplish this?Here is my code:
$(function() {
$("#ARTransferForm").submit(function() {
var msgsCount = 0;[code]....
The following will submit the form data to popup by clicking the submit button. I want it will submit the form automatically to the popup, there is no submit button in this page. Basically this page should not show up.
<script type="text/javascript">
function submitmyform(f) {
f.target = 'foo'
window.open('',f.target,'menubar=no,scrollbars=no, width=800,height=800');
f.submit();
return false;
}
</script>
<form name="myform" action="popup.asp" target="_blank" method="post"
onsubmit="return submitmyform(this);">
<input type="hidden" name="item" value="item"/>
<input type="submit" value="submit to popup"/>
</form>
I'm using mailchimps signup box (they don't have a decent forum to ask on) on my website and want to adjust the submit button and change it to a normal link. Here's the button that submit's the form:
[Code]...
I have a page with many forms that I need to change from a post to an ajax call. That part is working, no problem, but now I want to disable the submit button while it's waiting on the server response and then re-enable it when the response comes back.
Here's what I have:
$(function() {
$('form').each(function() {
$(this).submit(function(){
[code]...
I can't figure out what my selector should be to get the submit button of the form that's being submitted. What should I be using instead? Also, if the call errors out, I'd like to just post the form as usual.
I have a form without a submit button. It gets submitted programmatically with document.form.submit().
What I need is to be able to disable this form's submit capability on page load and then reenable it at some later point. Remember there is no 'input' button element.
Code:
What I've tried so far is like this:
savedSubmit = document.inputs.submit;
document.inputs.submit = None;
then later:
document.inputs.submit = savedSubmit;
but this does not work. How can I do this?
i am trying to build my form to submit without reloading the screen. I am using jquery and it inserts the data but it also reloads the screen and two database records get inserted. It appears that one is being inserted by jquery and the other is being inserted by the normal form submission. How do I remove/override the normal form submission? My html code is as follows:
[Code]...
I've got a pretty basic form that i need to be able to "submit" by simply loading it in the browser address, not click the submit button. is there a way to do this? Here is the code:
<form name="newform" action="/dir/submition.cfm?z=1&Myid=1" method="post">
<input type="hidden" name="userID" value="27566">
<input type="submit" value="Submit">
</form>
if i load just http://www.mysite.com/dir/submition.cfm?z=1&Myid=1, it doesnt do anything because it doesnt have the userID variable, but im not sure how to include that into the address...
I am making a webpage that can query a database table of employees
and return them in a html table. Currently I have submit buttons beside
each employee to open the page to edit their details. I have been using
php/HTML to do this, but wondered if it is safe to use javascript to
just click on the name of the employee to jump to the edit page. I've
been looking on the web for sometime about how to do this without any
luck. Any thoughts?
I have this as my Dialog Javascript
$( "#dialog-form<?php echo $diagnumber; ?>" ).dialog({
autoOpen: false,
height: 300,
[code]....
I have a problem in getting database query result to text field when button was clicked. So far I have two php files, one used to connect database and save the database query value to a variable. The second php file used to create the buttons and text field.
code for dbconnection.php
[code]
$con_str = "host=jojo1234 port=3214 dbname=postgres user=postgres password=postgres";
$con = pg_connect("$con_str") or die('Could not connect: ' . pg_last_error());
$queryresult1 = pg_query($query1);
$result11=array_sum(pg_fetch_array($queryresult1));
[code]
[Code]...
I have this PHP page that has the following functions...Insert Form into Table in MySQL DatabaseDelete Entry from Table in MySQL DatabaseA Link to a separate page where I Edit entries I have coded a PHP function for Delete. Looks like this:
// delete if provided if (isset($_GET['id'])) { // query to delete $query = "delete from student where STUDENT_ID =".$_GET['id']; //execute query if ($db->query($query)) {
Somewhere else I query the results and print them into a table. I added a column to this table with a Delete link for each row. Clicking on delete will execute the PHP function to delete.
The link looks like this: PHP Code:
echo "<td><a href=".$_SERVER['PHP_SELF']."?id=".$row[0]."></a></td>";
Likewise the first names also have links when clicked bring the user to another page with a form to edit form data.
The link looks like this: PHP Code:
echo "<td><a href='editstudent.php?id=".$row[0]."'>".$row[1]."</a></td>";
From here however I need to change those links to Buttons. One for Delete and another for Edit and use JavaScript functions to make them work as they originally did. So this is what I've done...
<td><form name='myform1'><input type='button' name='edit' value='Edit'
<td><form name='myform2'><input type='button' name='delete' value='Delete'
Next I need to write functions...this is where I am lost. Usually I'd know what to do when referencing elements in forms, i'd do something like document.formname.inputname, but my problem is those links or should I call them actions are replaced by those new buttons.
Should I put them into a form, or make them hidden, and then name the element? Or should my function look something like this...all I really want to do is to call the function, and have the function just do the action of loading the link (don't know if that sounds clear). I've tried to do this but it does not work.
<script>
function Delete()
{
document.myform2.action="href='$_SERVER['PHP_SELF']."?id=".$row[0]'";
document.myform2.submit();
}
</script>
Can anyone tell me what i am doing wrong, and could you please point me in the right direction?
getting the selected radio button and save it in the database. Here is my code:
[Code]....
I wanted to know which answers are selected and save those answers in the database with the corresponding question. I am using MYSQL as my database tool. I know how to use MYSQL and querying but I just don't know how to get the datas that I need.
I copied codes on internet and arranged them according to my knowledge of JQuery (which is very shallow).. The code works fine sometimes.... but not all the time. Can anyone help me to find out why the dialog box does not appear everytime I put wrong username and password?
Here is the JQuerycode:
[Code]...
Because I am using ui-darkness as the page theme, so at form tag, I don't have defaultbutton="but_submit" tag. If I add it back, I will get error while compiling..
I have a form where i use normal buttons with javascript so i can have multiple actions and I don't use a submit button. I can submit the form fine using document.adminform.submit(); but i have an onsubmit="return submitForm"; attached to my <form> tag.
If i had a submit button, it would call the javascript function, but seeing as i don't have a submit button, how can i call that function. here is the bits causing problems
PHP Code:
<form name="adminform" id="adminform" method="post" action="<?php echo $_SERVER['REQUEST_URI']; ?>"  onsubmit="return submitForm();">
<p><input type="button" class="back" id="backbutton" title="go back" onclick="performAction('back');" />
<input type="button" class="save" id="savebutton" title="save" onclick="performAction('save');" /></p>
[Code]...
when the save button is clicked, a javascript function is called which does the ocument.adminform.submit(); but i need to do call the submitForm() function at the same time
Below is the code for a Captive Portal Splash Page. I have tried to research how to fix the button at the bottom of my page but I can either add a link (which does not require checking the box to agree to terms) or simply how it came to me without the ability to send customer anywhere.
[Code]...
I have built search form that it retrieves information from database. for ex: when i click axtar (search) button it retrieves normally but when i click ENTER button via keyboard instead of axtar (search) button but it only displays results with white blank page
Here is web page address. [url]
Does the "submit" button always submit form data in non-ajax way?What are the best practices for submitting forms using xhr?
View 1 Replies View RelatedI am a graphic designer who has been stuck with creating a voting system for my son's school newspaper website.
Here is what I have so far: [url]
You will see that it is a 1-10 voting system where the user drags the choices into the order of his preference then hits submit. It is working on on a functional level, but I have no idea how to make it work on the backend. I need to somehow accomplish the following:
1. Either get rid of the numbers on the left or somehow stop them from all becoming "1" when dragged.
2. Upon hitting "submit" the votes should be added to a database. I can create a database but I have no idea how to get these numbers into it.
3. The 10 choices will change each week for at least 13 weeks so the database needs to adapt to this.
Here is a zip file with all of the files used on the voting page: [url]
I have a website containing 26 subdirectories 'a' to 'z'
On the home page I want a text field and submit button
If someone for example types 'j' it will go to the 'j' folder home page
Does anyone know where I might find code like this?
On the server side, I'm using a web framework that use submit button "name" attribute to decide which action to perform. On the client side, I have an script that submit forms in Ajax with the "serialize()" functions.
But as the JQuery documentation said, the serialize() function don't send the submit buttons names. I have tried to get the event.target object but this object is the form itself. So, I wonder to know if he is a way to discover obtains the name of the button that has triggered the form submission.
How can I disable a submit button for a form if JS is disabled. In other
words, JS must be enabled to submit the form.
how to create a drop down menu without the need of a submit button? Once user Choose his stuffs, it will auto submit the results without clicking submit?
View 4 Replies View Related