The code for example document.getElementById("id").value="value"; is not working. I have assigned it on a function to replace the current value of the textbox once the page is loaded.
here is my code
<script type="text/javascript">
function rplace(){
document.getElementById('idtext').value="New Value";
}
onload=rplace
</script>
<form>
<input type="text" id="idtext" name="idtext" value="">
</form>
use jquery insteat of javascript.
$(document).ready(function(){ $("#idtext").val("New Value")});
Refer this
Related
I have a form in codeigniter where I've an option to show the data in the front page or not.
<div class="form-group input-group">
<label>Show in Front</label>
<input type="checkbox" id="switch-change" name="show-infront" checked data-on-text="YES" data-off-text="NO" checked data-on-color="success" data-off-color="warning">
</div>
<script type="text/javascript">
$("[name='show-infront']").bootstrapSwitch();
</script>
When I print the value of the form in controller like echo $this->input->post('show-infront'), when 'yes' is checked, the output is 'on' and when 'no' is checked, there is no any output. So, I want that when I check yes, then the value should be 'yes' and vice-versa.
try: onSwitchChange ...
OK so if checkbox in not checked not send its value in POST, so you need a hidden field like this:
also need set its value at init and onChange
<input type="checkbox" id="switch-change" name="show-infront" data-on-text="YES" data-off-text="NO" checked data-on-color="success" data-off-color="warning">
<input type="hidden" name="show-infront-val"/>
<script type="text/javascript">
var ckbox = $("[name='show-infront']");
var ckbox_val = $("[name='show-infront-val']");
ckbox.on('switchChange.bootstrapSwitch init.bootstrapSwitch', function(event, state) {
if (this.checked) ckbox_val.val('Yes')
else ckbox_val.val('No')
});
ckbox.bootstrapSwitch('state', true);
</script>
Btw if you check php-s isset($_POST['show-infront']) also a good option for one checkbox
Documentation
On my website i have a button which when clicked takes you to one of two random youtube videos. However i would like to change this to a image in stead of a button.I have tried to change it to a INPUT type="image" but this doesn't work. Here is the code i am using.
<SCRIPT language="JavaScript">
<!--
function get_random()
{
var ranNum= Math.floor(Math.random()*2);
return ranNum;
}
function getaGame()
{
var whichGame=get_random();
var game=new Array(2)
game[0]= "https://www.youtube.com/watch?feature=player_detailpage&v=NcFQF3PZFRk#t=722s";
game[1]= "https://www.youtube.com/watch?v=klBAW4MQffU";
location.href = game[whichGame];
}
//-->
</SCRIPT>
<FORM name="form1">
<center>
<INPUT type="button" onClick="getaGame()" >
</center>
</FORM>
Thanks for any help
An onclick event can be fired from any element. Here are some examples!
i am a newbie programmer.I am stuck for two days with a simple code.I try to use jquery form plugin for submitting a form to another page and get a feedback from that page.The problem is the plugin is not working.the form is submitted as normaly without feedback.Here is the code
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
<div id='preview'></div>
<form action='ajaxcall.php' id='upload_pic' enctype='multipart/form-data' method='post'>
<input type='file' id='pic' name='picture'>
<input type='submit' id='sub'>
</form>
var options=
{
target:'#preview',
url:'ajaxcall.php'
};
$(document).ready(function(){
$("#sub").click(function(){
$('#preview').html("<img src='images/loader.gif' alt='Loading.....'/>");
$('#upload_pic').ajaxForm(options).submit();
});
});
Here is my ajaxcall.php page code
if(!empty($_FILES['picture']['size']))
{
echo "<img src='images/197.jpg'>";
}
Expectation was the echoed image would feddback but the page is simply redirected to ajaxcall.php page.ajaxForm() function is not working.But why?please help.Thanks in advance.
Just replace your submit button for a normal one, because you are already submiting the form programatically from your click handler, so replace your following html:
<input type='submit' id='sub'>
for this one:
<input type='button' id='sub'>
use these codes instead of your script codes. sorry for the late answer
$(document).ready(function(){
$("#sub").click(function(){
var options=
{
target:'#preview',
url:'ajaxcall.php'
};
$('#preview').html("<img src='images/loader.gif' alt='Loading.....'/>");
$('#upload_pic').ajaxForm(options).submit();
});
});
I am used following code hide DataPicker element only hide textbox does not hide image of calender.
HTML Code
#Html.Telerik().DatePickerFor(model=>model.Dateofpublisher)
Jquery
$('[name="Dateofpublisher"]').hide();
You dont need to write DatePickerFor.I always do it as following:
<input id="datepicker" name="date*" type="text" value="" />
*The name of your datepicker should be the same with the name of your model date filed.
And in script codes just write:
<script>
$(function() {
$( "#datepicker" ).datepicker();
});
</script>
Note:jqueryui files should be added first.
HTML
<div id="Date">
<input id="datepicker" name="date*" type="text" value="" />
</div>
Jquery
$(document).ready(function () {
$("Date").hide();
});
I am required to assign ViewData["file"]'s value to an html hidden control through javascript/jQuery
You need to have
<script>
$(document).ready(function(){
$("[name=hdnHiddenField]").attr("value",#ViewData["file"]);
});
</script>
and in your view:
<input type="hidden" name="hdnHiddenField"/>
Note that you cannot have this script in an external js file.