im new to salesforce integration.
i have a contact form in which i have a field called Optin/Optout filed which is checkbox. if the user tick that checkbox value should get captured in salesforce. but i could not do it. can anyone please help me in this.
here is my code what i have done so far.
index.php
<form action="salesforce.php">
<input type="checkbox" name="news" value="1" checked="checked"/>
</form>
salesforce.php
<?php
$newsletter = $_POST['news'];
if($newsletter == "1"){
$news_result = $newsletter;
}else{
$news_result = "0";
}
$req .= "&HasOptedOutOfEmail=" . urlencode($news_result);
?>
You miss the method in the form, if no method is defined it will be a get method and you use post the get the value
if you do this it will work:
<form action="salesforce.php" method="post">
<input type="checkbox" name="emailOptOut" id="emailOptOut" value="1" />
this solved my problem
Try using this it worked for me:
apex:inputCheckbox value="{!news.isSelected}"
selected="{!news.isSelected}" immediate="true"
Related
I've got this code from the web. It functions with my setup, calling the command all ok, i have the variables entered correctly.
But the if .. else statement redirects to only one URL, not the if statement, but only the else. so no matter what radio i select the https://thisdomain.com/1-2/ domain loads.
Here is the PHP/Html code:
<?php
$radio832112001 = intval($_POST['radio832112001']);
if ($radio832112001 === 2)
{
header("Location: https://thisdomain.com/2-2/");
die;
}
else
{
header("Location: https://thisdomain.com/1-2/");
die;
}
?>
<form action="https://thisdomain.com/code/action-900.php">
<input type="radio" id="no" name="radio832112001" value="2">
<label for="indoor-220">No</label>
<input type="radio" id="yes" name="radio832112001" value="1">
<label for="outdoor-220">Yes</label>
<input type="submit" value="Submit">
</form>
What am i missing?
Cheers in advance.
Steve
$radio832112001 = intval($_POST['radio832112001']); with:
$radio832112001 = intval($_GET['radio832112001']); or
add method="POST" to your form tag.
This worked. With updating the for="" code too. It was wrong. Now it all works as intended.
Thanks all
You are using GET method in your form by not declaring method keyword. Replace
$radio832112001 = intval($_POST['radio832112001']);
by
$radio832112001 = intval($_GET['radio832112001']);
or add method="POST" to your form tag.
Having issues with using a form's value in a different php file:
my firstpage.php
<form method="post">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
</form>
my secondpage.php is here
<?php
include("firstpage.php");
$result = $_POST['rdbbtn'];
if ($result == "1") {
echo 'thirdpage.php';
}
else {
echo 'fourthpage.php';
}
?>
problem:
Notice: Undefined index: rdbbtn in
how come I can't use "rdbbtn"? Should I have something like
$rdbbtn = $_POST['rdbbtn'];
in secondpage.php? Tried this but didn't solve my problem.
firstpage.php and secondpage.php are in the same directory.
Probably it's some pretty obvious thing that I don't see...thanks!
EDIT: I have accepted pradeep's answer as that helped me the most to figure what the problem should be. would like to say thank you for everybody else showing up here and trying to help!
When you change current page it reset the value and $_POST is empty.
You can try with set form action to next page . It will work
<form method="post" action="secondpage.php">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
<input type="submit" name="" value="Next">
</form>
Other wise you can make a function in a class and set each page action
to this function.
And set your each form data to session.
Finally when you change the page you read data form session.
Class FormAction{
public function setFormDataToSession(){
if(isset($_POST['rdbbtn']){
$_SESSION['rdbbtn'] = $_POST['rdbbtn'];
}
}
}
In your page simply get the session value.
echo $_SESSION['rdbbtn'];
Should be like this :
Check with isset method in
<?php
include("firstpage.php");
$result = isset($_POST['rdbbtn']) ? $_POST['rdbbtn'] : NULL;
if ($result == 1) {
echo 'thirdpage.php';
}
else {
echo 'fourthpage.php';
}
?>
and your form should be like this :
<form method="post">
<input type="radio" name="rdbbtn" value="1"> One
<input type="radio" name="rdbbtn" value="2"> Two
<input type="submit" name="submit" value="submit">
</form>
Sorry for not being able to comment in this post(less reputations). But seems like you are asking about storing the variables of the session. This way you can use the variables for a whole session. Just start the session by putting session_start() in the very beginning of secondpage.php file and then you can access the variables at any time during the session by simply calling $_SESSION['rdbutton] in any page like fourthpage.php or anything. Just make sure u put the session_start() at the top of each page where you want to use the variables. Don't forget the semicolons at the end. 😜 Hope this helps.
This is my first post and I'm complete beginner so please be gentle :)
I'm trying to create a form that after submitting an account name would check and return a CNAME of the host (account+domain.com)
The problem is that I want to do it all on the same website so it will either display the form if nothing has been posted or display the result otherwise.
This is what I've created, it seems that I'm not calling the POST correctly, but I can't really get what am I doing wrong.
Please help
<?php
if(isset($_POST[DomainSubmit])){
$AccountName = $_POST[ClientDomain];
$CName = dns_get_record($AccountName."domain.com", DNS_CNAME);
echo '<h1>'.$CName.'<h1>';
}
echo'<form action="index.php" method="POST" ">
<input type="text" name="ClientDomain">
<input type="submit" name="DomainSubmit">
</form>'
?>
Try to add else so it will be displayed one or another stuff
<?php
if(isset($_POST['DomainSubmit']) && isset($_POST['ClientDomain'])){
$AccountName = $_POST['ClientDomain'];
$CName = dns_get_record($AccountName."domain.com", DNS_CNAME);
echo '<h1>'.$CName.'<h1>';
} else {
echo'<form action="index.php" method="POST" ">
<input type="text" name="ClientDomain">
<input type="submit" name="DomainSubmit">
</form>'
}
?>
Edit:
You forgot to properly write array POST (missing quotes)
$_POST[DomainSubmit]
And it should be
$_POST['DomainSubmit']
I need to check whether checkbox is checked or not. Normally I would do it like this:
<?php
$checked = isset($_POST['checkbox']);
?>
But I don't know what is the name. More at screenshot (I'm using Laravel 4).
Screenshot
You simply can't. The data for unchecked checkboxes are not send to the server.
You could do a workaround with javascript where the JS appends some hidden fields before submit with the nonchecked boxes
Supposedly you should know what the list of checkboxes is/was that you asked the user to check. Checked checkboxes are submitted to the server, unchecked ones aren't. You can calculate the difference between these two lists.
if you are using jquery, and you know the id of the checkbox,
then, you can detect with following code:
var isChecked = $("#cbId").is(":checked");
<?php
if ( ! isset($_POST['checkbox_name']))
{
"Not checked";
}
?>
If checkbox didn't check - you will not have this variable in $_REQUEST.
<form action="">
<input type="checkbox" name="ch1"/>
<input type="checkbox" name="ch2"/>
<input type="checkbox" checked="checked" name="ch3"/>
<input type="submit" name="Post" value="Post">
</form>
When you click on "Post". In backend you'll see:
<?php
if(isset($_REQUEST['ch1']))
echo 'ch1 is checked!';
if(isset($_REQUEST['ch2']))
echo 'ch2 is checked!';
if(isset($_REQUEST['ch3']))
echo 'ch3 is checked!';
?>
In my case you'll see: "ch3 is checked!".
I have a while loop generating information with a checkbox, I would like to update the database with the new "completed" value. How can I select the specific checkbox that is generated. Please help with showing me how I can grab the specific value of a checkbox and the task_name.
Thanks, Ryan
while ($row = mysql_fetch_array($query)){
$task_name = $row['task_name'] ;
$task_description = $row['task_description'];
$task_completed = $row['completed'];
$tasks .= '<div id="tasksBody">
<form action="" method="post">Completed? <input name="completed" type="checkbox" '.
($task_completed == 1?'checked="checked"':'').
' /><input type="submit" value="Update"><br /><br />
<b>'.$task_name.'</b><br /><br />'.$task_description.'<hr><br /></form></div>';
}
}
echo $tasks;
You need to name your input with something unique for the row, such as the task_name, or better, a database record ID.
Then when the user submits the form, you will use $_POST["YourTaskNameOrIDHere"] to check the value.
What you have currently calls all the check boxes the same thing.
EDIT: I'm sorry, you're isolating all of these in their own forms, I just realized that.
What you can add is an <input type="hidden" value="$task_name" name="TaskName" /> to the form, so you can look what the checkbox is corresponding to. Then, when the user submits the form, use $_POST["TaskName"] to find out the name of the task.
Add a hidden field to each of your forms containing the task_id
<form action="" method="post">
Completed?
<input name="completed" type="checkbox" <?=($task_completed == 1?'checked="checked"':'')?> value="1" />
...
<input name="task_id" value="<?=$task_id"?> type="hidden" />**strong text**
</form>
After submit:
if (isset($_POST['task_id']) { // form has been submitted
$task_id = $_POST['task_id'];
$completed = $_POST['completed'];
$sql = "UPDATE task SET task_completed=$completed WHERE task_id=$task_id LIMIT 1";
// code for updating database
// better use PDO or mysqli-* instead of old and deprecated mysql_*
}