Reputation: 143
question above.
My code:
<form action="" method="post" class="text-center">
<select class="form-control">
<option value="work1">1 hours</option>
<option value="work2">2 hours</option>
<option value="work3">3 hours</option>
<option value="work4">4 hours</option>
<option value="work5">5 hours</option>
<option value="work6">6 hours</option>
</select>
<input type="submit" class="btn btn-success btn-block" value="Go!">
</form>
I would like to output the selected variable in the next step.
How to do it?
Thanks in advance.
Upvotes: 1
Views: 107
Reputation: 1190
Just check your $_POST.
<?php
if(isset($_POST['form-control'])){
$yourvar=$_POST['form-control'];
}
?>
<form action="" method="post" class="text-center">
<select class="form-control" name="form-control">
<option value="work1">1 hours</option>
<option value="work2">2 hours/option>
<option value="work3">3 hours</option>
<option value="work4">4 hours</option>
<option value="work5">5 hours</option>
<option value="work6">6 hours</option>
</select>
<input type="submit" class="btn btn-success btn-block" value="Go!">
</form>
Don't forget to validate the content of $_POST. You can not trust that it will always bring what you think is there.
Upvotes: 1
Reputation: 11987
First give name to your select box. Then use $_POST
in same file or in other file as per your requirement.
<?php
if(isset($_POST['abc'])){
$selected=$_POST['abc'];
}
?>
<form action="" method="post" class="text-center">
<select class="form-control" name="abc">
<option value="work1">1 hours</option>
<option value="work2">2 hours/option>
<option value="work3">3 hours</option>
<option value="work4">4 hours</option>
<option value="work5">5 hours</option>
<option value="work6">6 hours</option>
</select>
<input type="submit" class="btn btn-success btn-block" value="Go!">
</form>
Upvotes: 1