Issue
How to add 'active' on label class like after checking value from database with laravel {{$user->experience}} . if value is Entry-Level add 'active' on label class .
Here is my HTML Content
<div class="btn-group" data-toggle="buttons">
<label class="btn btn-success active" for="option1">
<input type="radio" name="experience" id="option1" value="Student">Student
<span class="glyphicon glyphicon-ok"></span>
</label>
<label class="btn btn-primary" for="option2">
<input type="radio" name="experience" id="option2" value="Entry-Level">Entry-Level
<span class="glyphicon glyphicon-ok"></span>
</label>
<label class="btn btn-info" for="option3">
<input type="radio" name="experience" id="option3" value="Mid-Level">Mid-Level
<span class="glyphicon glyphicon-ok"></span>
</label>
<label class="btn btn-default" for="option4">
<input type="radio" name="experience" id="option4" value="Senior-Level">Senior-Level
<span class="glyphicon glyphicon-ok"></span>
</label>
<label class="btn btn-warning" for="option5">
<input type="radio" name="experience" id="option5" value="Manager">Manager
<span class="glyphicon glyphicon-ok"></span>
</label>
<label class="btn btn-danger" for="option6">
<input type="radio" name="experience" id="option6" value="Director" >Director
<span class="glyphicon glyphicon-ok"></span>
</label>
</div>
Here is my css
.btn span.glyphicon {
opacity: 0;
}
.btn.active span.glyphicon {
opacity: 1;
}
My laravel value fetching
{{$user->experience}}
Mysql 'user' Table
- id name experience
- 1 john Entry-Level
Website : flutch
Solution
You can use ternary operators
to achieve your goal.
<label
class="btn btn-success {{$user->experience == 'Entry-Level' ? 'active' : ''}}"
for="option1"
>
Answered By - Tharaka Dilshan
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