在laravel中选择特定选项时显示输入

Question

I am using Laravel 5.2 . I have an input to display when a specific option is selected and it works fine.

But I want to use {{ old('field') }} to have the old results.

• When I select Yes , I have to display the input.
• When I select No , I have to hide the input.

When I have validation errors (server side) and I return to the page, then I have the old results (example: value='yes') but the input that was displaying is now hidden and I have to re-click on yes to display it again. I use the change event to display the input.

$("#etat_civil").change(function () {
    if ($(this).val() == "Marié") {
        $("#div_conjoint").show();
    } else {
        $("#div_conjoint").hide();
    }
});

Answer 1

You can check the old('field') value for it is set then you can set your element to be visible.

For example, you can do something like this.

<input type="text" name="my-field" value="abc" style="display: {{ old('field') ? 'block' : 'none' }}">

By doing this you are checking if old('field') is set then the field/block/div should be visible otherwise keep it hidden.

So in your case when the validation fails and it redirects you back to the form you will have old('field') value set and the element should be visible to you.

Update:

As suggested by other guys you can also do this by putting the JS code on page onload but I don't prefer that way as it sometimes shows a slight delay in loading the element. Not a great deal but I still prefer doing it the other way that I explained above.