Laravel（初学者） - 显示下拉列表中选定行的数据

Question

This is Controller.php

public function dropdown($id)
  {

    $stations = Station::lists('station_name','id');
    return view('stations.show')->withStation($station);
    return view('stations.dropdown',compact('stations', 'selectedStation','selectedStation'));

  }

This is the view.php

{{ Form::select('station',$stations,null,['class' => 'form-control']) }} 
            <br>
<?php
        if(isset($_GET['submit'])){


            $selectedStation = $_GET['station'];
            echo $selectedStation;echo "<br>";
        }
        else
        {
          echo "not working";
        }
    ?>

corresponding Database table.

This gives me a drop-down like in the below image.

When a station is selected it displays only "id" of the selected station. How to retrieve other columns of the selected station ?.

Answer 1

Many ways that you can tackle this and I have also made some assumptions. Your list (it would seem) only contains the name & the id.

I assume that you don't want to reload the page when you get the information for the specific item that you have chosen in the dropdown.

Here are the first 2 options that come to mind.

1.) AJAX - Use javascript framework like jQuery and monitor change() for the select, on change get the id and fire it to another route in Laravel which takes the ID and returns some more information in JSON. You can then use JS to display this to your users as you please.

2.) Preloaded JSON - Write new DB Query that gets all the information that you want and store it in JSON format inside some <script> tags. It wont be viewable to the user. Then you can use jQuery or other JS framework to grab that data again on change()

Part 2 will be faster, and if you have got 100,000s records i would advise that for perf.

Apologies if I totally misunderstood the question, it seems like you need a JS solution rather than Laravel/PHP solution. (Without a bad UX) if you did wan't more data you would need to post it and the fetch data from DB on new request.