Laravel-用派生值填充选择列表 [英] Laravel - Populating Select List with Derived Values

问题描述

我有一个雄辩的模型(Presenters)，该模型具有三列:id，first_name和last_name.我想做的是填充选择选项，以便id是值，而first_name . last_name是显示的内容.

I have an Eloquent model (Presenters) that has three columns: id, first_name and last_name. What I want to do is populate the select options so that the id is the value, and the first_name . last_name is what is displayed.

例如，如果我有这些主持人:

For instance, if I had these presenters:

id, first_name, last_name
1, Billy, Bob
2, Jose, Cuervo
3, Puff, the Magic Dragon

我想要与此输出类似的东西:

I want something similar to this output:

<select name="presenters">
    <option value="1">Billy Bob</option>
    <option value="2">Jose Cuervo</option>
    <option value="3">Puff the Magic Dragon</option>
</select>

我知道，如果我只关心选择列表中的姓氏或名字，我可以这样做:

I know that if I only care about the last name or first name in the select list, I can do this:

{{ Form::select("presenters", Presenter::lists("first_name", "id"), Input::old("presenters"), array( "class" => "form-control" )) }}

关于在Laravel中填充模型的选择输入，已经有许多问题找到了一个显示如何填充选择输入的选项，其中每个选项中都将显示派生值，而不仅仅是单个列.

There are many questions already regarding populating select inputs for models in Laravel, although I haven't found one yet that shows how to populate a select input where a derived value is to be shown in each of the options rather than just a single column.

我可以使用lists方法填充select元素，还是需要填充自己的值/选项文本数组?

Can I use the lists method to populate the select element, or will I need to populate my own array of values / option texts?

推荐答案

Presenter::select('id', DB::raw('CONCAT(first_name, " ", last_name) AS full_name'))->lists('full_name', 'id');

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