从数据库输出的数据进入模板Laravel

Question

有一个后端（Laravel），它创建页面并将其存储在数据库中。文章标题，文章文字，链接，图像等

问题如何将数据从数据库输出到Laravel模板，即文章标题和文本？

用于创建和存储文章的后端：

public function store(Request $request)
{
    $this->validate($request, [
        'channel_name' => ['required', 'exists:channels,name', new NotBannedFromChannel()],
        'type'         => 'required|in:link,img,text,gif',
        'title'        => 'required|string|between:7,150',
        'url'          => ['required_if:type,link', 'url', 'active_url', new NotBlockedDomain()],
        'photos_id'    => 'required_if:type,img|array|max:20',
        'gif_id'       => 'required_if:type,gif|integer',
    ]);

    // Make sure user is not overdoing it.
    if ($this->tooEarlyToCreate(2)) {
        return res(429, "Looks like you're over doing it. You can't submit more than 2 posts per minute.");
    }

    switch ($request->type) {
        case 'link':
            $data = $this->linkSubmission($request);
            break;

        case 'img':
            $data = $this->imgSubmission($request);
            break;

        case 'gif':
            $data = $this->gifSubmission($request);
            break;

        case 'text':
            $data = $this->textSubmission($request);
            break;
    }

    $channel = $this->getChannelByName($request->channel_name);

    try {
        $submission = Submission::create([
            'title'        => $request->title,
            'slug'         => $slug = $this->slug($request->title),
            'url'          => $request->type === 'link' ? $request->url : config('app.url').'/c/'.$channel->name.'/'.$slug,
            'domain'       => $request->type === 'link' ? domain($request->url) : null,
            'type'         => $request->type,
            'channel_name' => $request->channel_name,
            'channel_id'   => $channel->id,
            'nsfw'         => $request->input('nsfw', 0),
            'rate'         => firstRate(),
            'user_id'      => Auth::id(),
            'data'         => $data,
        ]);

        event(new SubmissionWasCreated($submission));
    } catch (\Exception $exception) {
        app('sentry')->captureException($exception);

        return res(500, 'Ooops, something went wrong.');
    }

    if ($request->type === 'img' || $request->type === 'gif') {
        $this->updateSubmissionIdForUploadedFile($request, $submission->id);
    }

    $this->firstLike($submission->id);

    return new SubmissionResource(
        Submission::find($submission->id)
    );
}