使用全文搜索以通过laravel查询构建器查找部分单词

Question

I would like to write search query that uses supplied phrase to search a table, return the value that matches any part of the phrase. this code is working, but results are nothing. For example, table have 'abcde', and I searched 'bcd', result is nothing.

protected function fullTextWildcards($term)
{
    return str_replace(' ', '*', $term) . '*';
}

public function index(Request $request, $slug = null)
{
    $query = $slug
        ? \App\Tag::whereSlug($slug)->firstOrFail()->articles()
        : new \App\Article;
    if ($keyword = request()->input('q')) {
        $raw = 'MATCH(title,content) AGAINST(? IN BOOLEAN MODE)';
        $query = $query->whereRaw($raw, [$this->fullTextWildcards($keyword)]);
    }

    $articles=$query->latest()->paginate(10);
    return view('articles.index',compact('articles'));
}

How to chain everything together so I achieve the desired result?
Thanks in advance.

Answer 1

You can just use like in your query to get any matches in a given column. Try this:

public function index(Request $request, $slug = null)
{
  $filter = ['keyword' => $request->q , 'slug' => $slug];

  $articles = \App\Tag::where(function ($q) use ($filter){

    //you can leave this I just put the so the code seems clean
    if ($filter['slug'] !== null) {
      $q->whereSlug($slug);
    }

    if ($filter['keyword']) {

      $keyword  = $this->fullTextWildcards($filter['keyword']);

      //you can use **like** this
      $q->where('title', 'like', "%".$keyword."%")->orWhere('content', 'like',  "%".$keyword."%");
    }

  })->latest()->paginate(10);


  return view('articles.index',compact('articles'));

}

Hope this helps