Laravel中如何使用查询结果和LIKE语句再次查询

Question

I am trying to use questions result and get question again if title has some char in it. If I query in condition_question table, I get results as expected.

public function showQuestions($category)
    {
        $myArray = array($category);

        $questions = Question::whereIn('question_id', function ($query) use ($myArray) {
            $query->select('question_id')
                ->from('condition_question')
                ->whereIn('condition_id', $myArray);
        })->orderBy('question_id', 'desc')->paginate(20);

        return QuestionLiteResource::collection($questions);
}

Question: How can I use now $questions result and query again with LIKE statement. So far I tried many thing, for example like this, but something is missing as I am getting errors:

public function showQuestions($category, $queryQuestion)
    {
        $myArray = array($category);

        $chary = $queryQuestion;

        $questions = Question::whereIn('question_id', function ($query) use ($myArray) {
            $query->select('question_id')
                ->from('condition_question')
                ->whereIn('condition_id', $myArray);
        })->get();

        $results = $questions->where('question_title', 'LIKE', "%{$chary}%")->get();

        return QuestionLiteResource::collection($results->values());
}

I know it is not my best, but need some help...It would be also cool to have paginated result at the end. So, how to get collection of questions from questions table where title has char. Any help would be most welcomed!

Answer 1

You might know that once you call get() function, you got the results and not able to query any further. Maybe this is gonna work:

public function showQuestions($category, $queryQuestion)
    {
        $myArray = array($category);

        $chary = $queryQuestion;

        $questions = Question::whereIn('question_id', function ($query) use ($myArray) {
            $query->select('question_id')
                ->from('condition_question')
                ->whereIn('condition_id', $myArray);
        })
        ->where('question_title', 'LIKE', "%{$chary}%")
        ->get();

        return QuestionLiteResource::collection($questions);
}

Answer 2

Since you have called get() on question query, you get the result as an Laravel Collection .

To filter through collection you can use filter() function.

Example Code

$results = $questions->filter(function($question) use ($chary) {
    return Str::contains($question->question_title, $chary);
});

Answer 3

i think you can use join():

        public function showQuestions($category, $queryQuestion)
        {
            $myArray = array($category);
            $chary = $queryQuestion;

            $query = Question::getModel()->newQuery();
            $questions = $query
                 ->join('condition_question', function (Builder $join) use ($myArray) {
                   $join->on('questions.question_id', '=', 'condition_question.question_id');
                   $join->whereIn('condition_question.condition_id', $myArray);
                 }) 
                 ->where('questions.question_title', 'like', $chary)
                 ->orderBy('questions.question_id', 'desc')
                 ->paginate(20)

            return QuestionLiteResource::collection($questions);
        }