PHP 将数据库对象的数组返回给 AJAX Yii2

Question

因此，当我尝试将数据库对象数组返回到我的 AJAX 并解析它时，它会变成一个字符数组。我的意思是我的json_encode结果是[{'id':38, 'first_name':jana}]，当尝试将其解析为 ajax 中的数组时，发生的事情是和字符数组 - ['[', '{', ''']等。这是我的 ajax：

function searchInput() {
    var $content = $('.jobboard-quick-search-form').serialize();
    $.ajax({
        url : '/admin/site/search',
        method : "GET",
        data : $content,
        success : function ( data ) {
            var arr = JSON.parse(data);
            console.log(arr);
        }
    });
}

和我的行动：

public function actionSearch()
    {
        $lang = frontendmodelsLang::getCurrent();
        $pageSize = 100;
        if(Yii::$app->request->isAjax)
        {
            $search = Yii::$app->request->get('search-label');
            $town = Yii::$app->request->get('towns-list');
            $startsWith = '%'.$search;
            $between = '%'.$search.'%';
            $endsWith = $search.'%';
            $joinDoctors = "SELECT `doctor`.`id`, `doctorLang`.`first_name`, `doctorLang`.`second_name`, `doctorLang`.`city`, `doctorLang`.`hospital_name`
                            FROM `doctor` LEFT JOIN `doctorLang` ON `doctor`.`id`=`doctorLang`.`doc_id`
                            WHERE `doctorLang`.`city`='$town' 
                            AND `doctorLang`.`language`='$lang->url' 
                            AND `doctor`.`active`=1
                            AND (`doctorLang`.`first_name` LIKE '$startsWith'
                            OR `doctorLang`.`first_name` LIKE '$between'
                            OR `doctorLang`.`first_name` LIKE '$endsWith'
                            OR `doctorLang`.`second_name` LIKE '$startsWith'
                            OR `doctorLang`.`second_name` LIKE '$between'
                            OR `doctorLang`.`second_name` LIKE '$endsWith'
                            OR `doctorLang`.`third_name` LIKE '$startsWith'
                            OR `doctorLang`.`third_name` LIKE '$between'
                            OR `doctorLang`.`third_name` LIKE '$endsWith')";
            $doctor =  Yii::$app->db->createCommand($joinDoctors)->queryAll();

            $joinHospitals = "SELECT `hospital`.`id`, `hospitalLang`.`title`, `hospitalLang`.`address`, `hospitalLang`.`description`
                            FROM `hospital` LEFT JOIN `hospitalLang` ON `hospital`.`id`=`hospitalLang`.`hospital_id`
                            WHERE `hospitalLang`.`city`='$town' 
                            AND `hospitalLang`.`language`='$lang->url' 
                            AND `hospital`.`active`=1
                            AND (`hospitalLang`.`title` LIKE '$startsWith'
                            OR `hospitalLang`.`title` LIKE '$between'
                            OR `hospitalLang`.`title` LIKE '$endsWith')";

            return json_encode($doctor);
        }
    }

提前谢谢你！

Answer 1

首先添加 ajax：

dataType: 'json'

$.ajax({
            url:<code>,
            type: <code>,
            dataType: 'json',
            data:<code>
        })

接下来尝试使用rest控制器

class DataController extends yiirestController
{
    public function actionSearch()
    {
        [.....]
        return Yii::$app->db->createCommand($joinDoctors)->queryAll();
    }
}