如何在sections.php上返回获取值

Question

DBOperations.php

 public function studentSection($uname) {
        $this->sections = array();
        $stmt = $this->conn->prepare("SELECT students.id, students.lastname, sections.name AS section_name, years.name AS year_level, academic_years.start, academic_years.end FROM students INNER JOIN section_student ON students.id = section_student.student_id INNER JOIN sections ON section_student.section_id = sections.id INNER JOIN years ON sections.year_id = years.id INNER JOIN academic_years ON sections.academic_year_id = academic_years.id WHERE students.s_id = ?");

        $stmt->bind_param("s", $uname);
        $stmt->execute();
        $result = $stmt->get_result();
        $num_of_rows = $result->num_rows;
        while ($row = $result->fetch_assoc()) {
            return $this->sections[] = array(
                'id' => $row['id'],
                'lastname' => $row['lastname'], 
                'section' => $row['section_name'],
                'year_level' => $row['year_level'],
                'school_year' => $row['start']."-".$row['end']);
        }
        $stmt->free_result();
        $stmt->close();
    }

section.php

if($db->studentLogin($uname)){
            $sections = $db->studentSection($uname);
            $response['error'] = false; 

        $response['id'] = $sections['id'];
        $response['lastname'] = $sections['lastname'];
        $response['section'] = $sections['section'];
        $response['year_level'] = $sections['year_level'];
        $response['school_year'] = $sections['school_year'];

    }else{
        $response['error'] = true; 
        $response['message'] = "No Data";          
    }

返回值Illegal string offset时出错。 它不会显示所有数据。 输出为{"error":false,"id":"[","lastname":"[","section":"[","year_level":"[","school_year":"["}

当我尝试return内的$this->sections时，它会显示等于1的所有数据，然后将下面的return更改为echo json_encode($this->sections)有用。但它只显示一行。它应该显示等于1的所有数据行。