如何将PHP数组值指定为逗号分隔的字符串

我有两个mysql表，sla和department，我正在创建sla，在创建sla的同时，可以选择多个部门

当用户提交SLA表单时，部门表中受尊敬的部门会获取SLA id

但在添加之后，我需要将添加的SLA与更新的部门连接起来

下面是我的控制器代码

if ($this->form_validation->run() !== FALSE) {
    $result = $this->model_admin->updatesla($sla_data); // Add SLA
    if (!$result) {
        $lastentry = $this->model_admin->LastEntrysla(); //Get last added SLA
        foreach($this->input->post('sla_department') as $department_id) {
            $sladepartment_data = array( 'sla_id' => $lastentry->sla_id);
            $this->model_admin->updatedepartmentsla($department_id,$sladepartment_data);//Update SLA to respective multiselected department
        }
        //Now getadded SLA with join of department
        foreach($this->model_admin->getaddedsla($lastentry->sla_id) as $returnsla) {
            $returnslajson = array(  'sla_id' => $returnsla->sla_id,
                'sla_name' => $returnsla->sla_name,
                'sla_days' => $returnsla->sla_days.'D '.$returnsla->sla_hours.'H '.$returnsla->sla_minutes.'M',
                'department_name' => $returnsla->department_name); //Here this SLA is assign to department X and Y, i want department_name = X, y
        }
        echo json_encode($returnslajson);
    }
}

我在每行都添加了一些注释

当SLA增加时，我只得到一个部门的输出

department_name: "X"
sla_days: "1D 1H 1M"
sla_id: "38"
sla_name: "qweqweq"

我怎样才能把这两个部门的名称都命名为X，Y

下面是数组的输出，$this->model_admin->getaddedsla（$lastentry->sla_id）

提前感谢

试试-

    //Now getadded SLA with join of department
    $department_name = $name = $days = $id = array();
    foreach($this->model_admin->getaddedsla($lastentry->sla_id) as $returnsla) {
        $name[] = $returnsla->sla_name;
        $id[] = $returnsla->sla_id;
        $days[] = $returnsla->sla_days.'D '.$returnsla->sla_hours.'H '.$returnsla->sla_minutes.'M';
        $department_name[] = $returnsla->department_name;
    }
    $returnslajson['department_name'] = implode(',', $department_name);
    $returnslajson['sla_id'] = implode(',', $id);
    $returnslajson['sla_name'] = implode(',', $name);
    $returnslajson['sla_days'] = implode(',', $days);
    echo json_encode($returnslajson);

试试-

    //Now getadded SLA with join of department
    $department_name = $name = $days = $id = array();
    foreach($this->model_admin->getaddedsla($lastentry->sla_id) as $returnsla) {
        $name[] = $returnsla->sla_name;
        $id[] = $returnsla->sla_id;
        $days[] = $returnsla->sla_days.'D '.$returnsla->sla_hours.'H '.$returnsla->sla_minutes.'M';
        $department_name[] = $returnsla->department_name;
    }
    $returnslajson['department_name'] = implode(',', $department_name);
    $returnslajson['sla_id'] = implode(',', $id);
    $returnslajson['sla_name'] = implode(',', $name);
    $returnslajson['sla_days'] = implode(',', $days);
    echo json_encode($returnslajson);

---

你想要什么输出？@sgtBOSE，我想要我的json输出-部门名称：“X，Y”你想要什么输出？@sgtBOSE，我想要我的json输出-部门名称：“X，Y”我需要sla天：“1D 1H 1M”sla天号：“38”sla天号：“qweqweq”也一样，我假设上面的一个将只回应X，我也需要sla天：“1D 1H 1M”sla天号：“38”sla天号：“qweqweq”，我假设上面的一个只会回显X，Y