Question

请问，当我想加入并处理2个查询时，有人可以帮我修复将数据显示为数组的代码吗？

这是我的控制器：

<?php
defined('TYPO3_MODE') || die('Access denied.');

call_user_func(
    function ($extKey) {
        \TYPO3\CMS\Core\Utility\ExtensionManagementUtility::addTypoScriptSetup('<INCLUDE_TYPOSCRIPT: source="FILE:EXT:extKeyHere/Configuration/TypoScript/setup.ts">');
        \TYPO3\CMS\Core\Utility\ExtensionManagementUtility::addTypoScriptConstants('<INCLUDE_TYPOSCRIPT: source="FILE:EXT:extKeyHere/Configuration/TypoScript/constants.ts">');

        \TYPO3\CMS\Extbase\Utility\ExtensionUtility::configurePlugin(
            'TYPO3.' . $extKey,
            'Integration',
            array(
                'Integration' => 'integration',
            ),
            // non-cacheable actions
            array(
                'Integration' => 'integration',
            )
        );
    },
    $_EXTKEY
);

$signalSlotDispatcher = \TYPO3\CMS\Core\Utility\GeneralUtility::makeInstance('TYPO3\CMS\Extbase\SignalSlot\Dispatcher');
$signalSlotDispatcher->connect(
    'In2code\Powermail\Controller\FormController', // namespace of Class: In2code\Powermail\Controller
    'createActionAfterMailDbSaved',
    'Typo3\extName\Controller\IntegrationController', //namespace of Class: Typo3\extName\Controller
    'integration',
    FALSE);
?>

这是我的模特：

public function pegawai()
{
    $data['pegawai'] = $this->Ajax_model->pegawai_get();
    $this->load->view('ajax/pegawai', $data);
}

但是我的数据看起来像这样：

//get modal search data
public function pegawai_get($pegawai_id = TRUE)
{
    $this->db->select('m_riwayat_kepangkatan.*, m_pegawai.nip, m_pegawai.nama, r_golongan.golongan, r_golongan.jenis_pangkat');
    $this->db->from('m_riwayat_kepangkatan');
    $this->db->join('m_pegawai', 'm_pegawai.id = m_riwayat_kepangkatan.pegawai_id');
    $this->db->join('r_golongan', 'r_golongan.id = m_riwayat_kepangkatan.golongan_id');
    $this->db->order_by("pegawai_id", "asc");

    $query1 = $this->db->get();
    return $query1->result_array();

    $this->db->select('m_riwayat_jabatan.*, m_pegawai.nip, m_pegawai.nama, m_opd.nama_opd, m_unit_kerja.unit_kerja');
    $this->db->from('m_riwayat_jabatan');
    $this->db->where("m_pegawai.id ='$pegawai_id'");
    $this->db->join('m_pegawai', 'm_pegawai.id = m_riwayat_jabatan.pegawai_id');
    $this->db->join('m_opd', 'm_opd.id = m_riwayat_jabatan.opd_id');
    $this->db->join('m_unit_kerja', 'm_unit_kerja.id = m_riwayat_jabatan.unit_kerja_id');
    $this->db->order_by("pegawai_id", "asc");

    $query2 = $this->db->get();
    foreach ($query2->result_array() as $row2){
        //$pegawai_id    = $row2["pegawai_id"];
        //$nip           = $row2["nip"];
        //$nama          = $row2["nama"];
        $nama_opd      = $row2["nama_opd"];
        $unit_kerja    = $row2["unit_kerja"];
    }
    return $query2->result_array();
}

它仅显示为query1的值。我也想为query2内部数据赋值，例如“ nama_opd”和“ unit_kerja”。看起来我缺少了一些东西。

Answer 1

看看return $query1->result_array();。您已在第二次查询之前将其返回

将数据作为数组联接代码初始化器传递

1 个答案: