Question

我正在为一个网站的搜索栏创建一个页面，但另一个我不断收到无效的参数错误。希望所需的一切都在下面。我正在使用codeigniter。请帮忙。感谢。

//不起作用//

private function getUsers () {
    $search = $this->input->post('search');

    if($search && filter_var($search, FILTER_VALIDATE_EMAIL)) {
        $where = 'WHERE source_adusers.ad_Email="'.$search.'"';
    } else if ($search) {
        $where = "WHERE source_adusers.ad_account='".$search."'";
    } else if (!$search) {
        $where = '';
    }

    $query = $this->db->query('SELECT *, COUNT(rollout_systems.EAM_USER) as systems FROM source_adusers 
        LEFT JOIN rollout_systems ON rollout_systems.EAM_User = source_adusers.ad_account '.$where.' GROUP BY source_adusers.ad_account LIMIT 0,50');

    $users = null;

    foreach ($query->result() as $row) {
        $data['account'] = $row->ad_account;
        $data['email'] = $row->ad_Email;
        $data['name'] = $row->ad_DispName;
        $data['systems'] = $row->systems;

        $users[] = $data;
    }

    if(isset($this->data['users'])) {
        $this->data['users'] = $users;
    } else {
        $this->data['users'] = $users;
    }
}

//是否工作//

private function getSystems () {
    $search = $this->input->post('search');

    if ($search) {
        $where = 'WHERE disc_systempool.ad_name="'.$search.'"';
    } else {
        $where = '';
    }

    $query = $this->db->query('SELECT *, COUNT(rollout_systems.sys_name) as scopes FROM disc_systempool LEFT JOIN rollout_systems 
        ON rollout_systems.sys_name=disc_systempool.ad_name '.$where.' GROUP BY disc_systempool.ad_name LIMIT 0,50');

    foreach ($query->result() as $row) {
        $data['model'] = $row->eam_Model;
        $data['account'] = $row->ad_name;
        $data['city'] = $row->eam_City;
        $data['scopes'] = $row->scopes;
        $data['search'] = $this->input->post('search');

        $systems[] = $data;
    }

    if(isset($this->data['systems'])) {
        $this->data['systems'] = $systems;
    } else {
        $this->data['systems'] = $systems;
    }
}

//我收到错误//

<tbody>
    <? foreach ($users as $user) { ?>                      <---- Line 18
    <tr>
        <td class="center"><?=$user['account']?></td>
        <td><?=$user['name']?></td>
        <td><?=$user['email']?></td>
        <td class="center"><?=$user['systems']?></td>
        <td class="center"><a href="#">View Details</a></td>
    </tr>
    <? } ?>
</tbody>

//错误//

A PHP Error was encountered
Severity: Warning
Message: Invalid argument supplied for foreach()
Filename: headquarters/users.php
Line Number: 18

Answer 1

它表明了正确的事情。我解决了这个问题。我误解了雇用我的人要求我做什么。他说搜索帐户名称。我是搜索显示名称，一直给我错误。

Answer 2

尝试在视图中制作var_dump($data);。如果是空的。只需将其打印出来，即可在控制器中执行相同的操作。如果您在$data数组中看到数据。确保您将$data数组一起发送到视图$this->load->view("view",$data);

除此之外，我可能想知道为什么要对控制器内部的行进行操作，而不是仅仅在视图中使用对象

return $query->result();

在视图中

foreach($users as $user) { print $user->account; }

希望你弄明白。最好的祝福纳斯

制作搜索栏时出现无效的参数错误

2 个答案: