Question

我正在使用cakephp 3.3 我在控制器中

    $employee = TableRegistry::get('employees');
    $allNames = $employee->find('list', array('employee_name' => array('employee_name') ) );
    $allNames = $allNames->toArray();
    $this->set('name', $allNames);

并在我的模板中

    <?= $this->Form->input('employee', array('type'=>'select','label' => false,
        'options' => $name,'value'=>$name));?>

唯一检索和显示的是我在数据库中的条目数，$ name只包含一个数组（1,2,3,4），它应该是实际的人名。

Answer 1

您是否阅读过documentation find('list')如何运作？'

$allNames = $employee->find('list', [
    'keyField' =>     'employee_name'
    'valueField' => 'employee_name'
]);
$this->set('name', $allNames);

<?= $this->Form->input('employee', ['type'=>'select','label' => false,
    'options' => $name]);?>

Answer 2

您可以在列表查询

中按keyField和valueField映射ID和值控制器中的

 $employee = TableRegistry::get('employees');
 $allNames = $employee->find('list', array('valueField' =>employee_name','keyField'=>'employee_id' )); // correct employee_id with your table id field.
 $this->set('name', $allNames);

在您的视图中

 <?= $this->Form->input('employee', array('type'=>'select','label' => false,
        'options' => $name));?>

我试图从数据库中检索带有cakePHP的选择框但是数据没有被正确检索

2 个答案: