Question

我试图用ajax将数据添加到数据库，但这是我第一次尝试使用ajax，这就是为什么我无法解决这个问题。

我有add（）动作，用于检查是否发送了表单。如果是，则向数据库添加记录。

public function add()
{
    if ($this->request->is('ajax')) {
        Configure::write('debug', 0);
    }
    $this->loadComponent('RequestHandler');
    $this->loadModel('Galleries');

    $entity = $this->Galleries->newEntity();
    if ($this->request->is('post'))
    {
        /*$photos = $this->request->data['multiple_photos'];
        for ($i = 0; $i < count($photos); $i++) {
            $photo = [
                'name' => $this->request->data['multiple_photos'][$i]['name'],
                'type' => $this->request->data['multiple_photos'][$i]['type'],
                'tmp_name' => $this->request->data['multiple_photos'][$i]['tmp_name'],
                'error' => $this->request->data['multiple_photos'][$i]['error'],
                'size' => $this->request->data['multiple_photos'][$i]['size']
            ];
            echo "<pre>"; print_r($photo); echo "</pre>";
        }*/

        $data = [
            'id' => '',
            'thumb' => 'thuburl',
            'highres' => 'highresurl',
            'gal_id' => 1
        ];

        $entity = $this->Galleries->patchEntity($entity, $data);

        $this->Galleries->save($entity);
    }

    $this->set(compact('entity'));
    $this->viewBuilder()->layout(false);
    $this->render();
}

在我看来：

<body>
<?php

    echo $this->Form->create($entity, ['type' => 'file', 'id' => 'formUpload', 'action' => 'add']);
    echo $this->Form->hidden('id', ['value' => '']);
    echo $this->Form->input('multiple_photos[]', ['type' => 'file', 'multiple' => 'true', 'label' => false]);
    echo $this->Form->hidden('thumb', ['value' => 'thumburl']);
    echo $this->Form->hidden('highres', ['value' => 'highresurl']);
    echo $this->Form->hidden('gal_id', ['value' => '1']);
    echo $this->Form->button('Wyslij', ['type' => 'submit']);
    echo $this->Form->end();

?>

<script>  
        $('#formUpload').submit(function() {

            var formData = $(this).serialize();
            var formUrl = $(this).attr('action');

            $.ajax({
                type: "POST",
                url: formUrl,
                data: formData,
                success: function(data,textStatus,xhr) {
                    alert(data);
                },
                error: function(xhr,textStatus,error) {
                    alert(textStatus);
                }
            });
            return false;
        });
</script>

</body>

结果是什么？它添加了一条记录，因此Ajax脚本可以正常工作，但我得到了错误＆＃34;警报。

如何修改动作或ajax脚本以使其正常工作？

提前致谢：）

CakePHP 3.1 - 获取Ajax

0 个答案: