我想创建用户可以将产品添加到数据库的表单。他们必须上传图片并从选择框中选择几个选项。到目前为止,我的控制器和视图工作正常 - 我验证了上传和选择框,设置值以选择框,如果有一些错误。
但是,我希望我的用户能够看到上传的图片,并在必要时进行一些操作(旋转,裁剪 - 我很确定我知道如何做到这一点,所以这不是问题)在提交孔形式之前
所需的步骤看起来像这样:
到目前为止,这是我的代码 控制器:
public function create()
{
$this->load->helper(array('form', 'url'));
$this->load->library('form_validation');
$this->form_validation->set_rules('health', 'Health', 'required');
$this->form_validation->set_rules('manufacturer', 'Manufacturer', 'required');
if (empty($_FILES['product']['name']))
{
$this->form_validation->set_rules('product', 'Document', 'required');
}
$data['manufacturer'] = $this->input->get('manufacturer');
if ($this->form_validation->run() === FALSE)
{
$data['head'] = $this->load->view('templates/head', NULL, TRUE);
$data['navmenu'] = $this->load->view('templates/navmenu', NULL, TRUE);
$this->load->view('products/create', $data);
}
else
{
$user_id = $this->tank_auth->get_user_id();
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['file_name'] = $user_id;
$config['max_size'] = 100;
$config['max_width'] = 1024;
$config['max_height'] = 768;
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload('product'))
{
$error = array('error' => $this->upload->display_errors());
$this->load->view('products/create', $error);
}
else
{
$this->products_model->set_products();
$id = $this->db->insert_id();
$image_data = $this->upload->data();
$image['image_url'] = $image_data['file_name'];
$this->db->where ('product_id', $id);
$this->db->update('products', $image);
$this->load->view('products/success');
}
}
}
并查看:
<html>
<head>
<?=$head?>
</head>
<body>
<?=$navmenu?>
<?php echo form_open_multipart('products/create'); ?>
<input type="file" name="product" size="20" />
<?php echo form_error('product'); ?><br />
<br /><br />
<label for="health">Health</label>
<select name="health">
<option disabled selected value> -- select a health option -- </option>
<option value="new">New</option>
<option value="used">Used</option>
</select>
<?php echo form_error('health'); ?><br />
<label for="manufacturer">Manufacturer</label>
<select name="manufacturer" >
<option disabled selected value> -- select an manufacturer -- </option>
<option value="manufacturer1" <?php echo set_select('manufacturer','manufacturer1', ( !empty($manufacturer) && $manufacturer == "manufacturer1" ? TRUE : FALSE )); ?> >Manufacturer1</option>
<option value="manufacturer2" <?php echo set_select('manufacturer','manufacturer2', ( !empty($manufacturer) && $manufacturer == "manufacturer2" ? TRUE : FALSE )); ?> >Manufacturer2</option>
<option value="manufacturer3" <?php echo set_select('manufacturer','manufacturer3', ( !empty($manufacturer) && $manufacturer == "manufacturer3" ? TRUE : FALSE )); ?> >Manufacturer3</option>
<option value="manufacturer4" <?php echo set_select('manufacturer','manufacturer4', ( !empty($manufacturer) && $manufacturer == "manufacturer4" ? TRUE : FALSE )); ?> >Manufacturer4</option>
</select>
<?php echo form_error('manufacturer'); ?><br />
<input type="submit" name="submit" value="Create product item" />
</form>
抱歉如果代码在某些地方不洁净 - 它正处于开发过程中。我试图将表单拆分为两个但实际上在控制器代码中丢失了。一般的想法,pseidocode - 任何能让我走上正轨的东西都会很棒!
答案 0 :(得分:0)
使用Jquery执行此操作
function ShowImage(input) {
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function (e) {
$('#imgpreview').attr('src', e.target.result);
}
reader.readAsDataURL(input.files[0]);
}
}
$("#imgupload").change(function(){
ShowImage(this);
});
HTML
<form>
<input type='file' id="imgupload" />
<img id="imgpreview" src="#" alt="your image" />
</form>