我现在有一个弹出菜单,可以打开并显示文本。我想要的是能够将 Controller 附加到弹出菜单。例如:我希望显示一个编辑配置文件弹出窗口,其中包含编辑配置文件所需的所有输入。
userprofile_view.php:
<div class="upload">
<?php
$data = array('id' => 'test');
echo form_open('', $data);
echo form_submit('upload', 'Upload');
echo form_close();
?>
</div>
<div id="popupbox">
<center>
<p class="head">Terms and Conditions</p>
<p class="term_full">By entering your email, you allow Musiclear to send you information regarding Musiclear and
related services <br><br> By entering your email, you also understand that you can break our hearts and unsubscribe
at anypoint. <br><br> And feel free kick us if we ever share your email address (you can trust us)</p>
</center>
</div>
这是按钮和弹出表单后面的 html。
main.js:
$( "#test" ).submit(function( event ) {
event.preventDefault();
document.getElementById('popupbox').style.visibility="visible";
});
在按钮上单击“我使表单可见”。
如何在弹出菜单中使用表单来实现此功能?
编辑:
上传 Controller (我想在弹出窗口中显示的内容):
function do_upload_profilepicture()
{
$this->load->model('model_users');
$userID = $this->model_users->getUserID($this->session->userdata('username'));
$config['upload_path'] = './img/profilepictures/';
$config['allowed_types'] = 'jpg|png';
$config['overwrite'] = TRUE;
$config['file_name'] = $userID;
$config['max_size'] = '500';
$config['max_width'] = '1920';
$config['max_height'] = '1028';
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
$this->load->view('upload_profilepic_form', $error);
}
else
{
$upload_data = $this->upload->data();
$resize['image_library'] = 'gd2';
$resize['source_image'] = $upload_data['full_path'];
$resize['maintain_ratio'] = FALSE;
$resize['width'] = 180;
$resize['height'] = 180;
$this->load->library('image_lib', $resize);
$this->image_lib->resize();
$this->image_lib->clear();
$this->model_users->setProfilePic($userID, $upload_data['orig_name']);
$this->create_thumb($upload_data['orig_name']);
redirect('userprofile/home/' . $userID);
}
}
userprofile_view.php(弹出框):
<div id="popupbox">
<?php echo form_open_multipart('upload/do_upload_profilepicture');?>
<input type="file" name="userfile" size="20" />
<br /><br />
<input type="submit" value="upload" />
</form>
</div>
最佳答案
将函数调用放在脚本标签内 -
<div id="popupbox">
<?php
echo '<script type="text/javascript">';
echo "form_open_multipart('upload/do_upload_profilepicture');";
echo '</script>';
?>
<input type="file" name="userfile" size="20" />
<br /><br />
<input type="submit" value="upload" />
</form>
</div>
关于javascript - AJAX - 如何在弹出菜单中运行脚本,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26469996/