我正在尝试将两个不同字段上的两种不同类型的文件发布到数据库中。我一直在寻找整个Stack Overvlow的答案,但仍然找不到。
这是我的控制器:
public function insertLamaran()
{
$pas_foto = ($_FILES['pas_foto']['name']);
$cv = ($_FILES['cv']['name']);
if ($pas_foto !== "")
{
$config['upload_path'] = './uploads/';
$config['log_threshold'] = 1;
$config['allowed_types'] = 'jpg|png|jpeg|gif';
$config['max_size'] = '100000'; // 0 = no file size limit
$config['file_name']='smallImage.jpg';
$config['overwrite'] = false;
$this->load->library('upload', $config);
$this->upload->do_upload('pas_foto');
$upload_data = $this->upload->data();
$data['pas_foto'] = $upload_data['file_name'];
}
if ($cv !== "")
{
$config['file_name']='bigImage.pdf';
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'pdf';
$config['max_size'] = '100000'; // 0 = no file size limit
$config['overwrite'] = false;
$this->load->library('upload', $config);
$this->upload->do_upload('filename1');
$upload_data = $this->upload->data();
$data['cv'] = $upload_data['file_name'];
}
$data['u_id'] = $this->input->post('u_id');
$this->M_upload->insertLamaran($data);
}
这是我的看法:
<form action="<?php echo base_url('Upload/insertLamaran')?>" method="post">
<input hidden type="text" name="u_id" value="<?php echo ucfirst($this->session->userdata('u_id')); ?>"></input><br>
<div class="row wowload fadeInLeftBig">
<div class="col-sm-6 col-sm-offset-3 col-xs-12">
<label for="pasfoto">Pas Foto (.jpg)</label>
<input id="pas_foto" type="file" name="pas_foto" multiple="true">
<label for="CV">CV (.pdf)</label>
<input id="cv" type="file" name="cv" multiple="true">
<button class="btn btn-primary"><i class="fa fa-paper-plane"></i> Send</button>
</div>
</div>
</form
错误说
未定义的索引:pas_foto。文件名:controllers / Upload.php行号:22
有人知道发生了什么吗,我该如何解决?
如我所见,您的表单类型正确,因此答案是pdf文件,但您忘记添加pdf文件名
改变这个
$this->upload->do_upload('filename1');
进入这个
$this->upload->do_upload('cv');
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我来说两句