我有一个代码,其中包含上传到文件夹中的文件数量,我希望将这些文件的名称、大小和 URL 存入数据库,但我的 Controller 无法正常工作。 (我正在使用 CakePHP 框架)。我想将这些文件数据添加到数据库中(所有文件数据),但出现错误。
错误:
Notice (8): Undefined index: tmp_namā€ā€‹ā€ā€‹e [APP\Controller\UploadFilesController.php, line 24]
我的 Controller
public function uploadFile() {
$filename = '';
if ($this->request->is('post')) { // checks for the post values
$uploadData = $this->request->data;
//print_r($this->request->data); die;
foreach($uploadData as $file){
$filename = basename($file['name']); // gets the base name of the uploaded file
$uploadFolder = WWW_ROOT. 'files'; // path where the uploaded file has to be saved
$filename = $filename; // adding time stamp for the uploaded image for uniqueness
$uploadPath = $uploadFolder . DS . $filename;
if( !file_exists($uploadFolder) ){
mkdir($uploadFolder); // creates folder if not found
}
if (!move_uploaded_file($file['tmp_name'], $uploadPath)) {
return false;
}
echo "Sa sisestasid faili: $filename<br>";
}
foreach($this->request->data['UploadFile']['file_upload'] as $file){
if (!empty($this->request->data) && is_uploaded_file($this->request->data['UploadFile']['file_upload']['tmp_name'])) { //THIS IS LINE 24
$fileData = fread(fopen($this->request->data['UploadFile']['file_upload']['tmp_name'], "r"), $this->request->data['UploadFile']['file_upload']['size']);
$this->request->data['UploadFile']['name'] = $this->request->data['UploadFile']['file_upload']['name'];
$this->request->data['UploadFile']['size'] = $this->request->data['UploadFile']['file_upload']['size'];
$this->request->data['UploadFile']['URL'] = $this->request->data['UploadFile']['file_upload']['tmp_name'];
$this->request->data['UploadFile']['data'] = $fileData;
$this->UploadFile->create();
$this->UploadFile->save($this->request->data);
}
}
}
}
}
这是我的 View 文件:
<?php
echo $this->Form->create('uploadFile', array( 'type' => 'file'));
?>
<div class="input_fields_wrap">
<label for="uploadFilefiles"></label>
<input type="file" name="data[]" id="uploadFilefiles">
</div>
<button type="button" class="add_field_button">+</button> <br><br>
<form name="frm1" method="post" onsubmit="return greeting()">
<input type="submit" value="Submit">
</form>
<?php
echo $this->Html->script('addFile');
如果有必要我也可以添加AddFile
脚本。
这是我的表 (upload_files) 结构:
最佳答案
我认为这是最好的方法:
public function uploadFile() {
$filename = '';
if ($this->request->is('post')) { // checks for the post values
$uploadData = $this->request->data ['UploadFile']['file_upload'];
//print_r($this->request->data); die;
foreach($uploadData as $file){
$filename = basename($file['name']); // gets the base name of the uploaded file
$uploadFolder = WWW_ROOT. 'files'; // path where the uploaded file has to be saved
$filename = $filename; // adding time stamp for the uploaded image for uniqueness
$uploadPath = $uploadFolder . DS . $filename;
if( !file_exists($uploadFolder) ){
mkdir($uploadFolder); // creates folder if not found
}
if (!move_uploaded_file($file['tmp_name'], $uploadPath)) {
return false;
}
echo "Sa sisestasid faili: $filename<br>";
$this->request->data['UploadFile']['name'] = $file['name'];
$this->request->data['UploadFile']['size'] = $file['size'];
$this->request->data['UploadFile']['URL'] = $file['tmp_name'];
$this->UploadFile->create();
$this->UploadFile->save($this->request->data);
}
}
}
我在我的主机上试过这个并且它工作正常。
关于php - 上传文件中的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27856966/