I'm new in cakephp and i'm trying to create a simple file upload with cakephp 2.3 here is my controller
public function add() {
if ($this->request->is('post')) {
$this->Post->create();
$filename = WWW_ROOT. DS . 'documents'.DS.$this->data['posts']['doc_file']['name'];
move_uploaded_file($this->data['posts']['doc_file']['tmp_name'],$filename);
if ($this->Post->save($this->request->data)) {
$this->Session->setFlash('Your post has been saved.');
$this->redirect(array('action' => 'index'));
} else {
$this->Session->setFlash('Unable to add your post.');
}
}
}
and my add.ctp
echo $this->Form->create('Post');
echo $this->Form->input('firstname');
echo $this->Form->input('lastname');
echo $this->Form->input('keywords');
echo $this->Form->create('Post', array( 'type' => 'file'));
echo $this->Form->input('doc_file',array( 'type' => 'file'));
echo $this->Form->end('Submit')
it saves firstname, lastname, keywords, and the name of the file in DB , but the file which i want to save in app/webroot/documents is not saving , can anyone help ? Thanks
Update
thaJeztah i did as u said but it gives some errors here is controller if i'm not wrong
public function add() {
if ($this->request->is('post')) {
$this->Post->create();
$filename = WWW_ROOT. DS . 'documents'.DS.$this->request->data['Post']['doc_file']['name'];
move_uploaded_file($this->data['posts']['doc_file']['tmp_name'],$filename);
if ($this->Post->save($this->request->data)) {
$this->Session->setFlash('Your post has been saved.');
$this->redirect(array('action' => 'index'));
} else {
$this->Session->setFlash('Unable to add your post.');
}
}
}
and my add.ctp
echo $this->Form->create('Post', array( 'type' => 'file'));
echo $this->Form->input('firstname'); echo $this->Form->input('lastname');
echo $this->Form->input('keywords');
echo $this->Form->input('doc_file',array( 'type' => 'file'));
echo $this->Form->end('Submit')
and the errors are
Notice (8): Array to string conversion [CORE\Cake\Model\Datasource\DboSource.php, line 1005]
Database Error Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Array' in 'field list'
SQL Query: INSERT INTO first.posts (firstname, lastname, keywords, doc_file) VALUES ('dfg', 'cbhcfb', 'dfdbd', Array)
and Victor i did your version too , it doesnt work too .
You seem to be using the wrong 'key' to access the posted data;
$this->data['posts'][....
Should match the 'alias' of you Model; singular and a captial first letter
$this->data['Post'][....
Also, $this->data
is a wrapper for $this->request->data
for backwards compatibility, so it's better to use this;
$this->request->data['Post'][...
To check the content of the posted data and understand how it's structured, you may debug it using this;
debug($this->request);
Just be sure to enable debugging, by setting debug
to 1
or 2
inside app/Config/core.php
I just noticed you're also creating multiple (nested) forms in your code;
echo $this->Form->input('keywords');
// This creates ANOTHER form INSIDE the previous one!
echo $this->Form->create('Post', array( 'type' => 'file'));
echo $this->Form->input('doc_file',array( 'type' => 'file'));
Nesting forms will never work, remove that line and add the 'type => file' to the first Form->create()
The "Array to string conversion" problem is cause by the fact that you're trying to directly use the data of 'doc_file' for your database. Because this is a file-upload field, 'doc_file' will contain an Array of data ('name', 'tmp_name' etc.).
For your database, you only need the 'name' of that array so you need to modify the data before saving it to your database.
For example this way;
// Initialize filename-variable
$filename = null;
if (
!empty($this->request->data['Post']['doc_file']['tmp_name'])
&& is_uploaded_file($this->request->data['Post']['doc_file']['tmp_name'])
) {
// Strip path information
$filename = basename($this->request->data['Post']['doc_file']['name']);
move_uploaded_file(
$this->data['Post']['doc_file']['tmp_name'],
WWW_ROOT . DS . 'documents' . DS . $filename
);
}
// Set the file-name only to save in the database
$this->data['Post']['doc_file'] = $filename;