database error: A PHP Error was encountered in codeigniter 3 - php

Question

My model:
class Admin_model extends CI_Model{
function add_blog($data){
$this->db->insert('blog',$data);
return true;
}
}
My controller function:
function add_blog(){
$this->form_validation->set_rules('heading','Heading', 'trim|required');
if ($this->form_validation->run() == FALSE) {
$this->session->set_flashdata('error',"Please check the required field");
redirect('admin/dashboard/add');
}else{
//for upload file
$config['upload_path'] = './assets/uploads/';
$config['allowed_types'] = 'pdf|doc|docx|txt|jpeg|png|ppt';
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
}
else
{
$image_path = base_url("uploads/".$post['raw_name'].$post['file_ext']);
$data['picture'] = $image_path;
$data_image = array('upload_data' => $this->upload->data());
}
//Upload file end
$data['picture'] = $data_image['upload_data']['file_name'];
$data['heading'] = ucwords($this->input->post('heading'));
$data['description'] = ucfirst($this->input->post('description'));
$data['type_id'] = ucwords($this->input->post('type_id'));
$data['price'] = $this->input->post('price');
$insert_data = $this->admin_model->add_blog($data);
if($insert_data){
echo '<script>alert("Note added sucessfully.");</script>';
redirect('admin/dashboard');
}else{
$this->session->set_flashdata('message',"There is problem in inserting data, please retry once");
echo "error at last ";
redirect('admin/dashboard/add');
}
}
}
My view:
<form role="form" name="myForm" method="post" enctype="multipart/form-data" action="<?php echo site_url('admin/dashboard/add_blog');?>">
<label for="">Category</label>
<select name="type_id" id="" class="form-control" >
<?php
if(count($product_list )>0 ){
foreach ($product_list as $key => $value) {
?>
<option value="<?php echo $value['type_id'];?>"><?php echo $value['type'];?></option>
<?php
}}?>
</select>
<label for=""> Heading </label>
<input type="text" name="heading"class="form-control">
<label for="">Price</label>
<input type="number" name="price" class="form-control">
<label for="">Description</label>
<textarea name="description" id="" cols="30" rows="5" class="form-control"></textarea>
</div>
<div class="form-group">
<label for="exampleInputFile">Upload pic</label>
<div class="input-group">
<div class="custom-file">
<!-- <input type="file" name="userfile" class="form-control"> -->
<?php echo form_upload(['name'=>'userfile']); ?>
</div>
</div>
<?php if(isset($upload_error)){echo $upload_error;}?>
</div>
</div>
<div class="card-footer">
<button type="submit" class="btn btn-primary">Submit</button>
</div>
</form>
</div>
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: data_image
Filename: admin/Dashboard.php
Line Number: 64
Backtrace:
File:
/opt/lampp/htdocs/shremad/application/controllers/admin/Dashboard.php
Line: 64 Function: _error_handler
File: /opt/lampp/htdocs/shremad/index.php Line: 315 Function:
require_once
A PHP Error was encountered
Severity: Notice
Message: Trying to access array offset on value of type null
Filename: admin/Dashboard.php
Line Number: 64
Backtrace:
File:
/opt/lampp/htdocs/shremad/application/controllers/admin/Dashboard.php
Line: 64 Function: _error_handler
File: /opt/lampp/htdocs/shremad/index.php Line: 315 Function:
require_once
A Database Error Occurred
Error Number: 1048
Column 'picture' cannot be null
INSERT INTO blog (picture, heading, description, type_id,
price) VALUES (NULL, 'Djjdjdj', 'Jjdjdj', '1', '')
Filename: models/Admin_model.php
Line Number: 5

Answer 1

Let's analyze the error message first.
Message: Undefined variable: data_image
Which means, somewhere in the code you're calling a variable named "data_image" which is undefined. Now, please notice that this is a NOTICE, so there might be cases that your code would work well BUT it's a bad practice and might cause troubles. (As in your case for instance).
Looking at your code, the following "block":
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
}
else
{
$image_path = base_url("uploads/".$post['raw_name'].$post['file_ext']);
$data['picture'] = $image_path;
$data_image = array('upload_data' => $this->upload->data());
}
//Upload file end
$data['picture'] = $data_image['upload_data']['file_name'];
The $data_image variable is defined only under the "else" statement.
So if the condition returns true, the $data_image variable is truly undefined.
In that case, the following command:
$data['picture'] = $data_image['upload_data']['file_name'];
which is based on the $data_image variable, would return another problem:
Message: Trying to access array offset on value of type null
so $data['picture'] now is NULL, which cause your last problem:
Column 'picture' cannot be null
The solution?
Make sure you have a value for $data_image even in the case that the condition returns true. Or, if this condition returns true - handle the things differently. Please also notice that there might be a logical problem with the function in the condition that returns an unexpected result.
(The condition: if ( ! $this->upload->do_upload()))