Insert image name to database problem

[zazu]

Hi there guys,

I've a little problem with inserting a file name into a database table. I can't see wich is the problem. The code is bellow and i think the problem is at INSERT INTO part.

    <?php
  $path = "./cv/";

$valid_formats = array("doc", "pdf");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
  $name = $_FILES['photoimg']['name'];
  $size = $_FILES['photoimg']['size'];
  if(strlen($name))
  {
    list($txt, $ext) = explode(".", $name);
    if(in_array($ext,$valid_formats))
    {
      if($size<(20480*20480)) // Image size max 20 MB
      {
        $actual_image_name = time().$id.".".$ext;
        $tmp = $_FILES['photoimg']['tmp_name'];
        if(move_uploaded_file($tmp, $path.$actual_image_name))
        {
          mysqli_query($mysqli,"INSERT INTO formular_client (client_cv = '$actual_image_name')");
        }
        else
          echo "failed";
      } 
      else
        echo "Image file size max 20 MB"; 
    }
    else
      echo "Invalid file format.."; 
  }
}
?>

<input type="file" name="photoimg" id="photoimg" />

Edited October 5, 2015 by zazu

[cobusbo]

 

Hi there guys,

I've a little problem with inserting a file name into a database table. I can't see wich is the problem. The code is bellow and i think the problem is at INSERT INTO part.

    <?php
  $path = "./cv/";

$valid_formats = array("doc", "pdf");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
  $name = $_FILES['photoimg']['name'];
  $size = $_FILES['photoimg']['size'];
  if(strlen($name))
  {
    list($txt, $ext) = explode(".", $name);
    if(in_array($ext,$valid_formats))
    {
      if($size<(20480*20480)) // Image size max 20 MB
      {
        $actual_image_name = time().$id.".".$ext;
        $tmp = $_FILES['photoimg']['tmp_name'];
        if(move_uploaded_file($tmp, $path.$actual_image_name))
        {
          mysqli_query($mysqli,"INSERT INTO formular_client (client_cv = '$actual_image_name')");
        }
        else
          echo "failed";
      } 
      else
        echo "Image file size max 20 MB"; 
    }
    else
      echo "Invalid file format.."; 
  }
}
?>

<input type="file" name="photoimg" id="photoimg" />

Check my topic

 

I had similar situation regarding Mime checking I think You should use mime types as well and not explode to get file ext.

 

 

I'm not an expert with mysqli but try 

mysqli_query($mysqli,"INSERT INTO `formular_client` (`client_cv`) = (`$actual_image_name`)");

[Barand]

The syntax is

INSERT INTO `formular_client` (`client_cv`) VALUES ('$actual_image_name')

or

INSERT INTO `formular_client` SET `client_cv` = '$actual_image_name'

[zazu]

 

The syntax is

INSERT INTO `formular_client` (`client_cv`) VALUES ('$actual_image_name')

or

INSERT INTO `formular_client` SET `client_cv` = '$actual_image_name'

 

Doesn't work, and i don't know why because the code seems to be correct.

[scootstah]

You need to be more specific than "doesn't work".

[zazu]

You need to be more specific than "doesn't work".

 

When i execute the script the file i choose is sent to the specified folder, but the file name is not stored to the database! To be more specific the INSERT function doesn't work well.

[benanamen]

Have you checked to see if there is any data in $actual_image_name?

 

In the following line you have $id, but I dont see it anwhere in your code gettting a value.

$actual_image_name = time().$id.".".$ext;   

Edited October 5, 2015 by benanamen

[mac_gyver]

you must ALWAYS check for and handle query errors. there are multiple possible reasons that any query can fail.

 

for some quick debugging, just echo mysqli_error($mysqli); on the next line after your mysqli_query() statement, assuming you have a valid database connection in $mysqli.

 

do you have php's error_reporting set to E_ALL and display_errors set to ON so that php would help you by reporting and displaying all the errors it detects?

[zazu]

 

Have you checked to see if there is any data in $actual_image_name?

 

In the following line you have $id, but I dont see it anwhere in your code gettting a value.

$actual_image_name = time().$id.".".$ext;   

 

Ok mate! Thanks for help i've manage to solve the problem. The working code is listed bellow:

mysqli_query($mysqli,"UPDATE evaluatori_users SET client_cv='$actual_image_name' WHERE id='$id'");