Can't set image upload variable to NULL

Question

So weird well annoying problem, I'm trying to upload images from a form (all works fine) However when there is no image it should set itself to NULL, but because i'm using $target_dir . basename it keeps taking the last part of the $target_dir instead of setting to NULL and inserting the word products into the database instead of nothing. But obv need the $target_dir for when the upload takes place to put in correct place. Please see code below, all help much appreciated.

$target_dir = "../images/products/";


      if (!isset ($_FILES["img1"]["name"])) {
        $target_file1 = NULL;
      } else {
        $target_file1 = $target_dir . basename($_FILES["img1"]["name"]);
          }

          if (!isset ($_FILES["img2"]["name"])) {
            $target_file2 = NULL;
          } else {
            $target_file2 = $target_dir . basename($_FILES["img2"]["name"]);
          }

          if (!isset ($_FILES["img3"]["name"])) {
            $target_file3 = NULL;
          } else {
            $target_file3 = $target_dir . basename($_FILES["img3"]["name"]);
          }

          if (!isset ($_FILES["img4"]["name"])) {
            $target_file4 = NULL;
          } else {
            $target_file4 = $target_dir . basename($_FILES["img4"]["name"]);
          }

SQL query and relevant info

<pre>
$image1 = basename($target_file1);
$image2 = basename($target_file2);
$image3 = basename($target_file3);
$image4 = basename($target_file4);

echo $image1;
echo $image2;
echo $image3;
echo $image4;

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
  die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO products (product_code, product_name, category, 
filter, description, specification, img1, img2, img3, img4, price) 
VALUES('$product_code', '$product_name', '$category', '$filter', 
'$description', '$specification', '$image1', '$image2', '$image3', 
'$image4', '$price')";

if (mysqli_query($conn, $sql)) {
  echo "New record created successfully";
} else {
  echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
              ?>
</pre>

K, changed it to set the values first and only assign value if isset

same thing, see screenshot of output below.

Can you add your SQL call too, cant see much of a problem with what you have given. — PHPmaker
– PHPmaker, Commented Feb 28, 2021 at 11:37
when i take $target_dir out of the equation it leaves the mysql value Null but doesn't upload the image to the directory — MsKazza
– MsKazza, Commented Feb 28, 2021 at 12:23
You can try to define them as null before if statements actually and add a validation before inserting. So if they are isset you add value to your target. — Met Br
– Met Br, Commented Feb 28, 2021 at 14:28

Delete · Accepted Answer · 2021-02-28 20:44:00Z

1

u can try

$target_dir = "../images/products/";


if (!isset ($_FILES["img1"]["name"])) {
    $target_file1 = NULL;
} else {
    if (!empty($_FILES["img1"]["name"])) {
        $target_file1 = $target_dir . basename($_FILES["img1"]["name"]);
    } else {
        $target_file1 = NULL;
    }
}

if (!isset ($_FILES["img2"]["name"])) {
    $target_file2 = NULL;
} else {
    if (!empty($_FILES["img2"]["name"])) {
        $target_file2 = $target_dir . basename($_FILES["img2"]["name"]);
    } else {
        $target_file2 = NULL;
    }
}

if (!isset ($_FILES["img3"]["name"])) {
    $target_file3 = NULL;
} else {
    if (!empty($_FILES["img3"]["name"])) {
        $target_file3 = $target_dir . basename($_FILES["img3"]["name"]);
    } else {
        $target_file3 = NULL;
    }
}

if (!isset ($_FILES["img4"]["name"])) {
    $target_file4 = NULL;
} else {
    if (!empty($_FILES["img4"]["name"])) {
        $target_file4 = $target_dir . basename($_FILES["img4"]["name"]);
    } else {
        $target_file4 = NULL;
    }
}

answered Feb 28, 2021 at 20:44

Delete

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1 Comment

MsKazza Over a year ago

thank you thank you thank you worked like a charm. now i just have to figure out how i broke the actual upload of the image lol.

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