Why I Get Error While Uploading Image into DB?

Question

I want to upload an image into database using PHP. I get the following error when trying to upload:

Error:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

dbname:img    
tablename:image   
column:img   
type:LongBLob.

I've already managed to connect to the database before and was able to insert anything except image contents. Here is my code:

<form method="post" enctype="multipart/form-data">
  <input type="file" name="file">
  <button>Go</button>
</form>

<?php
  $C = new mysqli("localhost","root","","img");

  if(!$C->error) {
    echo "Connected";
  } else {
    echo $C->error;
  }

  if(isset($_FILES['file'])) {
    $F = file_get_contents($_FILES['file']['tmp_name']);
    $Q = "insert into image (img) values('$F')";
    $R = $C->query($Q);
    if($R == true) {
      echo "ok";
    } else {
      echo $C->error;
    }
  }
?>

I noticed you don't check whether there's actually a connection to the database. See: php.net/manual/en/mysqli.examples-basic.php It would also be helpful if you reported the whole and exact error message. Or are you doing that already? — KIKO Software
– KIKO Software, Commented Sep 15, 2018 at 10:29

brendonofficial · Accepted Answer · 2018-09-15 13:38:23Z

1

You get this error because you're essentially taking the raw contents of the file, and dumping it into the SQL statement without any form of sanitizing, or encoding.

Try: addslashes

Simply using addslashes would escape any conflicting characters that would cause the SQL query to fail.

$F = file_get_contents($_FILES['file']['tmp_name']);
$data = addslashes($F);
$Q = "insert into image (img) values('$data')";

Try: base64_encode

An alternative may be to use base64_encode instead.

Note: When using this method, your img column should be of type TEXT

$F = file_get_contents($_FILES['file']['tmp_name']);
$encoded = base64_encode($F);
$Q = "insert into image (img) values('$encoded')";

When retrieving the value from the database, you'll need to base64_decode in order to get the raw data back again.

edited Sep 15, 2018 at 13:38

answered Sep 15, 2018 at 13:32

brendonofficial

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Comments

Piyush Kumar Sahay · Accepted Answer · 2018-09-15 10:44:42Z

0

Check first whether you are connected with the database by :

$C = new mysqli("localhost","root","","img");
if ($C->connect_errno) {
  echo "Errno: " . $mysqli->connect_errno . "\n";
  echo "Error: " . $mysqli->connect_error . "\n";
}

answered Sep 15, 2018 at 10:44

Piyush Kumar Sahay

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Comments

imtaher · Accepted Answer · 2018-09-17 06:25:41Z

Try This

<form method="post" enctype="multipart/form-data">
<input type="file" name="file">
<button type="submit">Go</button>
</form>



<?php
$db_host = "localhost"; 
$db_username = "root";  
$db_pass = "";  
$db_name = "img"; 
$conn= mysqli_connect("$db_host","$db_username","$db_pass", "$db_name") or die ("could not connect to mysql");

if(isset($_FILES['file'])) {
$target_dir = "uploads/";//folder name to mave uploaded file
$target_file = $target_dir . basename($_FILES["file"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
$check = getimagesize($_FILES["file"]["tmp_name"]);
if($check !== false) {
    echo "File is an image - " . $check["mime"] . ".";
    $uploadOk = 1;
} else {
    echo "File is not an image.";
    $uploadOk = 0;
}

// Check if file already exists
if (file_exists($target_file)) {
echo "Sorry, file already exists.";
$uploadOk = 0;
}
// Check file size
if ($_FILES["file"]["size"] > 20000000) {
echo "Sorry, your file is too large Max file size is 2.5MB.";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" 
&& $imageFileType != "gif" ) {
echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["file"]["tmp_name"], $target_file)) {

    $image_name = basename( $_FILES["file"]["name"]);


 } else {
    $image_name = "default.jpg";
 }
 }
 mysqli_query($conn, "INSERT INTO image (img) VALUES('$image_name ')"")or die  (mysqli_error($conn));

 echo '<p align="center">Successful</p>';    

 }
 ?>

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Why I Get Error While Uploading Image into DB?

3 Answers 3

Try: addslashes

Try: base64_encode

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3 Answers 3

Try: addslashes

Try: base64_encode

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