Image path upload and displaying image in mysql

Question

I need to upload an image to database and path should be saved in table. I am using the below code to upload the image and store the path in database table. I have created a directory "images" for storing image which is uploaded by user.

if(isset($_POST['submit']))
{
  $errors= array();
  $file_name=$_FILES['photo']['name'];
  $file_size =$_FILES['photo']['size'];
  $file_tmp =$_FILES['photo']['tmp_name'];
  $file_type=$_FILES['photo']['type'];
  $file_ext=strtolower(end(explode('.',$_FILES['photo']['name'])));
  $expensions= array("jpeg","jpg","png");
  if(in_array($file_ext,$expensions)=== false){
  $errors[]="extension not allowed, please choose a JPEG or PNG file.";
  }
  if($file_size > 2097152){
  $errors[]='File size must be excately 2 MB';
  }
  if(empty($errors)==true){
  move_uploaded_file($file_tmp,"images/".$file_name);
  echo '<script language="javascript">';
  echo 'alert("Success")';
  echo '</script>';
  }
  else
  {
    echo '<script language="javascript">';
    echo 'alert("Failed")';
    echo '</script>';
    print_r($errors);
  }
  }
  $query = mysql_query("insert into e(photo) values ('$file_name')");
  echo '<script language="javascript">';
  echo 'alert("Ok")';
  echo '</script>';
 }

Then the image should be displayed from database using the stored path in table. I am using the below code to display the image from database.

$connection = mysql_connect("localhost", "root", "");
$db = mysql_select_db("s", $connection);
$sql = "SELECT * FROM e";
$result = mysql_query($sql);
while ($o = mysql_fetch_assoc($result))
{
    echo "<tbody>";
    echo "<tr>";
    echo "<td>".'<img src="$photo;"' ."</td>";
    echo "</tr>";
    echo "</tbody>";
}
?>

The table contains only file name i.e image name. The image is not displayed while using the select query which is showed above. — Developer
– Developer, Commented Mar 30, 2017 at 10:57
Sorry it is $photo it is the table column name. It is not displaying the image. — Developer
– Developer, Commented Mar 30, 2017 at 11:04
Little Bobby says your script is at risk for SQL Injection Attacks.. Even escaping the string is not safe! — Jay Blanchard
– Jay Blanchard, Commented Mar 30, 2017 at 14:31

Akshay · Accepted Answer · 2017-03-30 11:07:31Z

2

Here you are making mistake

echo "<td>".'<img src="$photo;"' ."</td>";

Replace it with (Ensure that you are completing img tag)

echo '<td><img src="your_images_folder_path/'.$o['photo'].'"></td>';

Please stop using mysql_* it's depreciated since PHP5.6

answered Mar 30, 2017 at 11:07

Akshay

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3 Comments

Developer Over a year ago

Thank You but I am using old version in my system.

Developer Over a year ago

Yes now it is displaying

Akshay Over a year ago

Great! Happy coding! :D

Rotimi · Accepted Answer · 2017-03-30 11:16:42Z

0

$connection = mysql_connect("localhost", "root", "password", "s");
$sql = "SELECT * FROM e";
$result = mysql_query($sql);
while ($o = mysql_fetch_assoc($result))
{
    echo "<tbody>";
    echo "<tr>";
    echo "<td>".'<img src="path/'.$o['photo']."></td>";
    echo "</tr>";
    echo "</tbody>";
}
?>

Where photo is the column name from your database. Also, you can simply pass the database name in your connection declaration instead of using the mysql_select_db

edited Mar 30, 2017 at 11:16

answered Mar 30, 2017 at 11:09

Rotimi

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7 Comments

Developer Over a year ago

There is an error in the following statement. I think single quotation is missing.

Developer Over a year ago

Buddy still image is not displaying.

Rotimi Over a year ago

it wont display because you have to enter your path. I do not know where you save your images. Change that line to match your structure

Developer Over a year ago

echo "<td>".'<img src="images/'.$o['photo']."></td>";

Developer Over a year ago

This is not displaying

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