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Dec 8th, 2016
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  1. <?php
  2.  
  3. //Create connection variables
  4. $server = "localhost";
  5. $username = "Stu324";
  6. $password = "Stu8195!";
  7. $dbname= "stu324";
  8.  
  9. //create connection to database
  10.  
  11. $conn = new mysqli ($server, $username, $password, $dbname);
  12.  
  13. // Check connection
  14.  
  15. if($conn->connect_error)
  16. {
  17. die("Connection failed: " . $conn->connect_error);
  18. }
  19. if(isset($_POST['upload']))
  20. {
  21. $target = $_FILES['image']['name'];
  22. $sql = "INSERT INTO uploadimage (image) VALUE ('$target')";
  23. $conn->query($sql);
  24. $target_dir = "image/";
  25.  
  26. if (!file_exists($target_dir) )
  27. {
  28. mkdir ($target_dir, 0744);
  29.  
  30. }
  31.  
  32. $target = $target_dir.basename($_FILES['image']['name']);
  33.  
  34. if(move_uploaded_file($_FILES['image']['tmp_name'], $target))
  35. {
  36. echo " Image has been uploaded ";
  37.  
  38. }
  39. else
  40. {
  41. echo "Error while uploading";
  42.  
  43. }
  44.  
  45. if($conn->query($sql) === TRUE)
  46. {
  47. echo "New record created";
  48.  
  49. }
  50. else
  51. {
  52.  
  53. echo "Error: " . $sql . "<br>" . $conn->error;
  54. }
  55. }
  56.  
  57. ?>
  58.  
  59.  
  60.  
  61. <form action="uploadimg.php" method="post" enctype="multipart/form-data">
  62. Select image to upload:
  63. <input type="file" name="image">
  64. <input type="submit" value="Submit" name="upload">
  65. </form>
  66.  
  67. <?php
  68. $sql = "SELECT * FROM uploadimage";
  69. $result = mysqli_query($conn, $sql);
  70. while ($row = mysqli_fetch_array($result))
  71.  
  72. $img = $row['image'];
  73. echo "<img src='" . $target . "'>";
  74.  
  75. $conn->close()
  76.  
  77. ?>
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