Home/ Questions/Q 8496031
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-10T23:44:17+00:00

I’m making a script where the user can upload an image of his pet.

  • 0

I’m making a script where the user can upload an image of his pet. I have this code:

<?php
    session_start();
    $name = $_SESSION['myusername'];
    $petName = $_POST['picup']; // a pet name picked by a dropdown list

    $con = mysql_connect( "localhost", "lalala", "blabla" );
    mysql_select_db( "lalal_animal", $con );
    if ( @$_POST ['submit'] ) {
      $file = $_FILES ['file'];
      $name1 = $file ['name'];
      $type = $file ['type'];
      $size = $file ['size'];
      $tmppath = $file ['tmp_name'];
      if ( $name1 != "" ) {
        if ( move_uploaded_file( $tmppath, 'upload/' . $name1 ) ) {
          $query = "insert into pics(animalName,username,image) VALUES('$petName','$name','$name1')";
          mysql_query( $query ) or die( 'could not updated:' . mysql_error() );
          echo "Your image upload successfully !!";
        }
      }
    }
    ?>

    <html >
      <head>
        <title>Image Upload</title>
      </head>
      <body>
        <form name="form" action="" method="post" enctype="multipart/form-data">
          Photo <input type="file" name="file" />
          <input type="submit" name="submit" value="submit" /> 
        </form>
      </body>
    </html>

After running this the tabele column named: “animalName” is blank.

When I run this SQL command inside of phpmyadmin:

insert into pics(animalName,username,image) VALUES('Sparky','tester','sparky.jpg')

The tables are ok and the animalName column contains the right value.
The pet name is showing, if I do:

echo $petName;

The pet name is comming from another form. The user 1st have to choose the pet and then he’s redirected to the uload form.
Here is the dropdown you requested:

<table width="480" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<form name="upload" method="post" action="upload_form.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3"><strong>Upload image</strong></td>
</tr>
<td>Select a pet</td>
<td>:</td>
<td>

<?php
mysql_connect('localhost', 'blabla', 'lalal');
mysql_select_db('lalala_animal');

$sql = "SELECT name FROM animal where username='$name'";
$result = mysql_query($sql);

echo "<select name='picup'>";
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['name'] . "'>" . $row['name'] . "</option>";
}
echo "</select>";
?>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Upload"></td>
</tr>
</table>
</td>
</form>
</tr>
</table>

I suspect that there is somthing wrong with the HTML form below of the php code, but I can’t understand what.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    As you said that you pass the $_POST['picup'] from the first submit, just simply modify your second form and put the $_POST['picup'] value in the hidden field. Modify your form like this..

    <form name="form" action="" method="post" enctype="multipart/form-data">
      Photo <input type="file" name="file" />
      <input type="hidden" name="picup" value="<?php echo $petName; ?>" /> 
      <input type="submit" name="submit" value="submit" /> 
    </form>

    Hope this will work..

    Because when you submit second, that time your first post values not working.

    • 0