[SOLVED] Checking for a valid filetype? Code provided

[Mr Chris]

Hello,

 

I have a script i;m using to insert data into a database and at the same time check for required fields.  Now the below checks for required fields and the filesize of the uploaded image:

 

<?php 
if (isset($_POST['Submit'])) 
{ 
   $errors = array(); 
   $byte = "2097152"; 
   $your_name = trim($_POST['your_name']); 
   $your_email = trim($_POST['your_email']); 
      
   // Check for normal errors in name and email
   if ($your_name === '') 
   { 
      $errors[] = "You forgot to enter your name."; 
   } 
   
   if ($your_email === '') 
   { 
      $errors[] = "You forgot to enter your email address."; 
   } 
   
   // Check  to see if the uploaded file is empty and greater than the $byte variable set above, if it is bigger or there is nothing do the below
   if(empty($_FILES['ufile'])) 
   { 
      $errors[] = "No file was uploaded."; 
   } 
   
   else if($_FILES['ufile']['size'][0] > $byte) 
   { 
      $errors[] = "Your file is too big."; 
   }

  
   // If errors is equal to zero then set the image details and insert into the database!
   if(count($errors) == 0) 
   { 
   
      // set the image details and upload the file
      $picunique = md5(uniqid());
      $picture = "/home/************/*********/images/$picunique".$_FILES['ufile']['name'][0];
      move_uploaded_file ($_FILES['ufile']['tmp_name'][0], $picture);

      // and insert into the database  
      $result = mysql_query("Insert into image_test (your_name,your_email,picture) values('$your_name','$your_email','". basename($picture) ."')") or die(mysql_error());
      $picture_id=mysql_insert_id();
      $error = "Sucess - everything uploaded";
   } 
   else 
   { 
      $error = "<span style='color:#c00'>" . implode(' ', $errors) . "</span>"; 
   } 
} 
?>
<form action="" method="post" enctype="multipart/form-data" name="myform" id="myform">
<h3>Submit your Photo</h3>
<?php echo $error;?>
  <fieldset>
    <legend>Please fill in the details below: </legend> 
  <p>  
    <label for="Story">Your Name:</label> 
    <input type="text" name="your_name" id="your_name" value="" tabindex="1" />
  </p>  
  <p>  
    <label for="Story">Your Email:</label> 
    <input type="text" name="your_email" id="your_email" value="" tabindex="2" />
  </p>
  <div style="clear:left;"></div>
  <p> 
    <label for="Main Picture">Upload Image:</label> 
<input name="ufile[]" type="file" id="ufile[]" />
  </p>
  <input name="Submit" type="submit" value="Submit" />
  <input name="Reset" type="reset" value="Reset" />  
  </fieldset>
</form>
</body>
</html>

 

However, i'd also like the system to check the file is a .jpg and i've been trying to adapt the following in the above before the if(count($errors) == 0) , but i'm not too sure how to do it.

 

  $FileType = array (".jpg", ".jpeg", ".JPG", ".JPEG");    
   if(!in_array($FileType)){ 
      $errors[] = "Sorry, but this extension type is not supported."; 
   }  
  

 

Can anyone please help?

[zq29]

in_array() expects two parameters, a needle and a haystack. Try this:

<?php
if(!in_array(substr($filename,strrpos($filename,".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

[Mr Chris]

Sorry, might be being thick but where does $filename derive from.  Where is that variable created?

[zq29]

Sorry, might be being thick but where does $filename derive from.  Where is that variable created?

Sorry, replace $filename with $_FILES['ufile']['name'], I didn't write my example specifically for your code - It was just that, an example.

[Ayon]

@.Mr Chris:

this is where the filename is set.... SemiApocalyptic just used $filename instead of $picture.. $filename is the most normal to use on fileupload

$picture = "/home/************/*********/images/$picunique".$_FILES['ufile']['name'][0];

 

 

@SemiApocalyptic:

Sorry for going off topic here... but i got one question that i've been wondering...

is there actually a difference between this

<?php
if(!in_array(substr($filename,strrpos($filename,".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

 

and this?

<?php
if(!in_array(end(explode(".",$filename)),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

[Mr Chris]

Thanks, sorry to be a pain, but have the snippett like this now:

 

// Check the extension
$FileType = array (".jpg", ".jpeg", ".JPG", ".JPEG");  
if(!in_array(substr($_FILES['ufile']['name'],strrpos($_FILES['ufile']['name'],".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}

 

and get the message Warning: strrpos() expects parameter 1 to be string, array given?

[zq29]

@SemiApocalyptic:

Sorry for going off topic here... but i got one question that i've been wondering...

is there actually a difference between this

<?php
if(!in_array(substr($filename,strrpos($filename,".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

 

and this?

<?php
if(!in_array(end(explode(".",$filename)),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

 

Apart from it requiring less typing? Probably not - There's more than one way to skin a cat

[Ayon]

try

$FileType = array ("jpg", "jpeg", "JPG", "JPEG");

[zq29]

Thanks, sorry to be a pain, but have the snippett like this now:

 

// Check the extension
$FileType = array (".jpg", ".jpeg", ".JPG", ".JPEG");  
if(!in_array(substr($_FILES['ufile']['name'],strrpos($_FILES['ufile']['name'],".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}

 

and get the message Warning: strrpos() expects parameter 1 to be string, array given

 

Replace both instances of $_FILES['ufile']['name'] with $_FILES['ufile']['name'][0]. I didn't realise you were passing it as an array - Why have you chosen to do that?

[Ayon]

@SemiApocalyptic:

Sorry for going off topic here... but i got one question that i've been wondering...

is there actually a difference between this

<?php
if(!in_array(substr($filename,strrpos($filename,".")),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

 

and this?

<?php
if(!in_array(end(explode(".",$filename)),$FileType)) {
    $errors[] = "Sorry, but this extension type is not supported.";
}
?>

 

Apart from it requiring less typing? Probably not - There's more than one way to skin a cat

 

hehe that's what i thought i love less typing

 

-----

Back on topic... I belive he's editing a tutorial or premade script or something... am I right Chris?

[Mr Chris]

Thanks SemiApocalyptic, great help.  Just thought it would be easier that way.