Parse error: syntax error, unexpected T_ELSE

Question

i just need help regarding my code becuase i odnt have any idea where is the syntax error.

$file = $_FILES['image']['tmp_name'];

if (!isset($file)) 
   echo "Please select an image";
else
{
   $image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
   $img_name = addslashes($_FILES['image']['name']);
   $img_size = getimagesize($_FILES['image']['tmp_name']);
   if ($img_size == FALSE) 
      echo "select valid image.";
   else
   {
      if (!$insert = mysql_query("insert into testblob values('','$image','$img_name')"));
         echo "Problem uploading!";
      else
      {
         $lastid = mysql_insert_id();
         echo "Image uploaded.<p />Your image:<p /><img src=get.php?id=$lastid>";
      }
   }
}

i just need to know where is the error. thnaks for the help. more power ..

i know its possible to do if..else without curly braces, but you really should just always use curly braces even if its only one line... — celeriko
– celeriko, Commented Mar 21, 2014 at 15:27
also, if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")); should not have a ; at the end.. — celeriko
– celeriko, Commented Mar 21, 2014 at 15:27

echo_Me · Accepted Answer · 2014-03-21 15:31:16Z

You have a semi-colon ; in your conditional statement extra here:

if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')"));

Remove it and change it to:

if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')"))

why dont you use this better ?

   $file= $_FILES['image']['tmp_name'];
   if (!isset($file)){
      echo "Please select an image";
     }
   else{
        $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
        $img_name= addslashes($_FILES['image']['name']);
        $img_size = getimagesize($_FILES['image']['tmp_name']);

       if ($img_size==FALSE){
             echo "select valid image.";
              }
       else
         {

              if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")){
                   echo "Problem uploading!"; }
              else{

                $lastid = mysql_insert_id();
                    echo "Image uploaded.<p />Your image:<p /><img src=get.php?id=$lastid>";
                }

        }
 }

Reeno · Accepted Answer · 2014-03-21 15:27:37Z

I intended the code and added some more braces which makes it a lot easier to see an error. I also deleted a semicolon (;) after one of the if clauses which probably caused the error.

if (!isset($file)) {
    echo "Please select an image";
}
else {
    $image= addslashes(file_get_contents($_FILES['image']['tmp_name']));
    $img_name= addslashes($_FILES['image']['name']);
    $img_size = getimagesize($_FILES['image']['tmp_name']);

    if ($img_size==FALSE) {
        echo "select valid image.";
    }
    else {
        if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')")) /* HERE WAS A SEMICOLON */ {
            echo "Problem uploading!";
        }
        else {
            $lastid = mysql_insert_id();
            echo "Image uploaded.<p />Your image:<p /><img src=get.php?id=$lastid>";
        }

    }
}

Andrius Naruševičius · Accepted Answer · 2014-03-21 15:27:42Z

1

This line

if (!$insert=mysql_query("insert into testblob values('','$image','$img_name')"));

has an unnecessary ; at the end of it.

answered Mar 21, 2014 at 15:27

Andrius Naruševičius

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