Value still inserting, even though query encased in if statement which should stop it from doing so?

I really don't know what's going on here, I've been racking my brain for ages trying to figu开发者_JAVA技巧re this out;

I have the variable $overflow = $_POST['overflow']; and then the following code:

if($overflow != "" || $overflow != "Overflow link"){
    $query = "DELETE FROM links WHERE overflow='YES'";
    $result = mysql_query($query) or die(mysql_error());
    $query = "INSERT INTO links (link, overflow) VALUES ('$overflow','YES') ON DUPLICATE KEY UPDATE overflow=VALUES(overflow)";
    $result = mysql_query($query) or die(mysql_error());
}

The idea is if $overflow is empty or holds the default value, it isn't inserted, but if it's a valid value the already existing entry for it is deleted (because I can't make the column unique), and it's re-inserted.

Anyone have any idea was to why it's inserting on ANY value?

Any help would be greatly appreciated!

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"If not empty or not some value".

Well, it cannot be empty and "some value" at the same time, so this condition is always true. You're looking for:

if ($overflow != "" && $overflow != "Overflow link")

---

if(!empty($overflow) && $overflow != "Overflow link"){
    $query = "DELETE FROM links WHERE overflow='YES'";
    $result = mysql_query($query) or die(mysql_error());
    $query = "INSERT INTO links (link, overflow) VALUES ('$overflow','YES') ON DUPLICATE KEY UPDATE overflow=VALUES(overflow)";
    $result = mysql_query($query) or die(mysql_error());
}