Commit 050b0808 authored by Eli Schwartz's avatar Eli Schwartz Committed by Lukas Fleischer
Browse files

Fix more PHP 7.4 warnings

The try_login() function documents it returns an array containing an
'error' key, and our only caller *only* consults the 'error' key. Then
the function returns null instead of an array, if the login succeeded!

I question why we bother returning the new SID if we never use it,
surely we could either return the error or return default null. But, for
now, I'm just going to fix it to return what it's actually supposed to,
without changing the API.

Signed-off-by: Eli Schwartz's avatarEli Schwartz <>
Signed-off-by: Lukas Fleischer's avatarLukas Fleischer <>
parent 5ca1e271
......@@ -659,6 +659,7 @@ function try_login() {
header("Location: " . get_uri($referer));
$login_error = "";
return array('SID' => $new_sid, 'error' => null);
Supports Markdown
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment