azraelGG Posted October 30, 2012 Share Posted October 30, 2012 im trying to build simple login form with some if functions(currently without database) but i got stuck on line 77 if (isset($korisnickoIme)!=='admin' && isset(sha1($lozinka))!=='7110eda4d09e062aa5e4a390b0a572ac0d2c0220') { echo "Molimo unesite ispravne podatke."; } when i erase isset from isset(sha1($lozinka)).... then it works but i get error for variable not defined - any ideas? call if you need all code but rest of it works fine (notice that i have isset at $korisnickoIme and there it works Link to comment https://forums.phpfreaks.com/topic/270069-cant-use-function-return-value-in-write-context-in/ Share on other sites More sharing options...
Muddy_Funster Posted October 30, 2012 Share Posted October 30, 2012 You can't pass a sha1() into isset() as a parameter. As isset is a check in it's self I would give it it's own space in the if conditional, which should remove your problem aswell. i.e. if ((isset($korisnickoIme) && $korisnickoIme !=='admin') && (isset($lozinka) && (sha1($lozinka) !== '7110eda4d09e062aa5e4a390b0a572ac0d2c0220')) { Link to comment https://forums.phpfreaks.com/topic/270069-cant-use-function-return-value-in-write-context-in/#findComment-1388676 Share on other sites More sharing options...
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