peddel Posted August 12, 2008 Share Posted August 12, 2008 $dag = date("d", time()); $maand = date("m", time()); $jaar = date("Y", time()); $uur = date("H", time()); $minuten = date("i", time()); $sql = 'INSERT INTO `ivoo_voorontwerp`.`testtabel` (`id`, `dag`, `maand`, `jaar`, `uur`, `minuten`, `PR`, `tag`, `waarde`, `eenheid`, `verschil`, `bovengrensOperator`, `ondergrensOperator`, `bovengrensOpvolging`, `ondergrensOpvolging`, `ploeg`, `oorsprongData`, `periodiciteit`) VALUES (0,\'$dag\', \'$maand\', \'$jaar\', \'$uur\', \'$minuten\', \'PR1\', \'naam\', \'$naam\', \'\', \'\', \'\', \'\', \'\', \'\', \'$ploeg\', \'operator\', \'\');'; $result = @mysql_query($sql); if(!$result){print("Query mislukt [".mysql_error()."]");} if u execute this and trough phpmyadmin check the items in the table, i find that it has inputted this : id dag maand jaar uur minuten 1 $dag $maand $jaar $uur $minuten i dont know why it doesnt take the values of those variables, can someone help? Link to comment https://forums.phpfreaks.com/topic/119263-solved-problem-with-adding-variables-into-mysql-database/ Share on other sites More sharing options...
GingerRobot Posted August 12, 2008 Share Posted August 12, 2008 Variables inside single quotes are not evaluated. They are treated as literals. That is: $foo = 'bar'; echo $foo;//echos bar echo "$foo";//echos bar echo '$foo';//echos $foo So, wrap your $sql string in double quotes. It'll also then save you the headache of escaping all those single quotes inside your query. Link to comment https://forums.phpfreaks.com/topic/119263-solved-problem-with-adding-variables-into-mysql-database/#findComment-614342 Share on other sites More sharing options...
peddel Posted August 12, 2008 Author Share Posted August 12, 2008 thx for the answer, this did the job Link to comment https://forums.phpfreaks.com/topic/119263-solved-problem-with-adding-variables-into-mysql-database/#findComment-614407 Share on other sites More sharing options...
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