4
I am again looking for help to understand an error given in a call to mysql_result(), which I cannot solve this error:
Warning: mysql_result() expects Parameter 1 to be Resource, Object Given in C: wamp www\
I don’t know what parameter you ask for.
That is the code:
<?php
$visitas_total = mysqli_query($conexao,"SELECT Sum(visitas) AS visitas FROM lp_post")
or die(mysql_error());
if(@mysqli_num_rows($visitas_total) <= '0') echo '';
$views = 0;
$visitas = mysql_result($visitas_total, $views, 'Visitas') ;
?>
someone could help me?
please arrange the post.
– RFL
Sorry dear for my mistake. Thank you.
– saulo
Try to take out the third parameter of the
mysql_result
and say what was the return– RFL
Sorry Rafael Acioly, I deleted all comments without that goal. But I tell you that in the way in which you guided me, you eliminated the error, but you did not print for me the numbers but the word Array, can guide me how to do?
– saulo
except the third 'visits' error persists
– saulo