I want to store the result of an SQL query in a PHP variable but as an integer, this is my code:
$maxnroviaje=mysql_query("SELECT MAX(NROVIAJE) from viajes");
$nroviajenew=mysql_fetch_assoc($maxnroviaje);
and doing echo on those two variables:
<?php echo $maxnroviaje; ?>
<?php echo $nroviajenew; ?>
what throws me is the following:
Resource id # 15
Notice: Array to string conversion in C: \ xampp \ htdocs \ agent-nav-online \ add-trip.php on line 47 Array
I solved it in the following way:
$maxnroviaje=mysql_query("SELECT MAX(NROVIAJE) as maximo from viajes");
$nroviajenew=mysql_fetch_assoc($maxnroviaje);
$nroviajex=$nroviajenew["maximo"];
Of course, based on the answers posted here.
1.- $ maxnroviaje is an arrangement because it is the information that returns mysql from the query at a general level.
2.- $ nroviajenew is also an arrangement because it is the result of the conversion of the query to the requested fields.
You could try doing this:
$maxnroviaje = $db->query("SELECT MAX(NROVIAJE) as cuenta from viajes");
$nroviajenew = $maxnroviaje->fetch_assoc();
echo $nroviajenew["cuenta"];
//otra alternativa para mostrar el resultado de un vector
echo $nroviajenew->cuenta;
in this way if you would be showing the result you are looking for.
Greetings
NOTE: The mysql library is depreciated in PHP, you must use mysqli or pdo for your queries, that way you will avoid your system failing in the future.
$nroviajenew is an array, it is not a single value. You can use var_dump to see the structure of the array.
The correct thing in this case is to use $nroviajenew[0] .