selecting MAX(val) from a sql resource in php - php

Question

I have a sql result that looks like this:
MAX(val) | instance
17410742.00 | 0
I need to loop through it in php but I can't seem to select the Max(val) column. Usually I would do a something like this:
foreach ($sqlmax as $maxrow){
$myvar=$maxrow['instance'];
}
Which returns the 'instance' value, but I can't get the syntax to retrieving the Max(val) as
foreach ($sqlmax as $maxrow){
$myvar=$maxrow['Max(val)'];
}
doesn't work. I get the error Notice: Undefined index:
How do I select the Max(val) result in the same way I can select the 'instance' value? Thanks

Answer 1

Use AS in your SQL Query:
SELECT MAX(val) AS max_val, instance FROM table WHERE instance = 0;
Then you will be able to access the column as $myvar=$maxrow['max_val'];.
The query I wrote here is just an example, since you do not provide yours.
I will go ahead an assume you use mysql and will provide a link to MySQL docs:
A select_expr can be given an alias using AS alias_name. The alias is used as the expression's column name and can be used in GROUP BY, ORDER BY, or HAVING clauses. For example:
SELECT CONCAT(last_name,', ',first_name) AS full_name FROM mytable ORDER BY full_name;

Answer 2

Assign a column name in your query, so you can access the field easier:
SELECT MAX(val) AS max_value ....
Then you can access it with
$myvar=$maxrow['max_value']

Answer 3

Simply do it like
select max(val) as mval
and access it like
$maxrow['mval']

Answer 4

Just providing an alternate answer, in case you don't want (or are unable to) use AS.
The problem here was that the key of the array is case sensitive, so you have to use $maxrow['MAX(val)'] instead of $maxrow['Max(val)'].

Answer 5

SELECT MAX(val) AS instance
Access this
foreach ($sqlmax as $maxrow){
$myvar=$maxrow['instance'];
}