selecting MAX(val) from a sql resource in 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

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 2

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 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'];
}