returning MySQL Max()

[ravens_chance]

I can get my mySQL query to run fine in phpMyAdmin, but can't get anything when I put it into my PHP code.

 

When I run the query in MySQL in phpMyAdmin I get the result:

 

MAX(jobID)

43

 

But, when I put this query into my code (with or without a variable) I get nothing. And everything downcode from it doesn't show up. What am I missing?

 

Ideally what I want to work:

$result = mysql_query("SELECT MAX(jobID) From job WHERE jobEmail='$jobEmail->email'");
while ($row = mysql_fetch_object($result)) {
    echo $row->jobID;
}
mysql_free_result($result);

Second attempt at code:

$result = mysql_query("SELECT MAX(jobID) FROM job WHERE jobEmail='email@company.com'";
while ($row = mysql_fetch_array($result)) {
	echo $row['MAX(jobID)'];
}   

 

Thank you

 

[wildteen88]

Try using an alias instead

$result = mysql_query("SELECT MAX(jobID) as maxJobID FROM job WHERE jobEmail='email@company.com'";
$row = mysql_fetch_array($result);
echo $row['maxJobID '];

 

NOTE: if your query only returns one row/result, then you don't need to use a while loop.

[Maq]

This may be a typo but you're missing the terminating ')' for your mysql_query call.

[ravens_chance]

yes that was a typo...

 

Thank you for the Alias recommendation, that fixed it in both scenarios

[Maq]

yes that was a typo...

 

Thank you for the Alias recommendation, that fixed it in both scenarios

 

I always thought you could reference aggregate functions in the result set without an alias...?