Jason Petersen wrote:
On Wed, 2 Mar 2005 13:02:39 +0200, William Stokes <kalles@xxxxxxxxxxxxx> wrote:
Hello
Can someone explain this to me. I don't know how to read this.
if (!$variable = mysql_query("select id,sessid from users where ...
What is this "if(!"
This is a way to run a SQL query, capture the return value, and
perform an action if the query failed. For code readability, I would
probably write that as two statements:
$variable = mysql_query("SELECT * FROM blah");
if(!$variable) { error_handler(mysql_error()); }
in terms of readability an alternative if to write it like so:
if(! ($variable = mysql_query("SELECT * FROM blah")) {
// do stuff
}
what you have to realise is that the expression '$variable = mysql_query("SELECT * FROM blah")'
has a return value itself - which just happens to be the same as the value of $variable.
to put it another way, when you do:
echo $variable;
your not saying:
"output the contents of this variable"
but your actually saying:
"output the result of the expression '$variable'"
it just so happens that the result of the expression is this case
if the contents of $variable.
hopefully you get what I mean!, no doubt there are others that can explain it better!
See the documentation for information on mysql_query return values:
http://us2.php.net/manual/en/function.mysql-query.php
Jason
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
