| Re: cause of error is what? [message #181574 is a reply to message #181573] |
Wed, 22 May 2013 10:25   |
Thomas 'PointedEars'
Messages: 701
Registered: October 2010
Karma:
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Senior Member |
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Scott Johnson wrote:
> On 5/21/2013 6:44 PM, richard wrote:
>> $result = mysql_query("SELECT atitle,artist,avid FROM A$year WHERE id =
>> $number");
>> if (!$result) { echo 'Could not run query: ' . mysql_error(); exit; }
>> $vid = mysql_fetch_row($result);
>>
>> echo $vid[0];
>>
>> }
>>
>> echo "</div>";
>>
>> ?>
>>
>
> *JTOL*
-v
> I could be wrong but..
You are. RTFMs.
> Even if you left $number unset, that query should not of returned false
> unless 'id' is not an (int) in your table definition.
$ php -r 'echo "foo" . @$number . "bar\n";'
foobar
If they left $number unset, it would be evaluated to NULL, expanded to the
zero-length string, and the query would be syntactical in error starting
from the last “=”.
mysql_query(), which is *deprecated*, returns FALSE if that happens or the
query cannot be executed for some other reason (which can be found out with
mysql_errorno() and mysql_error()). It *never* returns the result set such
as a field value; that has to be obtained by passing the “resource” value
returned by it as parameter to mysql_fetch_row() etc.
<http://php.net/mysql_query>
Also, MySQL implicitly converts numeric values to string values and vice-
versa. Suppose the `id` column of a table `foo` would be of the common type
“UNSIGNED INT(10, 0)”, both
SELECT * FROM `foo` WHERE `id` = 42
and
SELECT * FROM `foo` WHERE `id` = '42'
would *always* yield the same result set.
<http://dev.mysql.com/doc/refman/5.6/en/type-conversion.html> p.
PointedEars
--
Prototype.js was written by people who don't know javascript for people
who don't know javascript. People who don't know javascript are not
the best source of advice on designing systems that use javascript.
-- Richard Cornford, cljs, <f806at$ail$1$8300dec7(at)news(dot)demon(dot)co(dot)uk>
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