Supplied argument is not a valid MySQL result resource?

Question

I get this error when I use this code for people to input data. The submit form won't be listed because it's not useful in this circumstance:

function some_more_custom_content() {

    $output="<BR>";

    ob_start();

    if ($_REQUEST['code'] != "") {
        $code = $_REQUEST['code'];
        $query="INSERT INTO `fc` (`code`,`datetime`) values ('" . mysql_real_escape_string($code) . "', now())";
        $result=mysql_query($query);

        while ($fetch_array = mysql_fetch_array($result)) {
            $seconds = time() - strtotime($fetch_array["datetime"]);

            if ((time() - $entry['datetime']) < 60*60) {
                echo ("The code " . htmlentities($code) ." was updated less than an hour ago.");
            } else {
                echo ("Inserted " . htmlentities($code) ." into the top.");
            }
        }
    }

Any idea why?

Random best practice comment: When using "echo" statements, don't concatenate your string. Use commas instead of periods. The echo function can take multiple strings as arguments. Example: echo("Inserted ", htmlentities($code), " into the top.") This speeds up execution time. — Mark, Aug 06 '09 at 02:10

score 4 · Answer 1 · answered Aug 06 '09 at 01:59

4

INSERT statements don't return a resource, they return TRUE on success or FALSE on failure. Thus there is no resource for mysql_fetch_array() to operate on.

(This is one of the main reasons why people complain about PHP -- its semantics are inconsistent at best.)

answered Aug 06 '09 at 01:59

Meredith L. Patterson

4,853
29
30

So, how can I fix t his and make it more clean? – Aug 06 '09 at 02:02
Well, first off, what are you trying to do with that while block? If you just want to find out whether the insert succeeded or failed, The Disintegrator's response below will take care of it. – Meredith L. Patterson Aug 06 '09 at 02:20
I agree that there are inconsistencies in php, but in this case? What kind of (useful) resource should mysql_query() return for an INSERT operation? – VolkerK Aug 06 '09 at 06:10
In this case I'd suggest a more consistent approach for the semantics of `mysql_query()`, i.e., having it return the same kind of result for all queries. Python's `.execute()` (method of a cursor) returns the number of rows affected, which is consistent for INSERT, SELECT, UPDATE and DELETE; the `.fetchone()` and `.fetchall()` methods may subsequently be called if the query was a SELECT. – Meredith L. Patterson Aug 06 '09 at 07:40
Then it would have to return an object, which is what the mysql extension does not, it's an "old" procedural module. You can do if(!$result) { errorhandling(); } for all types of statements. If you want "more" use another module like pdo, http://php.net/pdo – VolkerK Aug 06 '09 at 07:58

score 2 · Answer 2 · answered Aug 06 '09 at 02:09

2

Adding to Meredith Answer. You shoul use

if($result!==FALSE) {
    // the insert was ok
} else {
    //the inser failed
}

answered Aug 06 '09 at 02:09

The Disintegrator

4,147
9
35
43

score 0 · Answer 3 · answered Aug 06 '09 at 05:03

0

change:

$result=mysql_query($query);

with:

$result=mysql_query($query) or die(mysql_error());

you will see the error.

answered Aug 06 '09 at 05:03

inakiabt

1,953
1
16
28

Supplied argument is not a valid MySQL result resource?

3 Answers3