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mysql php problem: no error message despite error_reporting(E_ALL) line

Answer 1

A:

You have inverted single quotes and double quotes somewhere in your code (at the end of your request). Make sure the display_error php setting is on

greg0ire 2010-06-15 10:52:08

Answer 2

+1 A:

You have a fatal error, so the script simply cannot run, and therefore cannot report any errors. Look at this line:

$result = $mysqli->query("insert into stuff (data) values ('
        .$_POST['data']
        ."');

Should read:

$result = $mysqli->query("insert into stuff (data) values ('".$_POST['data']."')");

And escape your $_POST['data'] value before using it in the SQL statement

Mark Baker 2010-06-15 10:53:11

I fixed that. But still no results. The line now reads:$result = $mysqli->query("insert into stuff (data) values ('" . $_POST['data'] . "')";

Joshua Moore 2010-06-15 11:00:16

get into the habit of using an editor with syntax highlighting, and it becomes much easier to see errors such as this... if you look at your original posting, it's obvious even there.

Mark Baker 2010-06-15 11:05:17

Answer 3

+3 A:

you have a syntax error in action.php :

ini_set not ini_sit in line 2 !

Haim Evgi 2010-06-15 10:53:15

Answer 4

+1 A:

ini_sit

has to be

ini_set

;)

oezi 2010-06-15 10:58:03

Answer 5

A:

Ok, first off, you have a massive SQL Injection vulnerability in there. Second, you have no error checking. Third, your quotes are incorrectly nested (which will cause a fatal error, which is why you're not seeing anything)

Modify it to something like this:

$mysqli = new mysqli('localhost', 'root', 'password', 'website');
if ($mysqli->connecterror) {
    //There was an error connecting, handle it
}

$result = $mysqli->query("insert into stuff (data) values ".
    " ('".$mysqli->real_escape_string($_POST['data'])."')";

if ($result === false) {
    //Query error, handle it
}

Also, you're looping through the data without checking if the result is valid:

if($result = $mysqli->query("select data from stuff")){
    echo 'There are '.$result->num_rows.' results.';
    while ($row = $result->fetch_object()){
        echo 'stuff' . $row->data;
    }
}

ircmaxell 2010-06-15 11:07:43

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mysql php problem: no error message despite error_reporting(E_ALL) line

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