PHP - MySql Database info not storing

Question

I have been trying to figure this out for hours, I have created a database ( MySql/PHPMyadmin) and i am trying to get user input stored to be able to call back up, however the info is not making it/ saving it to the database, everything shows up okay except this part of code:

$registered = mysqli_affected_rows ($dbc);

        echo $registered. "Row is affected";

when run gives me a display of -1 row, I believe this to be a big part of the problem as everything else seems to work okay. I am a complete beginner so could you guys tell me how the best way of debugging this is.

$dbc = $dbc = mysqli_connect ($hostname, $username, $password, $dbname) OR die("Could not Connect");

To input the data to the db i have the following:

if ($_SERVER['REQUEST_METHOD'] == 'POST'){

        $comments = $_POST ['Comments'];

        if (!empty ($comments)){

        include ('mysqldb.php');

        mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");

        $registered = mysqli_affected_rows ($dbc);

            echo $registered. "Row is affected";

        }else {
                echo "Nothing Submitted";
            }

        }

$dbc is the variable i created in the connection file to connect to the database — Ricky
– Ricky, Commented Dec 28, 2014 at 21:53
Can you echo your SQL query, copy it and paste into phpmyadmin and see what's going on? — Ali1928
– Ali1928, Commented Dec 28, 2014 at 21:57

kRiZ · Accepted Answer · 2014-12-28 22:01:26Z

1

The line:

mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");

should be:

mysqli_query ($dbc,"INSERT INTO `User-Comments` (Comments) VALUES ('$comments')");

Notice the change in the apostrophe character ( ` ) around your table name.

answered Dec 28, 2014 at 22:01

kRiZ

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1 Comment

Ricky Over a year ago

I didnt notice this syntax error untill now, it was part of the problem

axiac · Accepted Answer · 2014-12-28 22:04:59Z

0

An excerpt from the documentation for function mysqli_stmt_affected_rows(): -1 indicates that the query has returned an error.

You should check the value returned by mysqli_query(). If it returns FALSE then you can get details about the reason (error message) by using function mysqli_error().

answered Dec 28, 2014 at 22:04

axiac

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2 Comments

axiac Over a year ago

Eagle-eye @kRiZ was able to identify the error without running the code. :-)

Ricky Over a year ago

The problem was this "INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')"); it should have been "INSERT INTO 'User-Comments' (Usercom) VALUES ('$comments')"); but yes wouldnt have noticed the backtick for a very long time after...

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