PHP and MySQL error when inserting user entered values into table

Question

everything functions on the side of data validation and error gathering and such, I get a problem specifically where $r is being executed. I always get the echo error. I've used this same bit of code before, but for some reason isn't working. Just need a second look of eyes at it perhaps.

if (empty($errors)){
    $q = "INSERT INTO testTable (test1, test2) VALUES ('Test', 'Test')";
    $r = @mysqli_query($dbc, $q);

    if($r){
        echo 'Complete!';
    }
    else{ 
        echo 'error';
        }

    mysqli_close($dbc);

    include('footer.html');
    exit();

I can enter the statement manually in MySQL and it will add it to the table, so I don't think its a syntax error there, I am just a little tired at this point.

Change echo 'error' to echo mysqli_error($dbc); so you see the reason for the error. — Barmar
– Barmar, Commented Apr 13, 2015 at 21:56
Get the real error or die(mysqli_error($dbc)) to mysqli_query() and remove the @ — Funk Forty Niner
– Funk Forty Niner, Commented Apr 13, 2015 at 21:56
Yeah, it seems strange to ask a question about an error you're getting when you've told it to not report any errors in your function call. — Phylogenesis
– Phylogenesis, Commented Apr 13, 2015 at 21:59
I have changed it and now I just get a blank page. No indication of anything. — Burninrock24
– Burninrock24, Commented Apr 13, 2015 at 22:03

Funk Forty Niner · Accepted Answer · 2015-04-13 22:40:59Z

2

"I'm getting a lot of warnings saying that parameter 1 of mysqli_xxx must be mysqli. So does that mean that my $dbc variable isn't working properly?"

The reason for that may very well be because your DB connection method is probably based on mysql_connect() or PDO to connect with.

Something you haven't posted in your question.

Those different MySQL APIs do not intermix with each other.

mysql_ + mysqli_ or PDO = no love. mysqli_ + PDO, same thing.

Refer to both manuals:

and use only "one" MySQL API, from beginning to end and not a "mix of".

You should also remove the @ symbol(s) from your code. Those are error suppressors and won't help you when debugging code.

So your connection for MySQLi_ would look something like this:

$dbc = mysqli_connect("myhost","myuser","mypassw","mybd") 
        or die("Error " . mysqli_error($dbc));

Unlike mysql_:

$dbc = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$dbc) {
die('Could not connect: ' . mysql_error());
}

edited Apr 13, 2015 at 22:40

answered Apr 13, 2015 at 22:29

Funk Forty Niner

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5 Comments

Burninrock24 Over a year ago

My $dbc looks exactly like your mysqli_connect example. But I looked back up higher and saw that my mysqli_select_db() had the parameters in the wrong order. I had no idea that mysqli and mysql were different. Thank you so much.

Funk Forty Niner Over a year ago

@Burninrock24 Ah, glad to know this one was solved, cheers and you're welcome.

Funk Forty Niner Over a year ago

@Burninrock24 You can use all 4 parameters in mysqli_connect() rather than using mysqli_select_db(). Saves you a bit of code too.

Burninrock24 Over a year ago

Thanks again, what happened is that I was mixing and matching elements from different scripts that I had completed independent of each other, not knowing that there were compatibility issues between them. Thank you again!

Funk Forty Niner Over a year ago

@Burninrock24 It happens and you're welcome. The old mysql_ method of connecting, required the connection variable being last, as opposed to mysqli_ being the first.

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