mysql_query("SHOW DATABASES") doesn't show my database

Question

I am testing a local page using xampp. Searching how to fix a "No database selected" mysql error lead me to testing this out:

mysql_connect('localhost') or die ("Connect error");

$res = mysql_query("SHOW DATABASES");
while ($row = mysql_fetch_row($res))
{
    echo $row[0], '<br/>';
}

But that query only shows two results: "information_schema" and "test". Which I don't understand cause on phpMyAdmin I can see all this databases:

cdcol, information_schema, mysql, performance_schema, phpmyadmin, test, xxxxxxxx_db (the one I created) and webauth

And (on phpMyAdmin) I can see that root has privileges to this database. Root has no password.

Thanks in advance.

EDIT:

I have an actual site online, I use mysql_connect and such to connect to the database... But I had to reinstall xampp and dreamweaver and everything on my PC and then had to download the database and the whole site cause I need to test some things locally. I can't afforrd to change the mysql querys and use PDO for now, maybe later when I have the time.

My main problem is the connection, it doesn't work, It prints "No database selected". I don't know why it doesn't find my xxxxxxxx_db database.

What does this print: echo ini_get('mysql.default_username'), ini_get('mysql.default_password');? — Daniel W.
– Daniel W., Commented Jan 12, 2014 at 13:03
@iBrazilian2 PDO printed my databases. I guess I'm gonna try conecting via PDO and not mysql_*... It used to work before though, a few months ago before I changed PCs. — Adocad
– Adocad, Commented Jan 12, 2014 at 13:39
@DanFromGermany That prints absolutely nothing :( Mr.Alien That shows the same results — Adocad
– Adocad, Commented Jan 12, 2014 at 13:40

Daniel W. · Accepted Answer · 2014-01-13 11:08:20Z

1

In the comments you said your default config is empty, so when you connect to mysql, you have to pass the username to the function:

$connection = mysql_connect('localhost', 'root', '') or die ("Connect error");

You should also keep the connectino resource in a variable and pass it to the query:

$res = mysql_query("SHOW DATABASES", $connection);
while ($row = mysql_fetch_row($res))
{
    echo $row[0], '<br/>';
}

If you cannot rewrite all the statements, well, you SHOULD sooner or later you will have even more trouble.

answered Jan 13, 2014 at 11:08

Daniel W.

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1 Comment

Adocad Over a year ago

Passing the username (root) works perfectly. Wow, I didn't know I was connecting to nothing, that's why It only showed two databases. I actually thought I was connecting to root. Thanks a lot. @JenniferDias solution also works.

Martin Bean · Accepted Answer · 2014-01-13 11:11:04Z

1

Please try this:

$link = mysqli_connect('localhost', 'root', '');

if ($res = mysqli_query($link, 'SHOW DATABASES')) {
    echo 'Connected.';
} else {
    echo 'Error occurred', mysqli_error($link);
}

while ($row = mysqli_fetch_row($res)) {
    echo $row[0], '<br/>';
}

edited Jan 13, 2014 at 11:11

Martin Bean

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answered Jan 12, 2014 at 13:33

Vaisakh Parannattil Cherul

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4 Comments

Adocad Over a year ago

It prints this: Connected Warning: mysql_fetch_row() expects parameter 1 to be resource, object given in C:\xampp\htdocs_____________________.php on line 13

Vaisakh Parannattil Cherul Over a year ago

try this one Or use mysqli_fetch_array()

Adocad Over a year ago

mysqli_fetch_array() works! thanks Jennifer. Problem solved, now on to the other problems :P Cheers

Daniel W. Over a year ago

Really great improvement Martin Bean.

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