Problem connecting to database

[Perkele]

Hej allihopa!

I'm trying to learn PHP/MySQL. I'm right now into learning how to connect to server and database.

When using this:

 <?php

 echo "trying to connect to the database server<br />";
 $link = mysqli_connect("localhost", "root", "password")
 or die("connection to server failed");

 echo "trying to connect to the database<br />";
 $result = mysqli_select_db($link, "klasbas")
 or die("connection to database failed");

 ?>

I get this in the browser:
trying to connect to the database server
trying to connect to the database

A connection to the database server has indeed opened. I know that, because if I, for example, change the hostname to "Obama", I get the "connection to server failed" message.

But it seems to ignore the request to connect to the database itself. At least that's my guess, because if I, for example, change the name of the connection from $link to $Clinton, I get no error message. It just seems to skip this request altogether.

I have no idea why, and don't know how to start troubleshooting.

Im 100% positive that the database klasbas exists, I can reach it using a tool i got from a book, but now I'm trying to learn how to write all this stuff myself.

Anybody?
All help most appreciated.

Regards
Klas

[halojoy]

Hej Perkele (finskt svärord). Det har varit lite kallt, eller hur ...

This is example from PHP Manual.
Notice how they catch specific error messages using
echo mysqli_error($link);

Here are all functions of mysqli http://www.php.net/manual/en/class.mysqli.php

You can both connect and select "dbname" in one statement:

<?php

$link = mysqli_connect("localhost", "my_user", "my_password", "dbname");

/* check connection */
if (mysqli_connect_error()) {
    die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
}

if (!mysqli_query($link, "SET a=1")) {
    printf("Errormessage: %s\n", mysqli_error($link));
}

/* close connection */
mysqli_close($link);

?>

To select database and get errormessage

mysqli_select_db($link, "dbname")
OR exit( mysqli_error($link) );

// or like this
echo "trying to connect to the database<br />";
$result = mysqli_select_db($link, "klasbas");
if( !$result ) die("Select database failed: " .mysqli_error($link));

[Perkele]

(Ja, kallt som Perkele. Och i Göteborg är det dessutom fuktig luft.)

There seems to be a bunch of things here for me to understand. But in order not to blow my brain up, I'm asking one thing at a time.

I'm now trying this version, partially influenced by your suggestions:

  <?php

  echo "trying to connect to the database server<br />";
  $link = mysqli_connect("localhost", "root", "password")
  or die("connection to server failed");

  echo "trying to connect to the database<br />"; 
  $result = mysqli_select_db($link, "klasbas")
  or die( mysqli_error($link) );

  ?>

With this code, i get this in the browser:

  trying to connect to the database server
  trying to connect to the database

And as I said before, It seems to establish contact with the server, since if I , for example, change the host name to "localhäst", i get this in the browser:

  trying to connect to the database server
  connection to server failed

As for the latter half of the code, i guess it IS connecting to the database, since when I change the database name (in the code, not in the database) to "klasbus", I get this:

  trying to connect to the database server
  trying to connect to the database 
  Unknown database 'klasbus'

BUT if I change the name of the "$link" variable to "$ullared" (on it's second appearance in the entire code) I get no error! Why?

Am I, with this code, really connected to both the server and the database, and how can I know for sure?

Regards
Klas

[halojoy]

Perkele;10939461 wrote:

Am I, with this code, really connected to both the server and the database, and how can I know for sure?

I think you are really connected. If no error message.
But how can you know for sure?

Maybe by try to creating a table in database.

$sql = "CREATE TABLE users (
id INT NOT NULL AUTO_INCREMENT, 
PRIMARY KEY(id),
 name VARCHAR(30), 
 age INT)";

query($sql)
OR exit( mysql_error() );

echo "Table Created!";

?>

[Perkele]

Tjena!
Thanks a lot for helping me!

If I put your code at the end of mine, like this:

<?php

echo "trying to connect to the database server<br />";
$link = mysqli_connect("localhost", "root", "password")
or die("connection to server failed");

echo "trying to connect to the database<br />";
$result = mysqli_select_db($link, "klasbas")
or die( mysqli_error($link) );

$sql = "CREATE TABLE users (
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
name VARCHAR(30),
age INT)";

query($sql)
OR exit( mysql_error() );

echo "Table Created!";

?>

And point my browser at it, I get an "500 Internal Server Error" exception. (and no table created, of course).

Maybe I was supposed to integrate the new code you provided with the old in some fashion, but I'm not knowledgeble enough, I'm afraid. Not yet anyway. 😉

Any further help appreciated!

Klas

[wuodpamela]

Try this:

$mysqli = new mysqli("localhost", "root", "password", "klasbas");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

[wuodpamela]

You can add the following snipet at the end[as an "else" branch statement];
else{

printf("Host information: &#37;s\n", $mysqli->host_info);

}

[halojoy]

query($sql) should of course be mysqli_query($link, $sql)
my wrong

I am not happy about "500 Internal Server Error"
This is usually not because of php script error, but is the SERVER.
I get this when I have something bad in Apache .htaccess or httpd.conf.

Does you other stuff run without error 500 in your website?

<?php

echo "trying to connect to database server<br />";
$link = mysqli_connect("localhost", "root", "password");
/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

echo "trying to select database<br />";
mysqli_select_db($link, "klasbas")
or die( mysqli_error($link) );

$sql = "CREATE TABLE users (
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
name VARCHAR(30),
age INT)";

mysqli_query($link, $sql)
OR exit( mysqli_error($link) );

echo "Table Created!";

?>