Resource id #31 ? Connecting to database?

**lord_of_the_ring** · Jun 27, 2007 16:42

I am a complete newbee and i am trying to connect to my sqldatabase.
Previously the connection to sql worked but when i try to connect to sql and my database & echo a confirm i get the following appear on my screen.
Resource id #31

What is this?

My code is below,

<?php
$connect_to_sql = mysql_connect("localhost", "username", "password")or die
("Could not connect");
if ($connect_to_sql)
{$sql_confirmed = "SQL Connected";
}

$connect_to_database = mysql_select_db("tablename") or die
("Could not connect to the database");
if ($connect_to_database)
{$database_confirmed = "Database Connected";
}
?>
<html>
<head>
<title>Untitled</title>
</head>
<body>
<?php
echo "$connect_to_sql";
echo "$connect_to_database";
mysql_close();
?>
</body>
</html>

**Servyces** · Jun 27, 2007 16:57

You are trying to echo the connect commands. This won't work, here's what you should be doing instead:

PHP Code:


<?php
$connect_to_sql = mysql_connect("localhost", "username", "password") or die("Could not connect");
if ($connect_to_sql) {
    echo "SQL Connected<br>";
}

$connect_to_database = mysql_select_db("tablename") or die("Could not connect to the database");
if ($connect_to_database) {
    echo "Database Connected<br>";
}
?>

If it connects, it echo's the OK message. If it doesn't connect it dies with the die message you already specified.

**Mittineague** · Jun 27, 2007 17:01

Hi lord_of_the_ring, welcome to the forums,
mysql_connect returns a resource http://us2.php.net/mysql_connect

Return Values
Returns a MySQL link identifier on success, or FALSE on failure.

if you echo your var and see "resource" that means the cnx was successful.

**hexburner** · Jun 27, 2007 17:06

First of all, the function mysql_connect() returns a resource, which is the connection to the database server. And the function mysql_select_db() returns a boolean to check if the database can be used or not to extract, alter or add data to the database.

Secondly, mysql_select_db() selects a database, not a table.
To get data from the database, you'll need to query the database.
For example:

Code PHP:

<?php
    // This query gets all rows and columns from the table TableName
    $result = mysql_query('Select * from TableName');
?>

The function mysql_query() returns another resource, which contains all the rows and columns returned by the database that meet your query.
To then use the returned results, you'll have to walk through the returned resource:

PHP Code:


<?php
    // This is the easiest way to learn
    for ($row = mysql_fetch_array($result)) {
       echo $row['FieldName1'];
        echo $row['FieldName2'];
    }
?>

The function mysql_fetch_array() returns an array with the fieldnames as keys of the array -you can get the values by their corresponding numeric keys too- and the values as of the fields as the value linked to their corresponding array key.
Every time you call mysql_fetch_array($result), you'll get one row from the resource the database returned. If there are no more rows available in the resource, the for-loop will be ended.

**lord_of_the_ring** · Jun 27, 2007 17:34

Thanks guys... i managed to get it working but have run into a problem trying to call the data from the table...

I followed hexburners advice but i get the following error,

Parse error: syntax error, unexpected T_STRING in db_connect_new.php on line 25

line 25 starts with "$result = mysql_query('SELECT *
FROM `tablename`);"

My code is below... any ideas what i have done wrong this time?

PHP Code:


<html>
<head>
<title>Untitled</title>
</head>
<body>
<?php
$connect_to_sql = mysql_connect("localhost", "username", "password")or die
("Could not connect");
if ($connect_to_sql)
{echo "SQL Connected";
}

$connect_to_database = mysql_select_db("tablename") or die
("Could not connect to the database");
if ($connect_to_database)
{echo "Database Connected";
}
$result = mysql_query('SELECT *
FROM `tablename`);
for ($row = mysql_fetch_array($result)) {
       echo $row['FieldName1'];
        echo $row['FieldName2'];
    }
?>
</body>
</html>

**Mittineague** · Jun 27, 2007 20:12

The quote delimiters should be in pairs.

PHP Code:


$result = mysql_query('SELECT * 

FROM `tablename`);

**hexburner** · Jun 28, 2007 01:19

Originally Posted by Mittineague

The quote delimiters should be in pairs.

PHP Code:


$result = mysql_query('SELECT * 
FROM `tablename`);

I don't think that's enough info for lotr

PHP Code:


// Some examples:
$result = mysql_query('Select * from tablename');
$result = mysql_query('Select * from \'tablename\'');
$result = mysql_query("Select * from 'tablename'")

**earl-grey** · Jun 28, 2007 04:23

Originally Posted by Mittineague

The quote delimiters should be in pairs.

PHP Code:


$result = mysql_query('SELECT * 

FROM `tablename`);

Your forgot a closing quote:

PHP Code:


$result = mysql_query('SELECT * 

FROM `tablename`');

**lord_of_the_ring** · Jun 28, 2007 09:41

I tried all 3 variations of that but i still get the same error...
Parse error: syntax error, unexpected ')', expecting ';' in db_connect_new.php on line 23

My code is now,

PHP Code:



<html>

<head>

<title>Untitled</title>

</head>

<body>

<?php

$connect_to_sql = mysql_connect("localhost", "username", "password")or die

("Could not connect");

if ($connect_to_sql)

{echo "SQL Connected";

}



$connect_to_database = mysql_select_db("tablename") or die

("Could not connect to the database");

if ($connect_to_database)

{echo "Database Connected";

}

$result = mysql_query("SELECT *

FROM 'tablename'");

for ($row = mysql_fetch_array($result)) {

       echo $row['FieldName1'];

        echo $row['FieldName2'];

    }

?>

</body>

</html>

**byron3@earthlink** · Jun 28, 2007 10:06

PHP Code:


while ($row = mysql_fetch_array($result)) {

       echo $row['FieldName1'];

       echo $row['FieldName2'];

 }

Use a while loop, a for each loop is expecting an array.

**lord_of_the_ring** · Jun 28, 2007 10:16

I did try that but i now i get the following error.. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in db_connect_new.php on line 23

**byron3@earthlink** · Jun 28, 2007 10:20

That message means your sql query is not returning a valid recordset. Is 'tablename' the name of the actual table you are retrieving data from ?

**lord_of_the_ring** · Jun 28, 2007 10:22

It is defintely the correct name... i can even insert into it but just can't seem to list the results using this code.

**lord_of_the_ring** · Jun 28, 2007 10:41

Its ok.. i got it working, it was to do with the line above.

Thanks

**byron3@earthlink** · Jun 28, 2007 10:41

Is your database named 'tablename' also ?

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