[Resolved] Cannot connect to database.

micsand

I have installed Apache, MySQL and PHP to my computer using EasyPHP 1.7. There wasn't a problem with setting it up and after making the necessary changes per their instructions, everything seems to be working fine. I can connect to the MySQL server without a problem. I've added my database and use a form that I have setup on my website and know that it will contact the database. Unfortunately, it is not contacting the database on my computer.

I know that this form works and have used it to contact the same database when I have had Apache 2, PHP 5, and MySQL installed using the PHP Home Edition 2 program. Okay, you may ask "why don't you use PHP Home Edition 2?" I would love to except that it is not designed for my OS (Windows ME), although, it does run on my system. It is designed to run as a service and the problem is that I cannot shutdown Apache properly from its interface or from a command line and I end up having to shut the entire computer down improperly. So, now EasyPHP does all that I require. Any thoughts to why I can't connect to the database?

This is the actual code that I use:

<?php

if($submit)

{

$db = mysql_connect("localhost", "root","********");

mysql_select_db("db_nacx****",$db);

$sql = "INSERT INTO mailinglist (firstname, lastname,email) VALUES ('$firstname', '$lastname', '$email')";

$result = mysql_query($sql);

echo ("Thank you $firstname $lastname! Your Information Has Been Entered. ");
}

else

{

echo ("$firstname $lastname! Your Information Has Not Been Entered. ");

}

As you can see the else statement is excuted since no connection can be made. My thought on the matter is that there is an issue with the password. Where else besides config.inc file that passwords are stored?

mdsent@hotmail.com

wilsonodk

Try putting a error catching command in there. Like this...

$db = mysql_connect("localhost", "root","********");
$db_select = mysql_select_db("db_nacx****",$db) or die(mysql_error());

That should generate an error that will indicate where the problem is.

I also like to store the result from a select database call in a variable, so I made that change as well.

micsand

Thanks. I will try that...

tsinka

Hi,

looking at your code I think the else statement gets executed because $submit isn't set

if ($submit) {
....
} else {
....
}

If that variable comes from a HTML form and register_globals is set to off then replace $submit by isset($_POST['submit']) and check if that changes anything.

Generally:

Use $POST['variable'] for POST variables and $GET['variable'] for GET variables.

Thomas

micsand

Thank everyone for responding to my post. I finally got the issue resolved. It was so simple. After rewriting the entire code so that it was reporting errors, I received the following message:

"Column count doesn't match value count at row 1"

In the original query it read:

mysql_query("INSERT INTO mailinglist VALUES ('$firstname', '$lastname', '$email', '$ad');

What was happening was that I had omitted the Primary Key. D'oh!!!

Here is the solution to the problem:

mysql_query("INSERT INTO mailinglist VALUES ('', '$firstname', '$lastname', '$email', '$ad')")
or die(mysql_error());

And the entire code looks ike this:

<?php

$firstname = $POST['firstname'];
$lastname = $POST['lastname'];
$email = $POST['email'];
$ad = $POST['adnum'];
$submit = $_POST['submit'];

//Connect to database
mysql_connect("localhost", "*", "") or die(mysql_error());
echo "Connected to MySQL ";
mysql_select_db("db_******") or die(mysql_error());
echo "Connected to Database ";

if (isset($_POST['submit']))
{
mysql_query("INSERT INTO mailinglist VALUES ('', '$firstname', '$lastname', '$email', '$ad')")
or die(mysql_error());

echo ("Thank you $firstname $lastname! Your Information Has Been Entered. ");
}

else

{

?>
<?

}

Thank you once again every one. 🙂

bpat1434

Out of curiosity, why did you not structure your SQL statement like below:

"INSERT INTO mailinglist (firstname, lastname, email, ad) VALUES ('$firstname', '$lastname', '$email', '$ad')"

That would ensure that you always put the correct information in the correct spot.

Would seem much easier to do that and you won't have to worry about primary keys & such....

just a thought.

~Brett

micsand

If you notice in my posting you will see that the original statement was structured that way and it worked perfectly on my remote server. However, there was problems with the code that allowed a download page when the information was not entered into the database. So, I needed to correct that problem and for some reason beyond my present capacity of knowledge, it refused to work on my computer and would not insert data into the database. So, after struggling with this for several weeks, including removing and reinstall Apache, PHP, and MySQL, I began to approach the problem differently. Since there was no need to initialize any error reporting for this script while it was on my remote server and I never ensured that a connection was established with the remote database, it had to be done here. Thus, the error message, "Column count doesn't match value count at row 1". Since the database table incorporates 5 columns, logic dictated that the primary had to be the problem. Now, I am in the process of investigating why this turned out to be an issue.