Q about the depreciated mysql_db_query

Anon

Man what I hard row to ho, I didnt know what I was getting into when I started messing with PHP.

Ive gotten to the part where Im connecting to databases, the part Im REALLY interested in, and wouldnt ya know it.....stuck!

The mannual I am using now uses the old function mysql_db_query, which is now gone bye bye. Could someone point me to an expamle of using the replacements functions, mysql_query and mysql_db_select? Complete with proper string variables and all that good stuff? Of if anyone knows off hand could they impart this knowledge to me in a reply.

For clarity sake, Ill include the code that has me stuck: (This code sends data to a database via form entries)
<HEAD>
<TITLE>Creating a Table</TITLE>
</HEAD>
<BODY>
<?php
$Host = "localhost";
$User = "administrator";
$Password = "";
$DBName = "NewDatabase";
$TableName = "Feedback";
$Link = mysql_connect ($Host, $User, $Password);
$Query = "CREATE table $TableName (id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
FirstName TEXT, LastName TEXT, EmailAddress TEXT, Comments TEXT)";
if (mysql_db_query ($DBName, $Query, $Link)) {
print ("The query was successfully executed! \n");
} else {
print ("The query could not be executed! \n");
} }
mysql_close ($Link);
?>
</BODY>

Sorry for the length =(

riyaz123

U can use this code.

<HEAD>
<TITLE>Creating a Table</TITLE>
</HEAD>
<BODY>
<?php
$Host = "localhost";
$User = "administrator";
$Password = "";
$DBName = "NewDatabase";
$TableName = "Feedback";
$Link = mysql_connect ($Host, $User, $Password);
$Query = "CREATE table $TableName (id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
FirstName TEXT, LastName TEXT, EmailAddress TEXT, Comments TEXT)";
if (mysql_query ($DBName, $Query, $Link)) {
echo ("The query was successfully executed! \n");
} else {
echo ("The query could not be executed! \n");
} }
mysql_close ($Link);
?>
</BODY>

I think this help u

Riyaz

grail

This code returns this error :
Warning: Wrong parameter count for mysql_select_db() in C:\Inetpub\wwwroot\php\HandleDBform.php on line 20

Where should the mysql_select_db() function be placed in the code, since its looking for it on line 20 Id assume there, but I cant seem to figure out the parameters it is requesting. The database's name is simply "NewDatabase" and this name has been assgined a variable, however placing that variable within the functions parentheses, does not seem to solve the problem. Thanks for the help Riyaz, you wouldnt happen to have any thoughts on this one would you?

riyaz123

Sorry.

Use this code.

<HEAD>
<TITLE>Creating a Table</TITLE>
</HEAD>
<BODY>
<?php
$Host = "localhost";
$User = "administrator";
$Password = "";
$DBName = "NewDatabase";
$TableName = "Feedback";
$Link = mysql_connect ($Host, $User, $Password);
$Query = "CREATE table $TableName (id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
FirstName TEXT, LastName TEXT, EmailAddress TEXT, Comments TEXT)";
mysql_select_db($DBName);
if (mysql_query ($Query)) {
echo ("The query was successfully executed! \n");
} else {
echo ("The query could not be executed! \n");
} }
mysql_close ($Link);
?>
</BODY>

Three steps are there.
mysql_connect();
mysql_select_db();
mysql_query();
I specified all parameter in ur code.
Hope it will work.

Riyaz T