Error inserting form data into database via PHP and SQL

I have a form that is using the following for action in a separate PHP file. It basically connects to the database and inserts the entered information from the form into the database.

The form fields are named: link,name,title,date. It is connecting fine to the database, however it always results to "Error Entering New News Link Code 1" when trying to fill out the form (add new entry to database). I'm pretty sure the syntax for the SQL Insert command is correct. Is there something I'm not doing right?

				$db = mysql_connect("localhost:8443","username,"password");


				mysql_select_db("database",$db);
						
				$date = mktime(0,0,0,$month,$day,$year);
				
				$Query = "INSERT INTO links " .
					"(url,name,description,date) " .
					"VALUES(\"$link\",\"$name\",\"$title\",$date)";

				if(!($result = mysql_query($Query,$db)))
					print("<p><font face=\"Arial, Helvetica, sans-serif\" class=\"header2\" align=center>Error Entering New News Link Code 1.</font></p>");
				else
					print("<p><font face=\"Arial, Helvetica, sans-serif\" class=\"header2\" align=center>News link added to database successfully.</font></p>");

Code (markup):

I can not replicate it, but once during my troubleshooting it claimed it had sucessfully added the link. However I then went in the database and it appeared that all fields for the entry were empty, except for the date which had a value of "-1".

 

replace your query to this.

$Query = "INSERT INTO links " .
"(url,name,description,date) " .
"VALUES(\"$link\",\"$name\",\"$title\",\"$date\")";

 

You are using an assignment operator instead of a comparison operator in your if statement. It should look like this:

if($result != mysql_query($Query,$db))
  print("<p><font face=\"Arial, Helvetica, sans-serif\" class=\"header2\" align=center>Error Entering New News Link Code 1.</font></p>");
else
  print("<p><font face=\"Arial, Helvetica, sans-serif\" class=\"header2\" align=center>News link added to database successfully.</font></p>");

Code (markup):

What you were doing is setting $result equal to mysql_query($Query, $db), so you were always taking the first path. If you use if($result != mysql_query($Query, $db)) then the comparison should work.

Check out http://us3.php.net/manual/en/language.operators.comparison.php for details on comparison operators.

 

Thanks. Changing it to a conditional statement solved the error message. It now says that it has submitted successfully (even though it doesn't).

The second problem is still an issue. When I check the database it did create a new entry with an ID#. However all fields (url, name, and description) have no values and the date is set to "-1".

 

if(!($result = mysql_query($Query,$db)))

That line is still correct, that is not a comparing process. It's an assigning the variable $result to the resultset that will return from the query.

pls post your table structure.

 

Here is the table structure.

Field Type Null Default
ID int(11) No
url varchar(255) No
name varchar(100) No
description varchar(255) No
archived smallint(6) No 0
date int(11) No 0

The form uses 3 fields:
textarea with name="title"
input name="link"
input name="name"

and three additional date fields.
select name=month
select name=day
select name=year

The form action is most of the code from the first post in an external php file.

 

I tried changing "VALUES(\"$link\",\"$name\",\"$title\",\"$date\")"; to "VALUES('$link','$name','$title','$date')";, however I am still getting null values entered into the database. Any suggestions?

 

Make sure that your id field is auto_increment

Your date data type is integer, so we don't need quote for date.

$Query = 'INSERT INTO links (url,name,description,date) VALUES("'.$link.'","'.$name.'","'.$title.'",'.$date.')';

 

Everytime form data was used as a variable, it claimed the variable was undefined even though the field's name was set in the form.

I found http://matthom.com/archive/2005/02/19/php-passing-variables-across-pages, so I added the following to my code. Everything works great now.

$title = $_POST['title'];
$link = $_POST['link'];
$name = $_POST['name'];
$month = $_POST['month'];
$day = $_POST['day'];
$year = $_POST['year'];
$result = "";

PHP: