Hi, I use somethin as $sql = 'INSERT INTO `text` (`Field`) VALUES (''aaa'')'; Also, check things with var_dump, use mysql_error, echo $sql/$insert and check whether the query is well formed. my best mario On Sat, 2007-01-27 at 19:55 -0600, Alexander wrote: > I fixed my code and it shows up fine now. It says it supposedly enters the > data when I input the fields but when I check the MySQL database, the info > isn't there. > > Here's the fixed code if you wish to test it: > > <html> > > <body> > > <?php > > function add_to_table( $picture, $thumbnail ) { > $picture = mysql_real_escape_string($picture); > $thumbnail = mysql_real_escape_string($thumbnail); > > $user = "user"; > $pass = "password"; > $database = "mydata"; > > $link = mysql_connect("localhost", $user, $pass); > if(!$link) { > echo 'Error in linking'; > } > > $db = mysql_select_db($database, $link); > > $insert = 'INSERT INTO domains ( picture, thumbnail ) values( > $picture, $thumbnail )'; > $indata = mysql_query( $insert, $link ); > > > } > > function write_form() { > echo "<form method='post' action='{$_SERVER['PHP_SELF']}'>"; > > echo "<p>Enter url of large picture: <input type='text' > name='picture' />"; > > echo "<p>Enter url of thumbnail: <input type='text' > name='thumbnail' />"; > > echo "<p><input type='submit' value='Add to Database' />"; > > echo "</form>"; > } > > if ( !empty( $_REQUEST['picture'] ) && !empty( > $_REQUEST['thumbnail'] ) ) { > $input = add_to_table( $_REQUEST['picture'], > $_REQUEST['thumbnail'] ); > if(!input) { > echo 'Error\!'; > } > else { > echo 'Data enter successfully\!'; > $idreturn = mysql_insert_id($input); > echo 'Image inserted to ID ' . $idreturn; > } > } > else { > write_form(); > } > > ?> > > </body> > > </html> > -- mario <chiari.hm@xxxxxxxxxxx> -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php