Nested querys not working

[surfgrommett]

Hey guys. This is driving me crazy. I can't see any code problems and I had another friend look and neither can he. EVERYTHING works PERFECTLY...all echos work right and everything and no errors get returned..the problem is $query4 and $query5 appear to do nothing..though i get no error..i get nothing pnput to the database either! i put a bunch of notes in the code so it would be easier to follow. thanks for the help ion advance.

<?php
include 'logincheck.php';
include 'document_type.php';

// get and/or set variables

$last_name=$_GET['item1'];
$first_name=$_GET['item2'];
$story_title=$_GET['item3'];
$chapter=$_GET['item4'];
$subchapter=$_GET['item5'];
$new=$_GET['item6'];
$credit=$_SESSION['username'];

// change story title to all lowercase and remove spaces
// then assign to another variable used to determin which table to use

if (!$function_check) {
$function_check = true;
function clean($string) {
	$non_alphanumeric_chars = "/([^a-zA-Z0-9_])/";
	$string = preg_replace( $non_alphanumeric_chars, "", $string );
	$string = strtolower($string);
	return $string;
}
};
$clean_title = clean($story_title);

// log in to database

$username="**********";
$password="***********";
$database="**************";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

// Check to see if this name is already in the database with a value of yes in the 'first' column

$query1="SELECT recnum FROM mastercharacterlist WHERE lname='$last_name' AND fname='$first_name' AND first='yes'";
$result1=mysql_query($query1) or die(mysql_error());
mysql_close();
if (mysql_num_rows($result1) > 0)
        
        // if $query1 returned a result then we go to the nested statements
        
        {
        $record_number1=mysql_result($result1,0,"recnum");
        if ($new=="yes")
                {
                
                //if the userinput that was retrieved via $_GET is yes then run this code
                
                mysql_connect(localhost,$username,$password);
                @mysql_select_db($database) or die( "Unable to select database");
                $query2="SELECT * FROM basicinfo WHERE recnum='$record_number1'";
                $result2=mysql_query($query2) or die(mysql_error());
                $n1=mysql_result($result2,0,"storyname");
                $n2=mysql_result($result2,0,"chapter");
                $n3=mysql_result($result2,0,"chapter_id");
                echo'
                '.$first_name.' '.$last_name.' has already been entered as having thier first appearance in '.$n1.' Ch. '.$n2.''.$n3.'.
                This character will not be entered at this time. Please make the appropriate corrections and try again. 
                ';
                mysql_close();
                }
                else
                {
                
                // if the userinput that was retrieved via $_GET is no then run this code
                
                mysql_connect(localhost,$username,$password);
                @mysql_select_db($database) or die( "Unable to select database");
                $query3="SELECT recnum FROM basicinfo WHERE storyname='$story_title' AND chapter='$chapter' AND  chapter_id='$subchapter'";
                $result3=mysql_query($query3);
                $record_number3=mysql_result($result3,0,'recnum');
                mysql_close();
                mysql_connect(localhost,$username,$password);
                @mysql_select_db($database) or die( "Unable to select database");
                $query4="INSERT INTO $clean_title (recnum,lname,fname,first,credit) VALUES ('$record_number3','$last_name','$first_name','$new','$credit')" or die(mysql_error());
                echo'
                The following data has been entered:
                <br>
                '.$first_name.' '.$last_name.'
                <br>
                '.$story_title.'
                <br>
                Chapter '.$chapter.''.$subchapter.'
                <br>
                New character? '.$new.'
                ';
                mysql_close();
                }
        } 
        else
        {
        
        // if the value of $query1 was NOT yes...then this is the code to run regardless of what the $new variable is
        $record_number1=mysql_result($result1,0,"recnum");
        mysql_connect(localhost,$username,$password);
        @mysql_select_db($database) or die( "Unable to select database");
        $query5="INSERT INTO $clean_title (recnum,lname,fname,first,credit) VALUES ('$record_number1','$last_name','$first_name','$new','$credit')" or die(mysql_error());
        echo'
        The following data has been entered:
        <br>
        '.$first_name.' '.$last_name.'
        <br>
        '.$story_title.'
        <br>
        Chapter '.$chapter.''.$subchapter.'
        <br>
        New character? '.$new.'
        ';
        mysql_close();
        }

// ending menu 

echo'
<br><br><a href="character_input_last_name.php" class="linkstyle1">Return to Character Input</a>
<br><a href="index.php" class="linkstyle1">Return to Main Timeline</a>
';
?>

[chrishea]

You created the query statements but you didn't actually call MySQL to take action on them. You need a mysql_query statement to execute each query.

[vibhaJ]

$query5="INSERT INTO $clean_title (recnum,lname,fname,first,credit) VALUES ('$record_number1','$last_name','$first_name','$new','$credit')";
mysql_query($query5) or die(mysql_error());

there is no php code to execute query in your given code.
same for $query4.

[surfgrommett]

i KNEW it was something stupid i was missing. i dunno why i didnt see that! thanks guys.

[metalix]

You also do not need to start and close your connection throughout the script. just close at the end