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NME

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Posts posted by NME

  1. ? what do u mean? am i making 2d arrays incorrectly? 

     

    im under the impression that to make 2d arrays in javascript, one must do the following: (ie for a 2x2 array).

     

    var stack = new Array(2);

     

    stack[1] = new Array(2);

    stack[2] = new Array(2);

    stack[1,1]="hello";

    stack[2,1]="world";

    document.write(stack[1,1] + " " + stack[2,1]);

  2. stack[1]=new Array(2);

    stack[2]=new Array(2);

    stack[1,1]="hello";

    stack[2,1]="world";

    document.write(stack[1,1] + " " + stack[2,1]);

     

    quick question.. why does the above write out the following:

     

    "world world";

     

    i need it to write out:

     

    "hello world";

     

    any ideas. thanks.

  3. hey guys, thanks for all the help so far.

     

    but the problem is still there.  here is the updated code:

     

    <html>

    <body>

     

     

    <!-- The data encoding type, enctype, MUST be specified as below -->

    <form enctype="multipart/form-data" action="uploadtest_script.php" method="POST">

        <!-- MAX_FILE_SIZE must precede the file input field -->

        <input type="hidden" name="MAX_FILE_SIZE" value="30000" />

        <!-- Name of input element determines name in $_FILES array -->

        Send this file: <input name="userfile" id="userfile" type="file" />

        <input type="submit" value="Send File" />

    </form>

    </body>

    </html>

     

    and its script:

     

    <?

     

    ini_set("memory_limit", "32M");

    ini_set("max_execution_time", 0);

    error_reporting(E_ALL & ~E_NOTICE);

     

     

    print "size:". $_FILES['userfile']['size'] . " ";

    print "name:". $_FILES['userfile']['name'] . " ";

    print "type:". $_FILES['userfile']['type'] . " ";

    print "tempname:" . $_FILES['userfile']['tmp_name'] . "";

     

    ?>

     

    and its result:

     

    size:0 name:2.mp3 type: tempname:

  4. Hey guys, im trying to upload mp3's onto my site. i know how to do it with images but I'm getting some problems when i select an mp3 file.  I cut the problem down to the following and made it very easy to read:

     

    <html>

    <body>

     

    <form action="uploadtest_script.php" method="post" enctype="multipart/form-data">

    <label for="file">Filename:</label>

    <input type="file" name="file" id="file" />

    <br />

    <input type="submit" name="submit" value="Submit" />

    </form>

     

    </body>

    </html>

     

     

    and the script is here. its a simple print out of all the FILES variables:

     

    <?

    print "name:". $_FILES['file']['name'] . " ";

    print "type:". $_FILES['file']['type'] . " ";

    print "size:". $_FILES['file']['size'] . " ";

    print "tempname:" . $_FILES['file']['tmp_name'] . "";

     

    ?>

     

    and these are the results im getting???:

     

    size:0 name:2.mp3 type: tempname:

     

    Basically, the variables are not being set to the FILES array.  any suggestions? thanks.

  5. Hey guys, I was just wondering if there was any command to jump to a different page without clicking a hyperlink.

    An example is suppose the user clicks a link that leads to a file where there are a few mysql commands, and then i want to go back to the initial page immediately after those commands are completed. I already know about include, but for some reason its not working when i do something like...

    include 'www.blah.com/blah.php?blah=$blahid'.
  6. hmm im still getting an error:

    You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''Info_User_19' ( 'index' INT( 25 ) NOT NULL AUTO_INCREMENT ,

    by the way, thanks so much for all your time poirot.
  7. Ok, this is what is says, but i still cant figure it out:

    You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'int(25) NOT NULL auto_increment, freinds int(25) NOT NULL,
  8. I am trying to dynamically create a table with a different name everytime. The following code is used, but it never seems to actually create a Table when i look at the database.

    [code]$dbhost = '<myhost>';
    $dbusername = '<myusername>';
    $dbpasswd = '<mypassword>';
    $database_name = '<mydatabasename>';

    $connection = mysql_pconnect("$dbhost","$dbusername","$dbpasswd")  
        or die ("Couldn't connect to server.");
    $db = mysql_select_db("$database_name", $connection)
        or die("Couldn't select database.");


    $sql = mysql_query ( "CREATE TABLE user_" . $userid . " (
            index int(25) NOT NULL auto_increment,
            count int(25) NOT NULL,
            PRIMARY KEY (index)
            )");
                    [/code]

    Any ideas why nothing is happening?
  9. This is my code:

    [code]$query= "SELECT userid, first_name FROM users2 WHERE username='$username' AND password='$password'";
    $result = @mysql_query($query);
    $row = mysql_fetch_array($result, MYSQL_NUM); //returns the row with that users information in the users2 table.

    if($row)
        {
            $_SESSION['userid']=$row[0];
            $_SESSION['first_name']=$row[1];
            $_SESSION['last_name']=$row[2];[/code]

    for some reason $row[2] returns nothing (meaning $_SESSION['last_name'] is nothing as well), although $row[1] and $row[0] return the values they intend from my SQL table. why won't row[3] or any subsequent rows show?

    thanks.
  10. [!--quoteo(post=375327:date=May 19 2006, 04:09 PM:name=Crayon Violent)--][div class=\'quotetop\']QUOTE(Crayon Violent @ May 19 2006, 04:09 PM) [snapback]375327[/snapback][/div][div class=\'quotemain\'][!--quotec--]
    [code]
    $sql3 = mysql_query("INSERT INTO '$list1' (user_id, fullname) VALUES ('$userid', '$name')") or die (mysql_error());
    [/code]
    try putting quotes around $list1
    [/quote]

    I already tried that. I get this error:

    You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ''' (user_id, fullname) VALUES ('1121', 'Jeff Gordon')'

    thanks tho
  11. My code looks like this :

    [code]$sql3 = mysql_query("INSERT INTO $list1 (user_id, fullname) VALUES ('$userid', '$name')") or die (mysql_error());[/code]

    However, it does not work and I get the following error:

    You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '(user_id, fullname) VALUES ('1121', 'Jeff Gordon')'

    can anyone tell me what I'm doing wrong. Any help would be greatly appreciated. Thanks
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