8 replies to this topic

[larrywcc]

Hello,

I tried without luck and my code well I scraped it. It was more of a mess then anything.
If someone can point me in the right direction, please.....

What I want to do is I do not want anyone to register on my site but I would like to send out an INVITE with a ID.

Email would be something like this.
Please use this code 1234567890 to register with our site etc.....

I have a database with the fields
invite,date,tries,active

The part I am having trouble with is the user entering the ID and updating the DB

I am able to connect to the data base and enter the code but updating the db is the issue. I need to check the DB to make sure the ID is still active return ID has been already used. If active then continue with script or go to register.php

invite - the id code
Tries just to see how many people have tried that code
date to when it went from active to not active.
active- active or not active

I only want the code to be only used once. As for entering the code into theDB from the admin I can do this by the backend and someday will write the code.

This sounds simple but just can not get it to work and like I said I would post the code but it would take more time to figure out what I am doing wrong then it is worth.

Thanks for any HELP, I tried looking for php code around the web and if I found something close it had more features but then I tried to take it out just lead to other issues.

[Alexander]

I've written really the base logic for what your script is wanting to do, I am unsure if it works as is, although it does show you how to update time, iterate tries and deactivate it once activated.

$usercode = mysql_real_escape_string( $_POST['invite'] );
$usercode_query = mysql_query( "SELECT id, active FROM table WHERE invite = '$usercode'" );

//if exists
if( mysql_num_rows( $usercode_query ) ) {
    $usercode_row = mysql_fetch_row( $usercode_query );
    //if active
    if( $usercode[1] == '1' ) { 
      echo "Your invite has been accepted";
      mysql_query( "UPDATE table SET active = 0 AND date = NOW() WHERE invite = '$usercode'" );
   } else {
   //if not active
      echo "The invite you have entered had already been used.";
      mysql_query( "UPDATE table SET tries = tries + 1 WHERE invite = '$usercode'" );
   }
} else {
  //if not exists
  echo "The invite you have entered was invalid";
}

You'll need to read up on datetime columns and also defaults (tries must be not null, and default to zero if you want it to iterate for example)

You could even generate invites in batches with your own makeshift hashing algorithm (such as half of a salted sha1 hashes of some known numbers beforehand, otherwise it is tedious)

[larrywcc]

Thanks, That looks better then mine :) I used to many if statements and believe that is the place I was lost.

[larrywcc]

Thanks for the code, have an issue with the line $usercode_row = mysql_fetch_row( $usercode_query );

The invite code does get passed on, so that is not the problem, the connecting to database has no errors. Not sure what to look at next. Thanks!


Get the error Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/public_html/register1.php on line 16

<?php


    $dbHost = "localhost";        //Location Of Database usually its localhost

    $dbUser = "XXXX";            //Database User Name

    $dbPass = "XXXX";            //Database Password

    $dbDatabase = "XXXX";    //Database Name

  $db = mysql_connect($dbHost,$dbUser,$dbPass)or die("Error connecting to database.");

  mysql_select_db($dbDatabase, $db)or die("Couldn't select the database.");

 

    

$usercode = mysql_real_escape_string( $_POST['invite'] );

$usercode_query = mysql_query( "SELECT id, active FROM table WHERE invite = '$usercode'" );




if( mysql_num_rows( $usercode_query ) ) {

    $usercode_row = mysql_fetch_row( $usercode_query );

    //if active

    if( $usercode[1] == '1' ) { 

      echo "Your invite has been accepted";

      mysql_query( "UPDATE table SET active = 0 AND date = NOW() WHERE invite = '$usercode'" );

   } else {

   //if not active

      echo "The invite you have entered had already been used.";

      mysql_query( "UPDATE table SET tries = tries + 1 WHERE invite = '$usercode'" );

   }

} else {

  //if not exists

  echo "The invite you have entered was invalid";

}  

?>

Edited by Orjan, 03 April 2011 - 03:55 PM.
use php or code tags around your code

[Mark Wylde]

Try using mysqli instead of mysql. I think I remember having trouble with this a while back.

<?php


$dbHost = "localhost"; //Location Of Database usually its localhost

$dbUser = "XXXX"; //Database User Name

$dbPass = "XXXX"; //Database Password

$dbDatabase = "XXXX"; //Database Name

$db = mysqli_connect($dbHost,$dbUser,$dbPass)or die("Error connecting to database.");

mysqli_select_db($dbDatabase, $db)or die("Couldn't select the database.");



$usercode = mysqli_real_escape_string( $_POST['invite'] );

$usercode_query = mysqli_query( "SELECT id, active FROM table WHERE invite = '$usercode'" );




if( mysqli_num_rows( $usercode_query ) ) {

$usercode_row = mysqli_fetch_row( $usercode_query );

//if active

if( $usercode[1] == '1' ) { 

echo "Your invite has been accepted";

mysqli_query( "UPDATE table SET active = 0 AND date = NOW() WHERE invite = '$usercode'" );

} else {

//if not active

echo "The invite you have entered had already been used.";

mysqli_query( "UPDATE table SET tries = tries + 1 WHERE invite = '$usercode'" );

}

} else {

//if not exists

echo "The invite you have entered was invalid";

} 

?>

[larrywcc]

Now I get the error

Warning: mysqli_select_db() expects parameter 1 to be mysqli, string given in /home2/public_html/jobs/register1.php on line 8
Couldn't select the database.

That is mysqli_select_db($dbDatabase, $db)or die("Couldn't select the database.");

[Orjan]

The PHP Manual is your best friend. Look the function up at PHP: Hypertext Preprocessor and you'll see that your parameters isn't in the right order...

[larrywcc]

Well looked at PHP Manual and many other examples that are close but I can not figure out what function is wrong. Like I say I do not know much about mysql or php but I do get things figure out after awhile except this small script. Just can not figure it. I tried so many options but always get different errors.

Thanks,

[Mark Wylde]

Go and have a look at the PHP website link below.
PHP: mysqli::query - Manual

Use this code and reword it to do what you need:

<?php

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");


/* check connection */

if (mysqli_connect_errno()) {

    printf("Connect failed: %s\n", mysqli_connect_error());

    exit();

}


/* Create table doesn't return a resultset */

if ($mysqli->query("CREATE TEMPORARY TABLE myCity LIKE City") === TRUE) {

    printf("Table myCity successfully created.\n");

}


/* Select queries return a resultset */

if ($result = $mysqli->query("SELECT Name FROM City LIMIT 10")) {

    printf("Select returned %d rows.\n", $result->num_rows);


    /* free result set */

    $result->close();

}


/* If we have to retrieve large amount of data we use MYSQLI_USE_RESULT */

if ($result = $mysqli->query("SELECT * FROM City", MYSQLI_USE_RESULT)) {


    /* Note, that we can't execute any functions which interact with the

       server until result set was closed. All calls will return an

       'out of sync' error */

    if (!$mysqli->query("SET @a:='this will not work'")) {

        printf("Error: %s\n", $mysqli->error);

    }

    $result->close();

}


$mysqli->close();

?>

Use MySQL Improve Extention and not the old one.