Strange SQL error

**amy.damnit** · Oct 11, 2009 16:37

I just ran my Account Activation scripts in PHP, and I haven't changed anything that I can think of since last night, but now got a *weird* SQL error...

An error occurred in script '/home/content/76/5060376/html/106_ActivateAccount.php' on line 72: Object of class mysqli_result could not be converted to string
Date/Time: 10-11-2009 15:15:41

Array
(
[GLOBALS] => Array
*RECURSION*
[_POST] => Array
(
)

[_GET] => Array
(
[x] => test@mail.com
[y] => f85e70af6307c75284af7b68453856dc
)

[_COOKIE] => Array
(
[PHPSESSID] => 9ghisl9pfjhks21cvfefo1sjm3
)

[_FILES] => Array
(
)

[_SESSION] => Array
(
[email] => test@mail.com
[first_name] => Amy
[account_results] =>

Here is my PHP code...

Code:

<?php # Script 16.7 - activate.php
  // This page activates the user's account.

  // Address error-handling.
  error_reporting(E_ALL & ~E_NOTICE);

  // Include Configuration Settings.
  require_once ('includes/config.inc.php');

  // Start output buffering.
  ob_start();

  // Initialize a session.
  session_start();

  // Validate $_GET['x'] and $_GET['y']:
  // Assume value are invalid.
  $x = FALSE;
  $y = FALSE;

  // Validate E-mail address.
  if (isset($_GET['x'])
        && preg_match ('/^[\w.-]+@[\w.-]+\.[A-Za-z]{2,6}$/', $_GET['x']) ) {
          $x = $_GET['x'];
  }

  // Validate Authorization Code.
  if (isset($_GET['y']) && (strlen($_GET['y']) == 32 ) ) {
          $y = $_GET['y'];
  }

  // Validate both URL values.
  if ($x && $y) {
      // Authorization values valid.

      // Path to DB Connection Script.
      require_once (MYSQL);

      //************************************************************************
      // Determine User's Name based on Email in URL.
      // Build query.
      $name = "SELECT first_name " .
              "FROM users " .
              "WHERE (email = '" . mysqli_real_escape_string($dbc, $x) .
              "') LIMIT 1";

      // Run query.
      $r1 = mysqli_query ($dbc, $name)
              or trigger_error("Query: $name\n<br />MySQL Error: " . mysqli_error($dbc));

      //************************************************************************
      // Remove Authorization String from database.
      // Build query.
      $auth = "UPDATE users " .
              "SET active = NULL " .
              "WHERE (email = '" . mysqli_real_escape_string($dbc, $x) .
              "' AND active = '" . mysqli_real_escape_string($dbc, $y) .
              "') LIMIT 1";

      // Run query.
      $r2 = mysqli_query ($dbc, $auth)
              or trigger_error("Query: $auth\n<br />MySQL Error: " . mysqli_error($dbc));

      //************************************************************************
      // Verify record was changed.
      if (mysqli_affected_rows($dbc) == 1) {
          // Insert succeeded.
          //echo "<h3>Your account is now active. You may now log in.</h3>";

          // Create "Account Validated" message.
          $_SESSION['validation_results'] =
            "<span id=\"congrats\">Congratulations, " . $r1 .
            ", your account is now active!!</span>" .
            "<span id=\"congrats_msg\">You may log in.</span>";

      } else {
          // Insert failed.
          //echo '<p class="error">Your account could not be activated. Please re-check the link or contact the system administrator.</p>';

          // Create Error message.
          $_SESSION['validation_results'] =
            "<span id=\"congrats\">Sorry, " . $r1 .
            ", your account could not be actived!!</span>" .
            "<span id=\"congrats_msg\">Please re-check the link or contact the system administrator.</span>";
      }

      // Close Database Connection.
      mysqli_close($dbc);

  } else {
      // Authorization values invalid.

      // Redirect.
      $url = BASE_URL . 'index.php'; // Define the URL:
      ob_end_clean(); // Delete the buffer.
      header("Location: $url");
      exit(); // Quit the script.

  }
?>

I don't see any code anywhere near Line 72 that looks wrong?!

UGH!!

Any idea what is wrong??

Amy

**amy.damnit** · Oct 11, 2009 16:50

Okay, I think I found the error...

Now I need help fixing it!!!

How do I assign the results of a query to a variable??

My SELECT query finds the person of the name activating their account, and I want to assign that name to $name so I can print a customized message.

Here is my code:

Code:

      // Determine User's Name based on Email in URL.
      // Build query.
      $name = "SELECT first_name " .
              "FROM users " .
              "WHERE (email = '" . mysqli_real_escape_string($dbc, $x) .
              "') LIMIT 1";

      // Run query.
      $r1 = mysqli_query ($dbc, $name)
              or trigger_error("Query: $name\n<br />MySQL Error: " . mysqli_error($dbc));

which apparently is wrong!

Amy

**r937** · Oct 11, 2009 17:33

wish i could help, but there's nothing here that looks like a mysql error

**amy.damnit** · Oct 11, 2009 17:41

Originally Posted by r937

wish i could help, but there's nothing here that looks like a mysql error

Sure there is?!

See my follow-up post.

I was trying to print out $r1 which is a pointer to my query results (i.e. an Object).

And in my follow-up post, I was asking if there is a simple way to assign the results set of a simple SELECT query to a single variable?

For instance, if...

SELECT name
FROM users
WHERE email='somebody@mail.com';

returned a single value of "Amy" then what is the easiest way to assign "Amy" to a variable like $name ???

I know there is...

mysqli_fetch_array ($r, MYSQLI_ASSOC);

but I don't have an array, per se, just a single value.

Follow me?

Amy

**r937** · Oct 11, 2009 17:57

Originally Posted by amy.damnit

Follow me?

not really, sorry

you're speaking php and this is the mysql forum

**amy.damnit** · Oct 11, 2009 18:05

Originally Posted by r937

not really, sorry

you're speaking php and this is the mysql forum

You know MySQL, but don't know PHP?!

Amy

**crmalibu** · Oct 11, 2009 18:20

Originally Posted by amy.damnit

what is the easiest way to assign "Amy" to a variable like $name ???

I know there is...

mysqli_fetch_array ($r, MYSQLI_ASSOC);

but I don't have an array, per se, just a single value.

A successful select result conceptually returns 1 or more rows. Even if it's only 1 row, and that row only has the value of a single column, that's the format.

I tend to usually just fetch things as an associative array because that's the most common format I want, and it's consistent.

PHP Code:


$row = mysqli_fetch_assoc($r);

echo $row['name'];

But if you like syntax,

PHP Code:


list($name) = mysqli_fetch_row($r);

// or 

extract(mysqli_fetch_assoc($r));

**amy.damnit** · Oct 11, 2009 18:46

Originally Posted by crmalibu

A successful select result conceptually returns 1 or more rows. Even if it's only 1 row, and that row only has the value of a single column, that's the format.

I tend to usually just fetch things as an associative array because that's the most common format I want, and it's consistent.

PHP Code:


$row = mysqli_fetch_assoc($r);

echo $row['name'];

But if you like syntax,

PHP Code:


list($name) = mysqli_fetch_row($r);

// or 

extract(mysqli_fetch_assoc($r));

Yah, here is what I ended up doing...

Code:

      // Determine User's Name based on Email in URL.
      // Build query.
      $qry01 = "SELECT first_name " .
              "FROM users " .
              "WHERE (email = '" . mysqli_real_escape_string($dbc, $x) .
              "') LIMIT 1";

      // Run query.
      $result01 = mysqli_query ($dbc, $qry01)
                      or trigger_error("Query: $qry01\n<br />MySQL Error: " . mysqli_error($dbc));

      // Assign query results to variable.
      list($name) = mysqli_fetch_row($result01);

Can you highlight the difference between your code...

$row = mysqli_fetch_assoc($r);
echo $row['name'];

and this...

mysqli_fetch_array ($r, MYSQLI_ASSOC);

Thanks,

Amy

**r937** · Oct 11, 2009 19:04

Originally Posted by amy.damnit

You know MySQL, but don't know PHP?!

this is correct

i've used coldfusion for many years, and see no need to learn an inferior language lol

**crmalibu** · Oct 11, 2009 19:17

mysqli_fetch_assoc($r);
is equivalent to
mysqli_fetch_array ($r, MYSQLI_ASSOC);

**amy.damnit** · Oct 11, 2009 19:25

Originally Posted by r937

this is correct

i've used coldfusion for many years, and see no need to learn an inferior language lol

Trying to expand your circle of friends, eh??

(I'll be sure not to stand next to you near the PHP forum!)

Amy

**amy.damnit** · Oct 11, 2009 19:26

Originally Posted by crmalibu

mysqli_fetch_assoc($r);
is equivalent to
mysqli_fetch_array ($r, MYSQLI_ASSOC);

So if my query returned 2 rows, what would happen to your code??

Code:

$row = mysqli_fetch_assoc($r);
echo $row['name'];

Amy

**crmalibu** · Oct 11, 2009 19:42

The first row would be fetched. The result set's hidden internal row pointer would be advanced to the next row.

**amy.damnit** · Oct 11, 2009 20:13

Originally Posted by crmalibu

The first row would be fetched. The result set's hidden internal row pointer would be advanced to the next row.

So if you had a query that returned only ONE Column, but MANY records, then you could use this code/function, but you would just have to use a loop or something, right?

And, for cases like mine, then you can use your coding approach or the way I did it with List ( ) , right?

Amy

**Dan Grossman** · Oct 11, 2009 21:41

PHP Code:


$name = mysql_result($r,0);

Or more simply

PHP Code:


$name = mysql_result(mysql_query("SELECT name..."),0);

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