funny coding error

[thedevilinu]

I am trying to run the following code

--------------------------

$result = mysql_query("SELECT * FROM indiatutors_profiles WHERE id='$id' ") or die(mysql_error());

while($row=mysql_fetch_array($result)){

$email =$row;

}

echo $email;

----------------------------

It gives me the following error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in website_path/profiles.php on line 7

 

The funny thing about the code is that inspite of giving the error, it outputs the email correctly based upon the id which it shouldn't have done if their was some error

[PFMaBiSmAd]

The actual code inside your while(){} loop is likely reusing and overwriting the $result variable, so when the while(){} loop is evaluated on the second pass through the loop, $result is no longer a result resource from the first query.

[Rifts]

$email =$row;

 

should be

 

$email =$row['email'];

[dragon_sa]

It should be echoed inside the while loop aswell if there is to be more than 1 result, if there is only supposed to be 1 result then there is no need for the while loop at all

[thedevilinu]

@PFMaBiSmAd -

I really expect the loop to run just once as their is only a single id associated with any email.

You are however right in saying that loop is running more than once. I can't fathom why it is running twice.

so changing the code to

------------------------------

$result = mysql_query("SELECT * FROM indiatutors_profiles WHERE id='$id' ") or die(mysql_error());

  while($row=mysql_fetch_array($result)){

      echo $email =$row;

  }

------------------

gives me error

----------------

email@gmail.com

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in website_path/profiles.php on line 7

----------

which confirms that loop runs twice. Bug was using another $result statement in the code due to which it was running again. Thanks for sorting this out for me.

Thanks again

 

@rifts - it works either way mate..my way is just time saving

@dragon - echo in or out of the loop .. either way it shouldn't give an error

 

 

 

[Pikachu2000]

$query = "SELECT * FROM indiatutors_profiles WHERE id=$id"; // Presumes that $id is a numeric value.
$result = mysql_query($query) or die("<br>Query string: $query<br>Produced error: " . mysql_error());
$row=mysql_fetch_assoc($result);
echo $row['email']; // And yes, the array index should be enclosed in quotes. Just because it works without them, doesn't mean it's right.