2

I'm curious as to why I'm getting an error on something I've done a million times before but am all of a sudden on a certain script getting an error 'Undefined variable: row'

Yet row seems defined to me...

$sql = 'SELECT * FROM table WHERE id="1" LIMIT 1 ';

$res = mysql_query($sql);

    if(mysql_num_rows($res) != FALSE) {

    $row = mysql_fetch_array($res);

    }

The above is pseudo sql... but I've checked that sql statement and I know its bringing out a result. I also know that $row is storing the data because if I go

echo $row[0];

I get the right data.

So to my knowledge, the $row variable is defined. Yet still - an error. Am I losing my mind or what am I missing here? Shouldn't this error/notice only occur if $row didn't exist?

edit

Sorry guys its all happening INSIDE the if statement:

$sql = 'SELECT * FROM table WHERE uID="' . $ID . '" LIMIT 1 ';

$res = mysql_query($sql);

if(mysql_num_rows($res) != FALSE) {

    $row = mysql_fetch_array($res);

$firstName = $row[0];

$lastName = $row[1];

$email = $row[2];

}

edit 2

if i do a print_r($row) I get the following:

Array
(
[0] => Robert
[firstName] => Robert
[1] => Nibbles
[lastName] => Nibbles
[2] => [email protected]
[email] => [email protected]
)
Undefined variable: row
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8
  • 2
    could you add the part of code that yields the mentioned error ? Commented Sep 27, 2009 at 7:23
  • 1
    Indeed, where is the error occurring? If it's outside the 'if' statement, then it's undefined. Commented Sep 27, 2009 at 7:37
  • inside the IF statement is the only place where $row is being used. Commented Sep 27, 2009 at 8:04
  • it certainly looks like you have a stray $row somewhere else... Commented Sep 27, 2009 at 8:29
  • thats what I thought as well but on doing a search in the file there is 4 instances and all 4 are in that if statement! i'm losing my marbles. Commented Sep 27, 2009 at 8:37

2 Answers 2

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8

If you don't initialize $row to something outside that if statement, then it's undefined.

Otherwise, if you don't want to initialize $row to some kind of null value (not entirely unreasonable), you might want to surround any code that checks $row outside of the if statement with something like

if (isset($row))
  doSomething();

It's a pain, but you've just got to remember that any variables you don't define explicitly, even to null, are undefined and can lead to a runtime error if referenced as an rvalue in code (except in isset etc.). So in general, either always initialize your variables or liberally apply code like the above.

I apologize if this turns out not to be the issue, but I can't think of anything more than this without seeing your code.

EDIT: Sorry, it's "isset" not "defined". Been a while since I've actualy worked with PHP. I tried to answer the question with a concept, not syntax. My mistake.

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2 Comments

defined doesn't do what you think here. You probably mean isset.
Damnation, that is precisely what I meant. Edit forthcoming in seconds.
1

Offtopic, but I recommend using mysql_fetch_assoc() instead of mysql_fetch_array, then you can use the actual field names in your code, instead of some arbitrary numbers.

print $row[2]

vs

print $row['email];
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2 Comments

cheers. but can't I do $row['email'] with mysql_fetch_array() as well can't I? I've just used $row[1] to for debugging purposes to make sure i haven't named the keys incorrectly or something
According to the mysql_fetch_array() page (us2.php.net/mysql_fetch_array), you can use either (the array has both numeric and associative indexing). Personally, I would still favor mysql_fetch_assoc() in case that reduces overhead by only allowing one form of indexing. But I don't know the ins and outs of PHP well enough to know if this is worthwhile or not.

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