Using 'or die()' to stop on errors in PHP

Question

Often in PHP, I see:

$result = mysql_query($query) or die();

Coming from python, I know why this should work, because or returns the first value if it is true in a boolean context, and the second value otherwise (see this).

But when I try the above technique in PHP in another context, for example something like:

$name = "John Doe";
echo $name or "Anonymous";

The or doesn't return the first value ("John Doe"), it returns 1.

Why does this work in the mysql_query() result case, but not in other cases? Is it bad to use in a mysql_query() case (ignore the fact that I am not returning a useful error to the user)?

Answer 1

In PHP, variable assignment (the equals sign) and functions both take precedence over the or operator. That means a function gets executed first, then the return value of the function is used in the or comparison. In turn when you use two values/variables together with an or operator, it compares the two values first then returns a Boolean value.

Therefore, the order of evaluation in this example is:

$result = mysql_query($query) or die();

1. mysql_query($query)
   Returns either a result set for DQL queries such as SELECT, or a Boolean value for DDL, DML or DCL queries such as CREATE, DROP, INSERT, UPDATE, DELETE and ALTER.

2. $result = mysql_query($query)
   The result of this query execution is assigned to the variable $result.

3. $result /* = ... */ or die();
   If it's either a result set or true, it's considered true (aka "truthy") so the or condition is satisfied and the statement ends here. Otherwise the script would die() instead.

---

echo is a language construct and therefore doesn't actually return a value, so it doesn't run like a function before the or comparison is made.

As $name or "Anonymous" is always true because the string "Anonymous" is non-empty and therefore truthy, the echo implicitly converts true to 1, hence that output.

The order of evaluation in this example is:

$name = "John Doe";
echo $name or "Anonymous";

1. $name = "John Doe";
   Pretty straightforward — assigns the string John Doe to $name.

2. $name or "Anonymous"
   PHP discovers that $name contains the string John Doe, so what ends up being evaluated is the following:

3. "John Doe" or "Anonymous"
   Since at least one string is non-empty here, it's considered truthy and the condition is satisfied. This evaluation then returns true.

4. echo true /* $name or... */;
   Converts true to 1 and prints the number 1.

Answer 2

The reason actually occurred to me shortly after asking the question. It's about operator precedence. The = happens before the or, so:

$result = mysql_query($query) or die();

is equivalent to:

($result = mysql_query($query)) or die();

not:

$result = (mysql_query($query) or die());

as it would be in Python. So this:

$a = false or true;

will set $a to false, not true, which is bound to catch me out at some point in the future.

Answer 3

Why should or return anything? or is a normal boolean operator. $a or $b is true if either $a or $b evaluates to true and false otherwise.

The difference between || and or is, that or has a lower operator precedance, even lower than =. This is why

$result = mysql_query($query) or die();

is same as

($result = mysql_query($query)) or (die());

whereas

$result = mysql_query($query) || die();

is same as

$result = (mysql_query($query) || die());

In your case

echo $name or "Anonymous";

gets

(echo $name) or ("Anonymous");

What you are looking for probably is the ternary operator:

echo $name ?: 'Anonymous';

The above will work as of PHP 5.3, if you have only PHP 5.2 use:

echo $name ? $name : 'Anonymous';