Arrays, substitution and double-quotes

[supersalo]

Why is it that I can do this:

$x[1] = 2;
$y[2] = "blah";

$a = $y[$x[1]];
print $a; // prints "blah"

---

but I can't print $y[$x[1]] directly:

$x[1] = 2;
$y[2] "blah";

print "$y[$x[1]]";

I get a 'Parse Error' (expecting `']'')

[jernhenrik]

print $y[$x[1]];

instead of print "$y[$x[1]]";

works, but as to why...err check the manual for print

hth

[supersalo]

I figured it out. You have to use:

print "{$y[$x[1]]}";

[Weedpacket]

Originally posted by supersalo
I figured it out. You have to use:

print "{$y[$x[1]]}";

You don't have to; as jernhenrik noted, you could just

print $y[$x[1]];

In fact, it's preferable that you do. By using

print "{$y[$x[1]]}";

you are:

1. creating a string,

2. looking through the string to see if there is a variable in it,

3. reading the variable,

4. getting the variable's contents,

5. converting those contents to a string if necessary,

6. replacing the variable's name in the string with those contents,

7. printing out the resulting string.

Which is a hell of a lot slower than

1. Get the contents of a variable

2. convert those contents to a string if necessary

3. print that string.

[supersalo]

The reason I'm not just using

print $y[$x[1]];

is because I'm trying to include the value as part of a string (which is actually part of an HTML form)

[laserlight]

You can concatenate strings instead of having the variable parsed within the string.

[Weedpacket]

print "Like this: " . $y[$x[1]];