Why is $b "b" in this logic when echoing out ${$b}
I'd have expected ${$b} to be 'a' but i开发者_JS百科t is 'b', why is this the case?
$a = "b";
$b = "a";
Sorry again I forgot to put ${$b} produces "b"
The variable variable expression ${$b}
takes the value of $b
for the variable name. So ${$b}
evaluates to ${"a"}
that is equivalent to $a
that then evaluates to "b"
.
If you use the ${$b} it is equal to $a. So that time it will print "b" only.
Only thing I can think of is that when you're getting its value you forgot the $
. This is a pretty common typo for new php programmers (and even for experienced ones that program in other languages).
It couldn't be.
$b = "a";
${$b} == $a;
$a = "b";
${$b} == "b"
It makes sense now with your addition
Sorry again I forgot to put ${$b} produces "b"
that works as intended: You are using $b
(containing "a") as a variable name. So as the end result, you are querying $a
.
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