Let's look at your original message:
--------------------
Using this string:
"$var1: $var2"
....of course it doesn't work as some might expect.
--------------------
It works exactly as expected. Within double-quotes, things like variables get evaluated. In this case $var1 and $var2 will be replaced by their values. That's what's "expected".
--------------------
But why in the name of [whatever, too many to list] doesn't this one
below work?
"\$var1: $var2"
If \ is an escape character why does it also appear in the string output?
Why does the above \ escape $var1 and not only the ?
--------------------
In the test I did, it behaves exactly as it's expected to behave. "\" escapes the next character. Or in the case of some characters, there may be more than one after the \ that get interpreted. For example "\x41" is hexdecimal value 41.
The \$var1 above is indicating you want to display a "$" literally and not interpret it despite being contained within double-quotes. The "var1" part, not having a "$" on the front of it, doesn't get interpretted as anything and is output literally as "var1".
Maybe rephrasing the question would help because reading your original message doesn't tell me anything except that there's a communication problem.
Unless "\$var1: $var2" does NOT output "$var1: <somevalue>", then everything is functioning as designed and expected.
-TG
= = = Original message = = =
Sorry tg, you missed the whole point. Read again.
> The escaping works fine for me.. using the code:
>
> $var1 = 1;
> $var2 = 3;
>
> echo "\$var1: $var2";
> print "\$var1: $var2";
> print ("\$var1: $var2");
>
> All output:
>
> $var1: 3
>
> as expected.
>
> Is there a way to re-define the escape character or something? I can't think of why that wouldn't escape the $ properly
>
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