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- "Late" variable substitution
November 10, 2006, 9:17 pm
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I want to know if this is possible.
I have a database with an attribute which has in it a SQL statement,
select blah from table where key = $x
In a PHP function, I read in the above string into $v.
In that routine, $x is defined.
However, if I print out the string I get the original string, with $x
not being substituted.
What I want to happen is the local value of $x to get substituted into
Is this possible? If so, how do you do it?
Re: "Late" variable substitution
Show code. In particular:
echo 'key = $x';
and echo "key = $x';
are *NOT* the same thing.
Um, what does this mean?
Do you mean you have something like:
$v = 'Up your $x with a $x';
(note: no substitution is done here)
and later you do:
$x = 'Hose';
and you want it to print
Up your Hose with a Hose
It might be possible with 'eval', if I have interpreted the