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Re: "Late" variable substitution

Posted by Gordon Burditt on 11/10/06 21:33

>I want to know if this is possible.
>
>I have a database with an attribute which has in it a SQL statement,
>e.g.
>
>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.

Show code. In particular:

echo 'key = $x';
and echo "key = $x';

are *NOT* the same thing.

>What I want to happen is the local value of $x to get substituted into
>$v.

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';
echo $v;

and you want it to print
Up your Hose with a Hose
?
>Is this possible? If so, how do you do it?

It might be possible with 'eval', if I have interpreted the
question correctly.

 

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