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Re: "Late" variable substitution

Posted by JoeT on 11/10/06 22:22

That did the trick!

Thanks much!

Joe


Gordon Burditt wrote:
> >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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