"Late" variable substitution

[JoeT]

Hi...

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.

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

Is this possible? If so, how do you do it?

TIA,
Joe

[Gordon Burditt]

>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.

[JoeT]

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.