"Late" variable substitution

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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

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Show code.  In particular:

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

are *NOT* the same thing.

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

Re: "Late" variable substitution

That did the trick!

Thanks much!


Gordon Burditt wrote:
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