[SOLVED] quite tricky, .. can you please help me ..

[jaikar]

 

Hi .. i want a code to do something... but not sure if its possible .. can you please help ...

 

<?php

$abc = '$test';

$test = "hello";

$xyz = $abc;

echo $xyz;

?>

 

from the above code .. the result should print " hello " ... is this possible ?????

 

my actual need is .. i need to store a query in the database where some conditions are dynamic. so it will like " select * from table where id=$id " now the whole thing should be inside the database so when i fetch the $id shoulld act as a variable ....

 

PLease adive..

 

Thanks

 

Jaikar

 

 

 

[wildteen88]

You can only call variables from within double quotes and not single quotes. If you are storing PHP code in a database then you will need to use eval() in order to process the PHP code in the database.

[flappy_warbucks]

Well...

 

lets look at the code shall we:

 

<?php
  $abc = '$test'; // litral string meaning it prints whats in the quotes and not whats in the varable, this is where your going wrong.
  $test = "hello"; // this is not a litral sting as it has the " and not the ' so that would print out what in the varible as well as the litral text.
  $xyz = $abc; // here you are assigning $xyz with the contents of $abc
  echo $xyz; // and here you are printing out the litral string: "$test"
?>

[per1os]

<?php
   $test = "hello"; // Variable positions is key.

   $abc = $test; // do not know why you are quoting it

   $xyz = $abc; // should be what it is.

   echo $xyz;

   // Could be done like this also:
   $test = "hello"; // Variable positions is key.

   $abc = "$test"; // shown as quoted, must be double to do the $

   $xyz = $abc; // should be what it is.

   echo $xyz;

?>

 

If you want it to be like how you have it you may need to look into reference/pointers. I never got the hang of pointers, but I am sure that is what you want if you want the variable $abc to change after declaration.

 

 

[jaikar]

Thanks a lot Friends.....

 

flappy_warbucks, ... your explanation is absolutly correct and that is what was in my mind .. what i need is .. when i print the literal "$test" i wanted it to act as a variable and should print the value inside the $test.. i think wildteen88 gave a good idea and now i will try it ..

 

 

frost110 , ..... i did not used the single quote because, if i store the $test in database ??... it will act as literal correct ?..

 

Thanks again !!....

 

Jaikar

[per1os]

If you want to store it in the DB as $test you want to use single quotes. If you want it to display the value you use double quotes.

 

$test = "Fun!";
$tst = "$test";
print $tst; // should print Fun!

$tst2 = '$test';

print $tst2; // should print $test

[jaikar]

thanks froast,

 

the concern is when retrieving the variable from database.... storing is not a problem... anyhow, the answer is using the eval()...

 

jaikar

[wildteen88]

That is correct. PHP will treat any PHP code coming out of a database literally. As when you fetch something from the database PHP will treat it as a string, depending on the data type you use for your columns.