inside a variable have multiple variables from database

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inside a variable have multiple variables from database

ApniGang

//$row[formula] fetching from database and the store value is: ($basic - $advance)

$basic=100;
$advance=75;
echo $result= $row[formula];

// my output comes like this "($basic + $advance)"
and i want it come like this (100 - 75)

please help

simplypixie

I am not entirely sure what you mean as you are not displaying all of your code and don't explain what $row[formula] does or how it is used.

To echo the variables:

echo '('.$basic.' - '.$advanced.')';

ApniGang

$basic=100;
$advance=75;
$row[formula]= ($basic - $advance) // this comes from database

when i echo "($basic - $advance)"; //my output comes (100 - 75)
but i echo "$row[formula]"; // then my output comes ($basic - $advance)

and i want when i echo $row[formula] my result will come=> (100 - 75);

NogDog

Make sure you are showing us the exact code, as each of these will give you different results (note the difference in how the expressions are quoted):

$foo = 8;
$bar = 9;

echo ($foo - $bar);   // -1
echo '($foo - $bar)'; // ($foo - $bar)
echo "($foo - $bar)"; // (8 - 9)

ApniGang

$foo = 8;
$bar = 9;

thanks for reply

but this one is store in a variable -> ($foo - $bar);
example $a=($foo - $bar);

now when i echo $a my result is coming ($foo - $bar)
but i want when i echo $a it will show like (9 - 8);

note ($foo - $bar); is store in a variable // a=($foo - $bar);

simplypixie

NogDog has just given you the answer to this! So to use in your case you need to use the version

echo "($basic - $advance)"; // (100 - 75)

ApniGang

$basic    =    500;
    $fetch[amount]= "($basic * 10 / 100)";
    $fetch[advance]="($basic * 10 / 100)  + ($advance * 5 / 100) ";

$advance = str_replace('$basic', $basic, $fetch['amount']);
$advance= calculate_string($advance);  // and is 50
echo $x = str_replace('$advance', $advance, $fetch['advance']);
// echo result (500 * 10 / 100) + ( * 5 / 100)

$advance is missing