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Thread: inside a variable have multiple variables from database

  1. #1
    Join Date
    May 2009
    Posts
    11

    Post inside a variable have multiple variables from database

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

  2. #2
    Join Date
    Aug 2012
    Posts
    155
    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:
    PHP Code:
    echo '('.$basic.' - '.$advanced.')'

  3. #3
    Join Date
    May 2009
    Posts
    11
    $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);

  4. #4
    Join Date
    Aug 2004
    Location
    Ankh-Morpork
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    19,312
    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):
    PHP Code:
    $foo 8;
    $bar 9;

    echo (
    $foo $bar);   // -1
    echo '($foo - $bar)'// ($foo - $bar)
    echo "($foo - $bar)"// (8 - 9) 
    "Please give us a simple answer, so that we don't have to think, because if we think, we might find answers that don't fit the way we want the world to be."
    ~ Terry Pratchett in Nation

    eBookworm.us

  5. #5
    Join Date
    May 2009
    Posts
    11
    $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);

  6. #6
    Join Date
    Aug 2012
    Posts
    155
    NogDog has just given you the answer to this! So to use in your case you need to use the version
    PHP Code:
    echo "($basic - $advance)"// (100 - 75) 

  7. #7
    Join Date
    May 2009
    Posts
    11
    PHP Code:
    $basic    =    500;
        
    $fetch[amount]= "($basic * 10 / 100)";
        
    $fetch[advance]="($basic * 10 / 100)  + ($advance * 5 / 100) ";
        
        
    $advance str_replace('$basic'$basic$fetch['amount']);
        
    $advancecalculate_string($advance);  // and is 50
        
    echo $x str_replace('$advance'$advance$fetch['advance']);
        
    // echo result (500 * 10 / 100) + ( * 5 / 100) 
    $advance is missing

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