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php Currentage not define in the scope

开发者 https://www.devze.com 2022-12-16 13:27 出处:网络
I am quite new to php. Can somebody please guide me what is wrong with the code. <?php if(!isset($_POST[\'submit\']) || $_POST[\'submit\']!=\"calculate\")

I am quite new to php. Can somebody please guide me what is wrong with the code.

<?php
    if(!isset($_POST['submit']) || $_POST['submit']!="calculate")
    {
       $_POST['Contrib']="";
       $_POST['Currentage']="";
       $_POST['Retireage']="";
       $Total =0;
       $AnnGain =7;
    }else{
       $AnnGain = $_POST['AnnGain'];
       $Years = $_POST['Retireage'] - $_POST['Currentage'];
       $YearCount = 0;
       $Total = $_POST['Contrib'];

       while ($YearCount < $Years)
       {
           $Total = (round($Total) *(1.0 + $AnnGain/100) +
               $_POST['Contrib']);
           $YearCount = $YearCount+1;

       }

    }
    ?>

           <b>A Retirement Saving calculator</b>


           <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
           <p> Your age now
           <inpu开发者_运维技巧t type="text" size = "5" name = "Currentage"
                  value="<?php echo $_POST['Currentage'];?>">
           <p> The age at which you want to retire
           <Input type="text" SIZE="6" name="Retireage"
                  value="<?php echo $_POST['Retireage']; ?>">
           <p> Annual Contribution
           <input type="text" size = "15" name = "Contrib"
                  value="<?php echo $_POST['Contrib'];?>">
           <p>Annual Return
           <input type = "text" size = "5" NAME = "AnnGain"
                  value="<?php echo$AnnGain; ?>">
           <BR><BR>
           <p><b>Nest Egg </b>: <?php echo $Total; ?>
           <p><Input type = "submit" Name = "submit" value = "calculate">
           </form>


In your code i see:

input type="test"

It is wrong, it should be:

input type="text"

Input type should be text if you mean a textbox.


   $_POST['Currentage']=="";
   $_POST['Retireage']=="";

You're checking if $_POST['Currentage'] is equal to "" instead of setting it to "". What you want is $_POST['Currentage'] = "";. You have the same problem with $_POST['Retireage'].


outside of the fact that modifying $_POST variables is bad practice (just assign those values to a variable and use that in your code)

  1. $_POST['Currentage']==""; should be $_POST['Currentage'] = '';
  2. $_POST['Retireage']==""; should be $_POST['Retireage'] = '';
  3. ALWAYS escape data using something like htmlentities() before you spit it out to the web browser to protect your page from injections. This is VERY important
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