PHP variable table

Question

So I have created a PHP page that will prompt the user to enter 2 Celsius temperatures (start & stop) and a number to increment between the two. Then it loops through the numbers and displays the Fahrenheit conversion of each number base on the increment in a table format. The problem i am currently having is that I am trying to put the table display in a variable so that i can display it below the number inputs but at the moment i am only able to display it above. I think the problem is that the variable isn't recognizing the table because when i try to echo it below it just says "1". I've tried putting the entire table in quotes and a bracket but it is still not working.

Here is the part of the code where the table comes in:

    if ($start_error == "" && $stop_error == "" && $inc_error == "") {

        $result = print "<table width=\"25%\" border =\"1\" style='border-collapse: collapse'>";
                  print "<tr><th>Celsius</th>";
                  print "<th>Fahrenheit</th></tr>";
                  for ($row=1; $row <= 1; $row++) { 
                      for ($c = $start; $c <= $stop; $c+=$inc) {
                      print "<tr> \n"; 
                      $f = $c * 1.8 +32;
                          for ($col=1; $col <= 1; $col++) { 
                                print "<td align=\"center\">$c&deg;</td> \n"; 
                                print"<td align=\"center\">$f&deg;</td> \n"; 
                           } 
                         print "</tr>"; 
                      } 
                   print "</table>";    
                   }

    }

}

?>

<h2><?php echo $start_error; ?></h2>
<h2><?php echo $stop_error; ?></h2>
<h2><?php echo $inc_error; ?></h2>


<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" >
    Starting Temperature:  <input type="text" name="start_number" value="<?php echo $start; ?>" size="5" /><br/>
    Stop Temperature:  <input type="text" name="stop_number" value="<?php echo $stop; ?>" size="5" /><br/>
    Temperature Increment:  <input type="text" name="inc_number" value="<?php echo $inc; ?>" size="5" /><br/>
    <input type="submit" value="Create Temperature Conversion Table" />


</form>

<?php echo $result; ?>

And here is an image of what the result looks like:

As documented print always returns 1. You don't use print for the return value. — Álvaro González
– Álvaro González, Commented Jun 28, 2017 at 16:59
@RiggsFolly I tried removing the prints but now it doesn't display anything. — Brett
– Brett, Commented Jun 28, 2017 at 17:02
And instead of ; at the ends use . to concatenate the strings to one (of course at the end of the last line you let ;) — derHugo
– derHugo, Commented Jun 28, 2017 at 17:04

RiggsFolly · Accepted Answer · 2017-06-28 17:01:09Z

2

Remove the print statements that directs output immediately and replace with text concatenation into the $result variable

if ($start_error == "" && $stop_error == "" && $inc_error == "") {

    $result = "<table width=\"25%\" border =\"1\" style='border-collapse: collapse'>";
    $result .= "<tr><th>Celsius</th>";
    $result .= "<th>Fahrenheit</th></tr>";
    for ($row=1; $row <= 1; $row++) { 
        for ($c = $start; $c <= $stop; $c+=$inc) {
            $result .= "<tr> \n"; 
            $f = $c * 1.8 +32;
            for ($col=1; $col <= 1; $col++) { 
                $result .= "<td align=\"center\">$c&deg;</td> \n"; 
                $result .= "<td align=\"center\">$f&deg;</td> \n"; 
            } 
            $result .= "</tr>"; 
        } 
        $result .= "</table>";    
    }
}

answered Jun 28, 2017 at 17:01

RiggsFolly

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2 Comments

RiggsFolly Over a year ago

Makes a nice change eh @Fred-ii-

Funk Forty Niner Over a year ago

Yes indeed @RiggsFolly

RToyo · Accepted Answer · 2017-06-28 17:04:16Z

The print function sends a string directly to the output buffer (the browser), and returns "1" to confirm it completed. By setting $result = print [...], you are having print perform its actions (sending the string to the browser), and then assigning the returned value ("1") to $result.

In other words, print is going to output the data to the buffer, rather than return a string for you to assign to your variable. The only value your $result variable has is what gets returned by print, which is "1".

You have two options:

1) Remove the call to print, and just set $result to your concatenated strings, as @RiggsFolly suggested.

2) You can keep the print as is, and remove the use of your $result variable, which currently only stores the returned value ("1") from your call to print.

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