PHP - If and else statements

Question

I have this code and there is a bug in it but I cannot see where I have gone wrong. Can someone please help? The problem is that I am trying to display a certain image to corrospond with the content of text files. I think i have that part sorted but when it comes to displaying the images there is always a bug (E.G it is always green even when the if statment says otherwize.Here is the code:

<?php
            if (empty($_GET['unit'])) {
            $output="Please Enter A Unit Number";
            echo $output;
            }
            else {
                $filepathhalf = "/data/";
                $file = "false";
                $date = date("Ymd");
                $unitnum = $_GET['unit'];
                $ext = ".txt";
                $filepath = $filepathhalf.$unitnum.$date.$ext;
                echo $filepath;
                echo $file;
                if(file_exists($filepath))
                {
                    $fh = fopen($filepath, 'r');
                    $file = fread($fh, 5);
                    fclose($fh);
                }
                echo $file; //This echo comes back as false as set but the green.png image still displays.
                if ($file = "true ")
                {
                    echo "<img src=\"images/green.png\" width=\"15\" height=\"15\" />";
                }                   
                else 
                {
                    echo "<img src=\"images/red.png\" width=\"15\" height=\"15\" />";
                }
                echo $_GET['unit']; 
            }
            ?>

Thankyou for pointing everyone pointing that out :) ill accept the 1st answer just cause I can only accpept one.. — Nick
– Nick, Commented Jul 13, 2012 at 5:27
While you shouldn't be mixing concerns in a single file, regardless, you may find that dropping in and out of PHP for HTML output advantageous: if($file): ?> <img src="<?= $file; ?>" /> <? endif; — Dan Lugg
– Dan Lugg, Commented Jul 13, 2012 at 5:56

Filip Roséen · Accepted Answer · 2012-07-13 05:30:24Z

4

There is a difference between comparing two instances and assigning one to the other.

See the below lines from your snippet and see if you might spot the error with the above clue:

if ($file = "true ")
{
  echo "<img src=\"images/green.png\" width=\"15\" height=\"15\" />";
}                   
else 
{
  echo "<img src=\"images/red.png\" width=\"15\" height=\"15\" />";
}

Otherwise hover with your mouse over the spoiler below!

^{If you want an explanation regarding the issue, do the same...}

$file = "true " will always evaluate to true, first it will assign the string "true " to $file and then the value of $file will be evaluated.

You are most probably looking for if($file == true), which will compare the value of $file to true.

edited Jul 13, 2012 at 5:30

answered Jul 13, 2012 at 5:24

Filip Roséen

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1 Comment

Filip Roséen Over a year ago

@Nicholas no problem mate, glad to be of service!

Ilmo Euro · Accepted Answer · 2012-07-13 05:24:46Z

3

You use a single =, which is used when assigning variables, not comparing them. When checking if two values are equal, use ==.

answered Jul 13, 2012 at 5:24

Ilmo Euro

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Comments

ManMohan Vyas · Accepted Answer · 2012-07-13 05:25:05Z

2

  if ($file == true)
  {
          echo "<img src=\"images/green.png\" width=\"15\" height=\"15\" />";
  }

hope that helps

answered Jul 13, 2012 at 5:25

community wiki

ManMohan Vyas

Comments

Rikesh · Accepted Answer · 2012-07-13 05:25:25Z

1

It should be == to check condition.

 if ($file != "false")
 {
   echo "<img src=\"images/green.png\" width=\"15\" height=\"15\" />";
 }

answered Jul 13, 2012 at 5:25

Rikesh

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Comments

Nir Alfasi · Accepted Answer · 2012-07-13 05:25:27Z

1

not only you're using a single "=" but also you compare it to "true " (with a concatenated space!). I would change the code to:

if ($file === true)
{
   echo "<img src=\"images/green.png\" width=\"15\" height=\"15\" />";
}

answered Jul 13, 2012 at 5:25

Nir Alfasi

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