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Asked: 2026-05-25T20:02:03+00:00

I’m trying to display an image based on a change in rank. If the

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I’m trying to display an image based on a change in rank. If the values of the array $rank_change are echoed before my if statements, the correct values are printed, such as:

0, 0, 0, 0, 0, 0, 0, -7, 1, 1, 1, 1, 1, 1, 1

However, if printed after the following if statements, side.gif is always shown:

if($rank_change[$i] > 0) { $rank_change[$i] = "<img src=\"up.gif\" width=\"16\" height=\"16\"/>"; }
if($rank_change[$i] < 0) { $rank_change[$i] = "<img src=\"down.gif\" width=\"16\" height=\"16\"/>"; }
if($rank_change[$i] == 0) { $rank_change[$i] = "<img src=\"side.gif\" width=\"16\" height=\"16\"/>";}

In the cases where the value of $rank_change are 7 and 1, why does the last statement still evaluate to true?

I realize it would be more efficient to use switch($rank_change[$i]) but, I still don’t understand why the final if statement is evaluating true on all values.

Any help would be greatly appreciated!

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1 Answer

  1. Editorial Team

    use an else-if instead of a simple if for the last two conditions.

    you are actually overwriting the value in $rank_change[$i] in the expression following each test. $rank_change[$i] = "<img src=\"up.gif\" width=\"16\" height=\"16\"/>" will change the value of $rank_change[$i], which will make one of the following test return true as well, resulting in another change of the value of $rank_change[$i]…

    use an else if construct, and only one test will succeed.

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