Php 值比较未按预期工作

我试图根据这两个值显示不同的颜色

首先，如果值相等，则显示绿色，效果良好

  <?php if (miletofurlong($pasthorse['distance']) == $todaysdistance)
    {
      $Horsedist = "<b><span style='color:#ff4500 '>". miletofurlong($pasthorse['distance'])."f</span></b>";
    } ?>

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您缺少括号，您的if案例有点混乱。让我们首先使用一个变量并在其中声明$pasthorse

$distance = miletofurlong($ pasthorse['distance']);
if ($distance == $todaysdistance)
    $Horsedist = "<b><span style='color:#ff4500 '>". $distance ."f</span></b>";
else if ($distance <= ($todaysdistance + 1) and $distance >= ($todaysdistance - 1)  )
    $Horsedist = "<b><span style='color:blue'>". $distance . "f</span></b>"  ($todaysdistance + 1). "and" . ($todaysdistance - 1);
else
    $Horsedist = $distance ."f" ;

$distance=miletofurlong（$passhorse['distance']）；
如果（$distance==$todaysdistance）
$Horsedist=”“$距离，“f”；
else if（$distance=（$todaysdistance-1））
$Horsedist=”“$距离。“f”（$todaysdistance+1）。“和”。（$todaysdistance-1）；
其他的
$Horsedist=$distance.“f”；

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在那里，它应该可以更好地工作，并且更具可读性。

您是否正确检查了此部分？“但是仍然在（$todaysdistance+1）和（$todaysdistance+1）之间”两个数字之间怎么可能有相同的东西？这可能有助于取代第二个条件（miletofurlong（$pasthorse['distance']）！=$todaysdistance&&miletofurlong（$pasthorse['distance']）=（$todaysdistance-1）））

$distance = miletofurlong($ pasthorse['distance']);
if ($distance == $todaysdistance)
    $Horsedist = "<b><span style='color:#ff4500 '>". $distance ."f</span></b>";
else if ($distance <= ($todaysdistance + 1) and $distance >= ($todaysdistance - 1)  )
    $Horsedist = "<b><span style='color:blue'>". $distance . "f</span></b>"  ($todaysdistance + 1). "and" . ($todaysdistance - 1);
else
    $Horsedist = $distance ."f" ;