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如何顯示此信息以便顏色代碼正確顯示文本?這只是一個小片段,但我似乎無法讓它工作。我試圖讓它成爲一個數組,但卡住了。顯示Var信息
if($pclass == "c1" or "c2" or "c3" or "c4" or "c5"){
$pcolor = '<font color="#0000FF">';
$pend = '</font>';
} else {
if($pclass == "w1" or "w2" or "w3" or "w4" or "w5"){
$pcolor = "<font color='#FF6600'>";
$pend = "</font>";
} else {
if($pclass == "r1" or "r2" or "r3" or "r4" or "r5"){
$pcolor = "<font color='#00FF00'>";
$pend = "</font>";
} else {
if($pclass == "h1" or "h2" or "h3" or "h4" or "h5"){
$pcolor = "<font color='#CC00CC'>";
$pend = "</font>";
}
}
?>
<div id="top">
<h1>
<?php echo "$requestname"; ?>
</h1>
<p class="small">
<?php
echo "$player's primary class is $pcolor $pclass $pend";
echo ", $player's secondary class is $sclass";
我不確定你在做什麼,只是回答標題:use [var_dump()](http://php.net/manual/en/function.var-dump.php )。 – HamZa 2013-04-28 20:58:12
$ pclass ==「c1」或「c2」或「c3」或「c4」或「c5」將始終爲真。試試$ pclass ==「c1」|| $ pclass ==「c2」|| ...或者in_array($ pclass,array('c1','c2','c3','c4','c5'))。還要考慮在這些名稱下定義CSS類,而不是使用字體元素。 – PleaseStand 2013-04-28 20:59:15
你有沒有終止的大括號。如果這不是複製/粘貼錯誤,我沒有在代碼中修復它。 – Gary 2013-04-28 21:01:58