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如何显示此信息以便颜色代码正确显示文本?这只是一个小片段,但我似乎无法让它工作。我试图让它成为一个数组,但卡住了。显示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";
我不确定你在做什么,只是回答标题: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”将始终为真。试试$ pclass ==“c1”|| $ pclass ==“c2”|| ...或者in_array($ pclass,array('c1','c2','c3','c4','c5'))。还要考虑在这些名称下定义CSS类,而不是使用字体元素。 – PleaseStand 2013-04-28 20:59:15
你有没有终止的大括号。如果这不是复制/粘贴错误,我没有在代码中修复它。 – Gary 2013-04-28 21:01:58