必须有更好的方法来写这个 [英] There must be a better way to write this

问题描述

我想编写一些代码来判断我的日期是否需要st，nd rd或

。 />
虽然下面的代码有效，但我确信有更好的方法....


if（（$ daterep ==" 1"）| |（$ daterep ==" 21"）||（$ daterep ==" 31"））{

$ letters =" st";

} elseif（（$ daterep ==" 2"）||（$ daterep ==" 22"））{

$ letters =" nd";

} elseif（（$ daterep ==" 3"）||（$ daterep ==" 23"））{

$ letters =" rd";

} else {

$ letters =" th";

}


我试过：


if（$ daterep ==" 1" ||" 21" ||" 31"）{... etc


但它没有工作...虽然我不明白为什么？

我把$ daterep投入一个字符串，但现在我发现我可以把它留下来

一个人... ..

任何建议如何让这段代码更顺畅？


Robert

Hi,

I was wanting to write some code to work out whether I needed st, nd rd or
th in my dates.
While the code below works, I am certain that there is a better way....

if (($daterep == "1") || ($daterep == "21") || ($daterep == "31")){
$letters = "st";
} elseif (($daterep == "2") || ($daterep == "22")){
$letters = "nd";
} elseif (($daterep == "3") || ($daterep == "23")){
$letters = "rd";
} else {
$letters = "th";
}

I tried:

if ($daterep == "1" || "21" || "31") { ... etc

but it did not work... I don''t understand why though?
I cast $daterep into a string, but now I find that I could have left it
alone....
Any suggestions as to how to get this code working a bit smoother?

Robert

推荐答案

daterep == 1）|| （

daterep == "1") || (

daterep ==" 21"）|| （

daterep == "21") || (

daterep ==" 31"））{

daterep == "31")){

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