我不知道我做错了什么。也许有人可以建议我。我正在尝试定义季节,然后在索引中使用结果。
...
带功能的文件:
function getCurrentTheme($for_area) {
// Definition of seasons
$spring='spring theme title';
$spring_season=array('03-21',......);
$summer='summer theme title';
$summer_season=array(......);
$autumn='autumn theme title';
$autumn_season=array(......);
$winter='winter theme title';
$winter_season=array(......);
// Today
$current_date=date('m-d');
// So what season is now?
if ($for_area==='some_area') {
if (in_array($current_date,$spring_season) {
$theme=$spring;
}
else if (in_array($current_date,$summer_season) {
$theme=$summer;
}
else if (in_array($current_date,$autumn_season) {
$theme=$autumn;
}
else if (in_array($current_date,$winter_season) {
$theme=$winter;
}
else {}
}
else if ($for_area==='other_area') {
// ...
}
else {}
return $theme;
}
...
索引:
$area='some_area';
getCurrentTheme($area);
// And here is the fault. Hope sometimes I will stop being retarded.
echo $theme;
// What should be printed?
summer theme title
提前致谢,请尝试了解我的清白。
答案 0 :(得分:1)
您需要存储该值,以便您可以回显它:
$area='some_area';
$theme = getCurrentTheme($area);
echo $theme;
答案 1 :(得分:1)
您没有存储从getCurrentTheme
函数返回的值。将其存储在变量中,然后echo
将其存储在
$theme = getCurrentTheme($area);
echo $theme;
或简单地回应呼叫而不必使用变量。
echo getCurrentTheme($area);
您的选择!