小编DT *_* DT的帖子

在TypeScript上还是新手,所以这个问题听起来可能对你们有些愚蠢.我有一个像这样的ToolTip类:

class ToolTip{
    public static show (str:string):void{
        console.log ("ToolTip show():" + str);
    }
    public static hide():void{
        console.log ("ToolTip hide()");
    }
}
export = ToolTip;

我想从另一个班级打电话来

import ToolTip = require ("app/view/common/Tooltip");

class Button  {
......
    private handleMouseEvent(event:MouseEvent):void {
        switch (event.type) {
            case "mouseover":
                ToolTip.show("tool tip string");
                break;
            case "mouseout":
                ToolTip.hide();
                break;            
        }
    }
......
}

export = MenuItem;

但它给了我这个错误:

Uncaught TypeError: Object app/view/common/Tooltip has no method 'show'

知道如何解决这个问题吗？

这是我的文件夹结构:

main 
- folder_1
-- folder_1_1
-- folder_1_2
-- folder_1_3
- folder_2
-- folder_2_1
-- folder_2_2
-- folder_2_3

这是我的代码:

<?php

$mainfolder ="main";

function readDirs($mainfolder){
    if(hasSubFolder($mainfolder)){
        echo("$mainfolder HAS sub folder");
    }else{  
        echo("$mainfolder DOESN'T HAVE sub folder");
    }   
}

function hasSubFolder($folder){ 
    $newPath="";
    if (is_dir($folder)) {  
        echo ("$folder IS a folder</br>"); 
        $handle = opendir($folder);
        while (false !== ($entry = readdir($handle))) {
            $newPath = $folder."\\".$entry;
            if (is_dir($newPath)){
                echo "$newPath IS a folder</br>";
            } else {                
                echo "$newPath IS NOT a folder</br>";
            }
        }       
    } …