我有以下情况:
if (event.status == AMFResultEvent.SUCCESS) {
var lev1:uint = 0;
var lev2:uint = 0;
var lev3:uint = 0;
var lev4:uint = 0;
var lev5:uint = 0;
var lev6:uint = 0;
for (var i:int = 0; i < event.result.length; i++) {
if (mainLevel == "1") {
lev1++;
}
if (mainLevel == "2") {
lev2++;
}
if (mainLevel == "3") {
lev3++;
}
if (mainLevel == "4") {
lev4++;
}
if (mainLevel == "5") {
lev5++;
}
if (mainLevel == "6") {
lev6++;
}
}
for (var j:int = 1; j < 7; j++) {
_row = new StatisticsRow(event.result[j], this);
_rowsPlace.addChild(_row);
_row.y = (_row.height +1) * j;
_row.codeLevel.htmlText = j; // works as it should
// need to access variables lev1 - lev6, called by something like "lev"+j here:
_row.amount.htmlText =
}
// traces correct amounts of mainLevels from the i loop:
trace ("level 1: " + lev1);
trace ("level 2: " + lev2);
trace ("level 3: " + lev3);
trace ("level 4: " + lev4);
trace ("level 5: " + lev5);
trace ("level 6: " + lev6);
}
我在这里遗漏了一些明显的东西,因为[“lev”] + j不起作用。如何动态访问j-loop中的lev1-lev6?正如底部的代码注释所示,这跟踪预期。
提前致谢!
答案 0 :(得分:1)
您可以使用括号,字符串连接和this
关键字访问它们。以下是如何在循环中使用括号表示法的示例:
for (var i:int = 0; i <= 6; i++) {
var currLev = this["lev"+i];
// do stuff to currLev
}
答案 1 :(得分:0)
感谢您的回答!
无论如何我对我的问题采取了糟糕的方法,并且应该立即使用数组:
var mainLevels:Array = new Array();
for (var n:int = 1; n < 7; n++) {
mainLevels[n] = 0;
}
if (event.status == AMFResultEvent.SUCCESS) {
for (var i:int = 0; i < event.result.length; i++) {
var data = event.result[i];
var correctCode:String = data["correct"];
var mainLevelFound:uint = uint(correctCode.substr(0, 1));
for (var k:int = 1; k < 7; k++) {
if (k == mainLevelFound) {
mainLevels[k]++;
}
}
}
for (var j:int = 1; j < 7; j++) {
_row = new StatisticsRow(event.result[j], this);
_rowsPlace.addChild(_row);
_row.y = (_row.height +1) * j;
_row.codeLevel.htmlText = j;
// Now this works as a reference to mainLevels[*] created above!
_row.amount.htmlText = mainLevels[j];
}
再次感谢您的努力:)