TA貢獻1802條經驗 獲得超6個贊
如果有N個問題，那么構建一個N維數組，通過查表計算arr[select_1][select_2][select_3][]...[select_n]來獲得結果。比如你的題目有2個問題，那么就構建一個二維數組，預先存儲一個表。賦值階段可以區分一下數組是不是足夠酥松如果酥松的話就把有值的地方賦好就行了：vararr=Array(2);arr[0]=Array(10);arr[1]=Array(10);arr[0][0]="123"arr[3][4]="123123"arr[5][10]="21312"functionshowAnswerForSelects(){returnarr[arguments[0]][arguments[1]];}showAnswerForSelects(3,4)->"123123"showAnswerForSelects(7,5)->undefined或者數據比較酥松，還可以構建一個switch邏輯，像這樣：functionshowAnswerForSelects(){varstr=arguments[0]+"-"+arguments[1];switch(str){case"0-1":return"123";case"3-4":return"123123";case"5-10":return"21312";default:returnnull;}}調用還是一樣的方式。如果比較密，那么就用字面量的方式直接賦值吧：vararr=[["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"],["123123","12312","123123"..."234134"]];functionshowAnswerForSelects(){returnarr[arguments[0]][arguments[1]];}嫌不好索引還可以用@kikong的對象字面量的方式來做。
TA貢獻1828條經驗 獲得超3個贊
這個就看你的選擇組合的內在邏輯了,簡單的就是把所有可能性放到一個映射對象中例如varanswer={"a1b1":"a1b1的結果描述~~~~","a1b2":"a1b2的結果描述~~~~"};在獲取用戶的選擇結果后,如結果userAnswer="a1b1";varuserAnswerDesc=answer[userAnswer];
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