カウンターを増やす場合に複数を短くする方法

Question

カウンタを使用して複数の他のショートを短くする方法を見つけましたが、自分の状況に関連するものは何も見つかりませんでした。

現在、増加するカウンタごとにif文が多すぎます。

これを達成する方法はありますか？事前にみんなで

$(".gift").each(function(){ 

     var i = 1; 

     $(this).on("click", function(){ 

      if(i===1){ 
       TweenMax.fromTo(this, 1, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===2){ 
       TweenMax.fromTo(this, 0.9, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===3){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===4){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===5){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===6){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===7){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===8){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===9){ 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
      } 

      if(i===10){ 
       $this = $(this); 
       TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x", onComplete:function(){ 
        $this.css("top", "500px"); 

        $(".gift").each(function(){ 
         $(this).off("click"); 
        }); 
       }}) 

      } 

      i++; 

      console.log(i); 

     }); 

    }); 

ありがとう：

以下は私のコードです！

Answer 1

コメントで示唆されているようにelse文がswitchステートメントを使用する場合、「複合体」に対する別の解決策：

$(".gift").each(function(){ 

     var i = 1; 

     $(this).on("click", function(){ 

      switch(i) { 
       case 1: 
        TweenMax.fromTo(this, 1, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
        break; 
       case 2: 
        TweenMax.fromTo(this, 0.9, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
        break; 
       case 3: 
       case 4: 
       case 5: 
       case 6: 
       case 7: 
       case 8: 
       case 9: 
        TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x"}) 
        break; 
       case 10: 
        $this = $(this); 
        TweenMax.fromTo(this, 2/i, {x:-1}, {x:1, ease:RoughEase.ease.config({strength:8, points:5, template:Linear.easeNone, randomize:false}) , clearProps:"x", onComplete:function(){ 
         $this.css("top", "500px"); 

         $(".gift").each(function(){ 
          $(this).off("click"); 
         }); 
        }}) 
        break; 


      } 

      i++; 

      console.log(i); 

     }); 

    });