AS2繪製山丘曲線 - 隨機工作

Question

我有一個數組，其中包含繪製多個山丘的點。當我畫直線時效果很好，但那是不自然的，所以我想讓山的頂部/底部彎曲。

for(i = 0; i < rPoints.length - 1; i++){ 
    gamebg.lineStyle(1,0x000000,100); 
    gamebg.moveTo(rPoints[i][0] + 45, rPoints[i][1]); //Doesn't directly move to a point so there is empty space for the curved parts. 
    if(rPoints[i+1][1] > rPoints[i][1]){ //Determines if it is the top part of a hill or a bottom part, compares y 
     gamebg.lineTo(rPoints[i+1][0], rPoints[i+1][1]); 
     gamebg.moveTo(rPoints[i+1][0], rPoints[i+1][1]); 
     //I didn't add a curveTo here because I only wanted to test it on one so I can make changes easier 
    } else { 
     gamebg.lineTo(rPoints[i+1][0], rPoints[i+1][1]); 
     gamebg.moveTo(rPoints[i+1][0], rPoints[i+1][1]); 
     gamebg.curveTo(rPoints[i+1][0]+22, rPoints[i+1][1]-25, rPoints[i+1][1]+45, rPoints[i+1][1]); 
    } 
} 

當我運行代碼時，它似乎工作了一些時間。

當它工作時，它只適用於第一個。

謝謝！

Answer 1

通過包含點的數組的順序來確定山的部分要容易得多。在這種情況下，你必須迭代3而不是1.我測試了下面的代碼，它工作正常。 * Nicholas

/* 
* ActionScript 2 
* 
* Drawing some Hills :-) 
* 
* The following Loop draws Hills using an Array containing 
* a multiple of 3 Points 
* 
*   p5 
* p2  /\ 
* /\ /\ 
* /\ / \ 
* p1 p3 p4 p6 
* 
*/ 

var rPoints = [ 
       // First Hill 
       [50,200], 
       [100,10], 
       [150,200], 
       // Second Hill 
       [100,200], 
       [150,80], 
       [200,200], 
       // Will not be drawn 
       [500,200] 
     ]; 


for(i = 0; i <= rPoints.length-3; i+=3){  

    gamebg.lineStyle(1,0x000000,100); 

    // Move to the 1st Point 
    gamebg.moveTo(rPoints[i][0], rPoints[i][1]);     
    // Curve to 2nd Point (For a smooth curve the Control Point should be in the Top Left Corner) 
    gamebg.curveTo(rPoints[i][0], rPoints[i+1][1], rPoints[i+1][0], rPoints[i+1][1]); 
    // Curve to 3rd Point (For a smooth curve the Control Point should be in the Top Right Corner) 
    gamebg.curveTo(rPoints[i+2][0], rPoints[i+1][1], rPoints[i+2][0], rPoints[i+2][1]); 
    // Close the shape (If you want to) 
    gamebg.lineTo(rPoints[i][0], rPoints[i][1]);  
}