矩阵游戏,当用户选择2个相同颜色的球时,它应该破坏2个相同颜色的图案

matrix game, condition when user selects 2 same colored balls it should destroy the 2 same color patterns

本文关键字:2个 颜色 选择 游戏 用户      更新时间:2023-09-26

我正在尝试一个矩阵游戏,条件是:

  • 当用户选择2个相同颜色的球时,将破坏2个相同颜色的图案。

我已经做了横向&垂直选择正确。但是当我尝试交叉选择(对角线)它不工作,我想,我已经做了一个错误的对角线选择条件。

这是我的编码,交叉选择不匹配相同的颜色模式。这是我的对角线选择编码,以下条件是否正确?

    onCheckPattern: function(pPattern) {
    if (pPattern != null) {
        this.mPromptTimerTally = 0;
        this.mPromptMarkSpr.setPosition(-1000.0, -1000.0);
        if (this.mFirstCheckPattern === null) {
            this.mFirstCheckPattern = pPattern;
            this.mCheckMarkSpr.setPosition(this.mPatternsPos[this.mFirstCheckPattern.m_nRowIndex][this.mFirstCheckPattern.m_nColIndex]);
        } else {
            this.mSecondCheckPattern = pPattern;
            if (this.mSecondCheckPattern === this.mFirstCheckPattern) {
                //                    this.mSecondCheckPattern = null;
                //                    return;
            }
            var isAdjacent = false;
            //HORIZONTAL& VERTICAL
            if (this.mFirstCheckPattern.m_nRowIndex == this.mSecondCheckPattern.m_nRowIndex) {
                if (this.mFirstCheckPattern.m_nColIndex > 0 &&
                    this.mFirstCheckPattern.m_nColIndex - 1 == this.mSecondCheckPattern.m_nColIndex)
                    isAdjacent = true;
                else if (this.mFirstCheckPattern.m_nColIndex + 1 < this.m_nMatrixCol &&
                    this.mFirstCheckPattern.m_nColIndex + 1 == this.mSecondCheckPattern.m_nColIndex)
                    isAdjacent = true;
            } else if (this.mFirstCheckPattern.m_nColIndex == this.mSecondCheckPattern.m_nColIndex) {
                if (this.mFirstCheckPattern.m_nRowIndex > 0 &&
                    this.mFirstCheckPattern.m_nRowIndex - 1 == this.mSecondCheckPattern.m_nRowIndex)
                    isAdjacent = true;
                else if (this.mFirstCheckPattern.m_nRowIndex + 1 < this.m_nMatrixRow &&
                    this.mFirstCheckPattern.m_nRowIndex + 1 == this.mSecondCheckPattern.m_nRowIndex)
                    isAdjacent = true;
            }
            //       

            //DIAGONAL SELECTION
            else if (this.mFirstCheckPattern.m_nRowIndex + 1, this.mFirstCheckPattern.m_nColIndex - 1 && this.mSecondCheckPattern.m_nRowIndex, this.mSecondCheckPattern.m_nColIndex)
            {
                isAdjacent = true;
            } else if (this.mFirstCheckPattern.m_nRowIndex - 1 == this.mSecondCheckPattern.m_nRowIndex && this.mFirstCheckPattern.m_nColIndex - 1 == this.mSecondCheckPattern.m_nColIndex) {
                isAdjacent = true;
            }


            if (isAdjacent) {
                this.mCheckMarkSpr.setPosition(-1000.0, -1000.0);
                this.swapTwoPattern(this.mFirstCheckPattern, this.mSecondCheckPattern, false);
                this.mFirstCheckPattern = null;
                this.mSecondCheckPattern = null;
            } else {
                this.mCheckMarkSpr.setPosition(this.mPatternsPos[this.mSecondCheckPattern.m_nRowIndex][this.mSecondCheckPattern.m_nColIndex]);
                this.mFirstCheckPattern = this.mSecondCheckPattern;
                this.mSecondCheckPattern = null;
            }
        }
    }
},

            else if (this.mFirstCheckPattern.m_nRowIndex + 1, this.mFirstCheckPattern.m_nColIndex - 1 && this.mSecondCheckPattern.m_nRowIndex, this.mSecondCheckPattern.m_nColIndex)

看起来很可疑。

略为缩写,比如

            else if (a + 1, b - 1 && c, d)

由于JavaScript逗号操作符的工作方式,它被解释为else if (d),因此如果d不为零,isAdjacent将被设置为true

            else if (this.mFirstCheckPattern.m_nRowIndex + 1 == this.mSecondCheckPattern.m_nRowIndex && this.mFirstCheckPattern.m_nColIndex - 1 == this.mSecondCheckPattern.m_nColIndex)