動態突出顯示表中的td

Question

問題

如何動態突出顯示選定的TD？

Codepen

筆在這里

碼

地圖是隨機生成的二維數組，如下所示：

map = [[1,1,1,1,0],
       [1,0,0,0,0],
       [1,0,1,1,1],
       [1,0,0,0,1],
       [1,1,1,0,1]]

我可以每回合移動玩家3個方格，其中一個方格是他的實際位置。 我用這個函數調用了這個動作：

function movements(character){
    var possibleMovement=3;
    let coord=character.actualPosition; 
        let row = $($("#tableGame tr")[coord.row]);
        let cell = $($("td", row)[coord.cell]);

    forward(row, cell, possibleMovement, character);
    backward(row, cell, possibleMovement, character);
    goUp(row, cell, possibleMovement, character);
    goDown(row, cell, possibleMovement, character);
};

並且通過下面的功能，我嘗試突出顯示角色實際可以移動的單元格。

function forward(row, cell,possibleMovements, character){
    for(var i = 0; i<possibleMovements; i++){
        cell = $($("td", row)[coord.cell+i]);
        var tile = $(".tile", cell).addClass('possibleSteps');
    };
};

function backward(row, cell, possibleMovements, character){
    for(var i = 0; i>=possibleMovements; i--){
        console.log('sei qua');
        cell = $($("td", row)[coord.cell+i]);
        var tile = $(".tile", cell).addClass('possibleSteps');
    };
};

任務

我需要突出顯示角色附近的瓷磚：

• 在character.actualPosition之上的兩個瓦片

• 下面有兩塊瓷磚

• 在他右邊的兩個瓷磚

• 左邊有兩個瓷磚

  

這是兩個“測試功能”

function forward(row, cell,possibleMovements, character){
    for(var i = 0 ; i<possibleMovements; i++){
        cell = $($("td", row)[coord.cell +i]);
        var tile = $(".tile", cell).addClass('possibleSteps');
        console.log([coord.row] + "<<<row" + [coord.cell+i] + "<<<cell");
    };
};

function backward(row, cell, possibleMovements, character){

    possibleMovements= possibleMovements*-1;

    for(var i = 0 ; i>possibleMovements; i--){
        cell = $($("td", row)[coord.cell+i]);
        var tile = $(".tile", cell).addClass('possibleSteps');
        console.log([coord.row] + "<<<row" + [coord.cell-i] + " <<<cell");
    };
};

Answer 1

最后我找到了答案。 我要感謝這篇文章，因為它幫助我找到了一個有效的解決方案。

突出角色附近的瓷磚的最終功能是這些

function movements(){
    let possibleMovement=3;
    let row = character.actualPosition.row;
    let cell = character.actualPosition.cell;
    forward(possibleMovement, row, cell);
    backward(possibleMovement, row, cell);
    goUp(possibleMovement, row, cell);
    goDown(possibleMovement, row, cell);

};

function forward(possibleMovements, row, cell){

    let charRow = row;
    let charCell= cell;
    var table = $("table")[0];

    for(var i = 0; i<possibleMovements; i++){
        let cell = table.rows[charRow].cells[charCell+i]; // This is a DOM "TD" element
        let $cell = $(cell);
        $(cell).addClass('possibleSteps');
    };
};

function backward(possibleMovements, row, cell){
    let charRow = row;
    let charCell= cell;
    var table = $("table")[0];

    for(var i = -1; i>(possibleMovements*-1); i--){
        let cell = table.rows[charRow].cells[charCell+i]; // This is a DOM "TD" element
        let $cell = $(cell);
        $(cell).addClass('possibleSteps');
    };
};

function goUp(possibleMovements, row, cell){
   let charRow = row;
    let charCell= cell;
    var table = $("table")[0];

    for(var i = -1; i>(possibleMovements*-1); i--){
        let cell = table.rows[charRow+i].cells[charCell]; // This is a DOM "TD" element
        let $cell = $(cell);
        $(cell).addClass('possibleSteps');
    };
};

function goDown(possibleMovements, row, cell){
    let charRow = row;
    let charCell= cell;
    var table = $("table")[0];

    for(var i = -1; i<possibleMovements; i++){
        let cell = table.rows[charRow+i].cells[charCell]; // This is a DOM "TD" element
        let $cell = $(cell);
        $(cell).addClass('possibleSteps');
    };
};

主要的解決方案是理解這4行代碼，以及我如何迭代它們：

    var table = $("table")[0];
    let cell = table.rows[charRow].cells[charCell+i]; // This is a DOM "TD" element
    let $cell = $(cell);
    $(cell).addClass('possibleSteps');