Javascript 使用两个';对于';循环
我试图通过按球员循环搜索结果数组来更新a联赛表格。用户预先通过html表填充结果Javascript 使用两个';对于';循环,javascript,arrays,for-loop,Javascript,Arrays,For Loop,我试图通过按球员循环搜索结果数组来更新a联赛表格。用户预先通过html表填充结果 players=[]; var players=["A","B","C"]; Results=[]; var Results=[ ["Home","F","A","Away"], ["A",,,"B"], ["A",,,"C"], ["B",,,"C"], ["B",,,"A"], ["C",,,"A"], ["C",,,"B"], ]; League=[]; var League=
players=[];
var players=["A","B","C"];
Results=[];
var Results=[
["Home","F","A","Away"],
["A",,,"B"],
["A",,,"C"],
["B",,,"C"],
["B",,,"A"],
["C",,,"A"],
["C",,,"B"],
];
League=[];
var League=[
["Team","P","W","D","L","F","A","Pts"],
["A",,,,,,,],
["B",,,,,,,],
["C",,,,,,,]
];
我尝试使用两个for循环,如下所示:
var pld=0;
var wins=0;
var draws=0;
var loses=0;
var goalsF=0;
var goalsA=0;
var pts=0;
for (p = 0; p <= players.length; p++)
{
for (i = 1; i < Results.length; i++)
{
if (Results[i][1]!= "")
{
if (Results[i][0]==players[p])
{
pld++;
if (Results[i][1]>Results[i][2])
{
wins++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
pts= +pts + 3;
}
else if (Results[i][1]<Results[i][2])
{
loses++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
}
else
{
draws++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
pts++
}
}
}
}
League[p][1]=pld;
League[p][2]=wins;
League[p][3]=draws;
League[p][4]=loses;
League[p][5]=goalsF;
League[p][6]=goalsA;
League[p][7]=pts;
}
var-pld=0;
var=0;
var=0;
var=0;
var goalsF=0;
var goalsA=0;
var-pts=0;
对于(p=0;p结果[i][2])
{
wins++;
goalsF=+goalsF++结果[i][1];
goalsA=+goalsA++结果[i][2];
pts=+pts+3;
}
否则,如果(结果[i][1]感谢到目前为止每个人的响应,我设法找到了一个解决方案——我在更新联赛后立即重置计数变量,但有更好的解决方案吗
var pld=0;
var wins=0;
var draws=0;
var loses=0;
var goalsF=0;
var goalsA=0;
var pts=0;
for (p = 0; p <= players.length; p++)
{
for (i = 1; i < Results.length; i++)
{
if (Results[i][1]!= "")
{
if (Results[i][0]==players[p])
{
pld++;
if (Results[i][1]>Results[i][2])
{
wins++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
pts= +pts + 3;
}
else if (Results[i][1]<Results[i][2])
{
loses++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
}
else
{
draws++;
goalsF=+goalsF + +Results[i][1];
goalsA=+goalsA + +Results[i][2];
pts++
}
}
}
}
League[p][1]=pld;
League[p][2]=wins;
League[p][3]=draws;
League[p][4]=loses;
League[p][5]=goalsF;
League[p][6]=goalsA;
League[p][7]=pts;
// Reset the 'count' variables here:
var pld=0;
var wins=0;
var draws=0;
var loses=0;
var goalsF=0;
var goalsA=0;
var pts=0;**
}
var-pld=0;
var=0;
var=0;
var=0;
var goalsF=0;
var goalsA=0;
var-pts=0;
对于(p=0;p结果[i][2])
{
wins++;
goalsF=+goalsF++结果[i][1];
goalsA=+goalsA++结果[i][2];
pts=+pts+3;
}
否则如果(结果[i][1]为什么不使用更面向对象的方法来实现这一点?谢谢octameter。我对不同的方法持开放态度,但仍然是一个新手。你能分享任何链接吗?你能给你的表格添加标题,以便我们知道你想显示什么吗Chris。请原谅标题的位置,但希望更清楚。