【发布时间】:2018-01-29 00:47:45
【问题描述】:
我有一个问题,如果有人能帮我找到答案,我会很高兴的。
我有一个对象:
myObj = {
"red":{
"10729,10730,10732":{
"-2.75":["+1.75","+1.50","+1.25","+1.00","+0.75","+0.50","+0.25","-0.25"],
"-2.50":["+2.00","0.00","-0.50","-0.75","-1.00","-1.25"],
"-2.25":["+2.00","0.00","-1.50","-1.75","-2.00"],
"-2.00":["+2.00","0.00","-2.00","-2.25"],
"-1.75":["+2.00","0.00","-2.25"],
"-1.50":["+2.00","0.00","-2.25"],
"-1.25":["-2.25"],
"-1.00":["+2.00","-2.25"],
"-0.75":["+2.00","-2.25"],
"-0.50":["+2.00","-2.25"],
"-0.25":["-2.25"],
"0.00":["+1.75","0.00","-2.25"],
"+0.25":["-2.00","-2.25"],
"+0.50":["+1.75","-1.75"],
"+0.75":["+1.75","0.00","-1.50","-1.75"],
"+1.00":["0.00","-1.50"],
"+1.25":["+1.75","0.00","-1.25"],
"+1.50":["+1.50","0.00","-1.00"],
"+1.75":["0.00","-0.75","-1.00"],
"+2.00":["+1.50","-0.50","-0.75"],
"+2.25":["+1.50","-0.25","-0.50"],
"+2.50":["+1.50","+0.25","0.00","-0.25"],
"+2.75":["+1.50","+0.75","+0.50","+0.25"],
"+3.00":["+1.75","+1.50","+1.25","+1.00","+0.75"],
"+3.25":["+1.75","+1.50"]
}
},
"gray":{
"10730,10731":{
"-1.00":["+0.25","0.00","-0.25","-0.50"],
"-0.75":["+0.75","+0.50","+0.25","-0.50","-0.75"],
"-0.50":["+0.75","-0.75"],
"-0.25":["+0.75","-0.75"],
"0.00":["+3.25","+3.00","+2.75","+2.50","+2.25","+1.00","+0.75","-0.50","-0.75","-1.50","-1.75"],
"+0.25":["+1.00","-0.50"],"+0.50":["+1.00","-0.50"],"+0.75":["+1.00","-0.50"]
}
}
};//end of myObj
我正在遍历它并将值作为选项附加到选择标记中 这是我的代码:
for(h in myObj){ //h is color
for (x in myObj[h]) { //x is diameter
for(y in myObj[h][x]){ //y is cylinder
$(".cyl").append('<option value="">' + y + '</option>');
for(z in myObj[h][x][y]){ //// z is array of sph,and myObj[h][x][y][z] is the array elements
$(".sph").append('<option value="">' + myObj[h][x][y][z] + '</option>');
} ////sph
}////cyl
}
}
循环完美运行。我不想推送所有元素,而是推送数组值。或类似依赖选项列表的东西。
例如,用户选择第一个圆柱体"-2.75",第二个选择标签将仅显示其数组中的值
["+1.75","+1.50","+1.25","+1.00","+0.75","+0.50","+0.25","-0.25"]
有什么想法吗? 提前致谢。
【问题讨论】:
标签: javascript jquery select dynamic