Javascript 满足if条件时数组和和和的数组中断

我有一个像这样的js数组

const test_arr = [ [ 20, 7, 23, 19, 10, 15, 25 ],
 [ 20, 7, 23, 19, 10, 15, 8 ],
  [ 20, 7, 23, 19, 10, 15, 13 ],
  [ 20, 7, 23, 19, 10, 25, 8 ],
  [ 20, 7, 23, 19, 10, 25, 13 ],
  [ 20, 7, 23, 19, 10, 8, 13 ],
  [ 20, 7, 23, 19, 15, 25, 8 ],
  [ 20, 7, 23, 19, 15, 25, 13 ],
  [ 20, 7, 23, 19, 15, 8, 13 ],
  [ 20, 7, 23, 19, 25, 8, 13 ],
  [ 20, 7, 23, 10, 15, 25, 8 ],
  [ 20, 7, 23, 10, 15, 25, 13 ],
  [ 20, 7, 23, 10, 15, 8, 13 ],
  [ 20, 7, 23, 10, 25, 8, 13 ],
  [ 20, 7, 23, 15, 25, 8, 13 ],
  [ 20, 7, 19, 10, 15, 25, 8 ],
  [ 20, 7, 19, 10, 15, 25, 13 ],
  [ 20, 7, 19, 10, 15, 8, 13 ],
  [ 20, 7, 19, 10, 25, 8, 13 ],
  [ 20, 7, 19, 15, 25, 8, 13 ],
  [ 20, 7, 10, 15, 25, 8, 13 ],
  [ 20, 23, 19, 10, 15, 25, 8 ],
  [ 20, 23, 19, 10, 15, 25, 13 ],
  [ 20, 23, 19, 10, 15, 8, 13 ],
  [ 20, 23, 19, 10, 25, 8, 13 ],
  [ 20, 23, 19, 15, 25, 8, 13 ],
  [ 20, 23, 10, 15, 25, 8, 13 ],
  [ 20, 19, 10, 15, 25, 8, 13 ],
  [ 7, 23, 19, 10, 15, 25, 8 ],
  [ 7, 23, 19, 10, 15, 25, 13 ],
  [ 7, 23, 19, 10, 15, 8, 13 ],
  [ 7, 23, 19, 10, 25, 8, 13 ],
  [ 7, 23, 19, 15, 25, 8, 13 ],
  [ 7, 23, 10, 15, 25, 8, 13 ],
  [ 7, 19, 10, 15, 25, 8, 13 ],
  [ 23, 19, 10, 15, 25, 8, 13 ] ]

var组合_之前=[20,7,23,19,10,15,25,8,13]
这些是从九个选项中选择七个选项的结果

我只想在元素之和为100时返回一些元素

如果条件满足，如何生成reducer？
您可以使用
filter
对数组进行筛选，使用
reduce
对嵌套数组求和，并在筛选器中检查哪个数组和等于100
const test_arr=[[20,7,23,19,10,15,25]，
[ 20, 7, 23, 19, 10, 15, 8 ],
[ 20, 7, 23, 19, 10, 15, 13 ],
[ 20, 7, 23, 19, 10, 25, 8 ],
[ 20, 7, 23, 19, 10, 25, 13 ],
[ 20, 7, 23, 19, 10, 8, 13 ],
[ 20, 7, 23, 19, 15, 25, 8 ],
[ 20, 7, 23, 19, 15, 25, 13 ],
[ 20, 7, 23, 19, 15, 8, 13 ],
[ 20, 7, 23, 19, 25, 8, 13 ],
[ 20, 7, 23, 10, 15, 25, 8 ],
[ 20, 7, 23, 10, 15, 25, 13 ],
[ 20, 7, 23, 10, 15, 8, 13 ],
[ 20, 7, 23, 10, 25, 8, 13 ],
[ 20, 7, 23, 15, 25, 8, 13 ],
[ 20, 7, 19, 10, 15, 25, 8 ],
[ 20, 7, 19, 10, 15, 25, 13 ],
[ 20, 7, 19, 10, 15, 8, 13 ],
[ 20, 7, 19, 10, 25, 8, 13 ],
[ 20, 7, 19, 15, 25, 8, 13 ],
[ 20, 7, 10, 15, 25, 8, 13 ],
[ 20, 23, 19, 10, 15, 25, 8 ],
[ 20, 23, 19, 10, 15, 25, 13 ],
[ 20, 23, 19, 10, 15, 8, 13 ],
[ 20, 23, 19, 10, 25, 8, 13 ],
[ 20, 23, 19, 15, 25, 8, 13 ],
[ 20, 23, 10, 15, 25, 8, 13 ],
[ 20, 19, 10, 15, 25, 8, 13 ],
[ 7, 23, 19, 10, 15, 25, 8 ],
[ 7, 23, 19, 10, 15, 25, 13 ],
[ 7, 23, 19, 10, 15, 8, 13 ],
[ 7, 23, 19, 10, 25, 8, 13 ],
[ 7, 23, 19, 15, 25, 8, 13 ],
[ 7, 23, 10, 15, 25, 8, 13 ],
[ 7, 19, 10, 15, 25, 8, 13 ],
[ 23, 19, 10, 15, 25, 8, 13 ] ];
const result=test_arr.filter（arr=>arr.reduce（（a，b）=>a+b，0）==100）；
控制台日志（结果）
const test_arr=[[20,7,23,19,10,15,25]，
[ 20, 7, 23, 19, 10, 15, 8 ],
[ 20, 7, 23, 19, 10, 15, 13 ],
[ 20, 7, 23, 19, 10, 25, 8 ],
[ 20, 7, 23, 19, 10, 25, 13 ],
[ 20, 7, 23, 19, 10, 8, 13 ],
[ 20, 7, 23, 19, 15, 25, 8 ],
[ 20, 7, 23, 19, 15, 25, 13 ],
[ 20, 7, 23, 19, 15, 8, 13 ],
[ 20, 7, 23, 19, 25, 8, 13 ],
[ 20, 7, 23, 10, 15, 25, 8 ],
[ 20, 7, 23, 10, 15, 25, 13 ],
[ 20, 7, 23, 10, 15, 8, 13 ],
[ 20, 7, 23, 10, 25, 8, 13 ],
[ 20, 7, 23, 15, 25, 8, 13 ],
[ 20, 7, 19, 10, 15, 25, 8 ],
[ 20, 7, 19, 10, 15, 25, 13 ],
[ 20, 7, 19, 10, 15, 8, 13 ],
[ 20, 7, 19, 10, 25, 8, 13 ],
[ 20, 7, 19, 15, 25, 8, 13 ],
[ 20, 7, 10, 15, 25, 8, 13 ],
[ 20, 23, 19, 10, 15, 25, 8 ],
[ 20, 23, 19, 10, 15, 25, 13 ],
[ 20, 23, 19, 10, 15, 8, 13 ],
[ 20, 23, 19, 10, 25, 8, 13 ],
[ 20, 23, 19, 15, 25, 8, 13 ],
[ 20, 23, 10, 15, 25, 8, 13 ],
[ 20, 19, 10, 15, 25, 8, 13 ],
[ 7, 23, 19, 10, 15, 25, 8 ],
[ 7, 23, 19, 10, 15, 25, 13 ],
[ 7, 23, 19, 10, 15, 8, 13 ],
[ 7, 23, 19, 10, 25, 8, 13 ],
[ 7, 23, 19, 15, 25, 8, 13 ],
[ 7, 23, 10, 15, 25, 8, 13 ],
[ 7, 19, 10, 15, 25, 8, 13 ],
[ 23, 19, 10, 15, 25, 8, 13 ] ];
var resultar=[]；
对于（变量i=0；ilog（resultar）
此时，您可以采用不同的方法，采用所有组合，只返回按长度和总和匹配条件的组合


函数getCombinations（数组、长度、和）{
功能iter（阵列、温度）{
如果（温度长度===长度）{
如果（温度降低（（a，b）=>a+b）==总和）{
结果：推送（温度）；
}
返回；
}
如果（！array.length | | array.length+temp.length
作为控制台包装{max height:100%！important；top:0；}
发布的问题似乎根本不包含解决问题的方法。StackOverflow希望您能这样做，因为您的尝试有助于我们更好地了解您的需求。请编辑问题以显示您已尝试的内容，以便说明您遇到的具体障碍。有关详细信息，请参阅并使用。好的，我稍后不会删除我的尝试。~当然，但时间限制存在，请稍等。太晚了，使用filter by is Artyom更好！