Angular - 动态操作过滤条件

Question

我有 3 个数组。

    var severityFilters = ["High", "Medium", "Low"]

 var ruleFilters = ["true", "false"] 

var allResults = [

{
    "name": "Test1",
    "severity": "High",
    "clean": true
},

{
    "name": "Test2",
    "severity": "Low",
    "clean": false
},

{
    "name": "Test3",
    "severity": "Medium",
    "clean": false
},

{
    "name": "Test4",
    "severity": "High",
    "clean": true
}

]

severityFilters 数组和ruleFilters 是在用户选中和取消选中某些过滤器复选框时动态创建的。因此，如果没有选中 severityFilters 复选框，则该数组将为空。如果只检查了High，那么severityFilters 将包含High。与ruleFilters 复选框相同。

我想根据severityFilters 和ruleFilters 中的值编写一个函数来返回一些过滤结果。 filterResults 函数在选中每个过滤器复选框时被调用。

var filteredResults = []

    filterResults (severityFilters, ruleFilters, allResults)
    {
       for (let i = 0; i < allResults.length; i++) {
            if (severityFilters.length > 0 && ruleFilters.length > 0)
            {
              if (ruleFilters.includes(this.allResults[i].clean.toString()) && 
severityFilters.includes(allResults[i].severity)
              )
              {
                this.filteredResults = [...this.filteredResults, ...allResults[i]];
              }
            }
    }
    } 

考虑到severityFilters 和ruleFilters 都已指定，此函数有效。

如何检查以下场景的过滤：

1. ruleFilters 可能为空
2. severityFilters 可能为空。
3. 定义了一个或多个severityFilters，但不存在ruleFilters，反之亦然。

如果不写冗长的 if 条件，有没有办法调整函数，以便过滤结果，无论 severityFilters 和 ruleFilters 数组的长度如何。

Answer 1

您可以先检查 filter 数组的 length，然后再检查它们是否包含属性值。如果severityFilters 为空，!severityFilters.length 将返回 true。因此，由于||，它不会检查第二个includes 条件。如果severityFilters 确实有一些值，那么severityFilters.includes(o.severity) 将被检查，只有severityFilters 中具有severity 值的对象将被过滤

const severityFilters = ["High"]
const ruleFilters = ["true", "false"]
var allResults=[{name:"Test1",severity:"High",clean:true},{name:"Test2",severity:"Low",clean:false},{name:"Test3",severity:"Medium",clean:false},{name:"Test4",severity:"High",clean:true}];

const output = allResults.filter(o =>
  (!severityFilters.length || severityFilters.includes(o.severity)) &&
  (!ruleFilters.length || ruleFilters.includes(o.clean.toString()))
)

console.log(output)