javascript - 如何使用 indexOf() 检查我的数组中的字符串是否部分匹配另一个数组中的字符串？

Question

我有一个功能可以过滤特定项目的表格。在我的应用程序中，我有许多带有不同过滤器的表，因此我编写了一个可以使用动态数据过滤表的函数。一些过滤器使用用户必须输入查询的输入。如果用户的查询仅部分匹配表数组之一中的字符串，则该表不会返回预期结果。

例如

filterArr = [null, null, "246", null]
TableFilter = [
    {"name": Matthew Shapiro", "claim_number":"2469135-6", "type": "Medical", "date": "05/04/2018", "doctor": "John Smith", "hospital": "University Hospital"},
    {"name": "John Connor", "claim_number": "1352460-2", "type": "Medical", "date": "01/01/2018", "doctor": "John Smith", "hospital": "University Hospital"},
    {"name": "Sam Heist", "claim_number": "1270120-9", "type": "Medical", "date": "02/11/2018", "doctor": "William Rodgers", "hospital":  "Springfield Hospital"}
]

在此示例中，即使在“2469135-6”和“1352460-2”中可以找到 246，该表也不会返回前两行。如果我将过滤器中的查询更新为任一数字的完整字符串，它会。

如果 filterArr 中的字符串部分匹配 TableFilter[i] 中的字符串，我如何更新我的函数以返回适当的 TableFilters？

下面是我的函数以及我的一张表的屏幕截图。

let rowArr = []
let filteredResults = []

this.rows.map(element => {
    rowArr = Object.values(element); // Object.values of each row
    rowArr.filter(e => { 
        if (rowArr.indexOf(filterArr) !== -1) {
            // check for substring match. Currently not working
            console.log('match');
        }

        if (filterArr.includes(e)) { // check if filter has value that matches row
            filteredResults.push(element);
            this.rows = filteredResults;
        }
    })
})

标签： javascriptarraysfiltersubstringindexof

Answer 1

您想要过滤TableFilter 数组中某些字段值包含某些filterArray 值的行。

var filterArr = [null, null, "246", null];
var TableFilter = [
{"name": "Matthew Shapiro", "claim_number":"2469135-6", "type": "Medical", "date": "05/04/2018", "doctor": "John Smith", "hospital": "University Hospital"},
{"name": "John Connor", "claim_number": "1352460-2", "type": "Medical", "date": "01/01/2018", "doctor": "John Smith", "hospital": "University Hospital"},
{"name": "Sam Heist", "claim_number": "1270120-9", "type": "Medical", "date": "02/11/2018", "doctor": "William Rodgers", "hospital":  "Springfield Hospital"}
];

var foundResults = TableFilter.filter(row => {
  return Object.keys(row).some(fieldName => {
    return filterArr.some(filt => {
      return row[fieldName].indexOf(filt) > -1;
    });
  });
});

console.log(foundResults);