查找属性的第一次出现(不需要完全匹配)

Question

您能告诉我如何使用Underscore.js查找属性的第一次出现(不需要精确匹配)吗？

我试过this.But，它只返回完全匹配的属性。假设在我的列表中，它有townName，比如Rocky Point，.But，我需要选择Rocky also.So，我怎么能做到呢？

var schoolDistrict = _.find(vm.schoolDistricts, function (d) { return d.townName 
== result.data[0].PropertyData.School; }); 

(例如)

vm.schoolDistricts = [townName:'Rocky Point',...]

result.data[0].PropertyData.School = 'Rocky'

Answer 1

OP的反馈:我已经使用了如下所示的，并且运行良好:)

var schoolDistrict = _.find(vm.schoolDistricts, function (s) { return 
s.townName.toLowerCase().indexOf(result.data[0].PropertyData.School.toLowerCase(
)) !== -1; });

另一种方式:

如果其中一个字符串是另一个字符串的子字符串，则可以执行基本检查。还可以将字符串小写，以便进行更好的搜索：

​

var schoolDistricts = [
  {townName:'Rocky Point'}, 
  {townName: 'Roc'}
];

var searchString = 'Rocky'.toLowerCase();

var res = _.find(schoolDistricts, function(s) {
  var townName = s.townName.toLowerCase();
  return townName.indexOf(searchString) !== -1 || searchString.indexOf(townName) !== -1; 
});

console.log(res);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

​

Answer 2

使用RegExp模式测试名称是否包含文本，并使用i标志忽略大小写：

​

var schoolDistricts = [{
  townName: 'South Rock'
}, {
  townName: 'Red Herring'
}, {
  townName: 'Rocky Point'
}, {
  townName: 'Whatnot'
}];

var school = 'Rocky'; // replacement for result.data[0].PropertyData.School

var pattern = new RegExp(school, 'i');

var result = _.find(schoolDistricts, function(d) {
  return pattern.test(d.townName);
});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

​