我已经实现了可重复使用的排序功能,通过数字和文字工作正常排序,但它未能按日期排序。
orderBy(array: Array<any>, fieldName: string, direction: string) {
return array.sort((a, b) => {
let objectA: number|string = '';
let objectB: number|string = '';
[objectA, objectB] = [a[fieldName], b[fieldName]];
let valueA = isNaN(+objectA) ? objectA.toString().toUpperCase() : +objectA;
let valueB = isNaN(+objectB) ? objectB.toString().toUpperCase() : +objectB;
return (valueA < valueB ? -1 : 1) * (direction == 'asc' ? 1 : -1);
});
}
如何通过日期,文本数字和特殊字符进行排序。
尝试这个:
orderBy(array: Array<any>, fieldName: string, direction: string) {
return array.sort((a, b) => {
let objectA: number|string|Date = '';
let objectB: number|string|Date = '';
[objectA, objectB] = [a[fieldName], b[fieldName]];
// I assume that objectA and objectB are of the same type
return typeof objectA === 'string' ? objectA.localeCompare(objectB) : objectA - objectB;
});
}
如果Date
类型无法识别,则可能需要添加es6
进入你的compilerOptions
,看到这个回答更多细节
UPDATE
如果要排序的所有值都是字符串试试这个:
orderBy(array: Array<any>, fieldName: string, direction: string) {
return array.sort((a, b) => {
let objectA: number|string|Date = '';
let objectB: number|string|Date = '';
// map function here will convert '15/12/2018' into '2018/12/15'
// so we can compare values as strings, numbers and strings
// will remain unchanged
[objectA, objectB] = [a[fieldName], b[fieldName]].map(i => i.split('/').reverse().join('/'));
return isNaN(+objectA) ? objectA.localeCompare(objectB) : +objectA - +objectB;
});
}
我认为这可能是最好的,因为localCompare 功能 后返回正面和前底片和0的equals(在这个例子中,我比较.NAME至极的对象A和B的属性附加伤害是在this.array)
this.array.sort((a, b) => {
return a.name.localeCompare(b.name);
});
如果您使用的脚本类型比你可以像这样,还没有尝试,但你可以试模
public orderby(fieldName : string)
{
switch (typeof obj[fieldName].constructor.name) {
case "String":
array.sort();
case "Number":
array.sort(function(a,b){
return a-b);
});
case "Date"://you can check type and add accordingly
//as suggested by @Andriy its going to be object
array.sort(function(a,b){
return new Date(b.date) - new Date(a.date);
});
default:
throw new Error("Type of T is not a valid return type!");
}
} else {
throw new Error("Key '" + key + "' does not exist!");
}
}
日期排序我喜欢这一点,在日期转换值,并获得返回零减运算值,加上或负vlaue
array.sort(function(a,b){
return new Date(b.date) - new Date(a.date);
});