Question

我有下面的对象，需要从结果中删除列，我将动态获取列名。您能帮我如何基于列名{"columnname":"couln2", "datatype":null}删除列及其对应的对象数组是：

{
"tabl1":
{"tablename":"tabl1","tablecolumns":"yes","patternCheckStatus":true,
"columns": [{"columnname":"column1","datatype":"Numeric","patternregex":"jjj"},{"columnname":"column2","datatype":"UpperCase","patternregex":"hkl;;"}]},
"table2":{"tablename":"table2","tablecolumns":"yes","patternCheckStatus":null,
"columns":[{"columnname":"t2column","datatype":"Alphabetic"}]
}}

let arr = 
    {"tabl1":{"tablename":"tabl1","tablecolumns":"yes","patternCheckStatus":true,"columns":[{"columnname":"column1","datatype":"Numeric","patternregex":"jjj"},{"columnname":"column2","datatype":"UpperCase","patternregex":"hkl;;"}]},"table2":{"tablename":"table2","tablecolumns":"yes","patternCheckStatus":null,"columns":[{"columnname":"t2column","datatype":"Alphabetic"}]}}


   

    const result = arr.reduce((a, {tablename, tablecolumns, columnname, datatype})   => {
        a[tablename] = a[tablename] || {tablename, tablecolumns, columns: []};
        if (columnname)
          a[tablename].columns.push({columnname, datatype});
        return a;
      },{})
    console.log(Object.values(result));

Answer 1

我无法正确理解您的问题，但我认为您想从columnname中删除具有特定arr值的特定对象。您可以这样过滤arr：

function deleteColumn (column) {
  let newArr = arr.filter(item => {
     return item.columnname !== column
  })
  return newArr
}

然后您可以运行：

deleteColumn('tabl2_colu') // Will return an array without object having any columnname = 'tabl2_colu'

Answer 2

如果要摆脱数组中的某些列，则可以在将Object.entries的数组放入另一个数组时使用map：

let propertyName = 'tablename';
arr.map(s=> Object.fromEntries(Object.entries(s).filter(([k, v]) => k!= propertyName)))

一个例子：

let arr = [
    {"tablename":"table1","tablecolumns":"yes"},
    {"tablename":"table1","columnname":"col1","datatype":"Alphabetic"},
    {"tablename":"table2","tablecolumns":"yes"},
    {"tablename":"table2","columnname":"tabl2_colu","datatype":null},
    {"tablename":"table2","columnname":"tab2_col2","datatype":"Numeric"}
];

const result = arr.reduce((a, {tablename, tablecolumns, columnname, datatype})   => {
    a[tablename] = a[tablename] || {tablename, tablecolumns, columns: []};
    if (columnname)
        a[tablename].columns.push({columnname, datatype});
    return a;
    },{})

let propertyName = 'tablename';
console.log(Object.values(result)
    .map(s=> Object.fromEntries(Object.entries(s)
        .filter(([k, v]) => k!= propertyName))));

更新：

您可以基于columnname过滤数组：

let columnname = 'column2';
obj.tabl1.columns = obj.tabl1.columns.filter(f=> f.columnname != columnname);

一个例子：

let obj = { "tabl1":
  { "tablename": "tabl1", "tablecolumns": "yes", "patternCheckStatus": true,
  "columns": [{ "columnname": "column1", "datatype": "Numeric", "patternregex": "jjj" },
  { "columnname": "column2", "datatype": "UpperCase", "patternregex": "hkl;;" }] }, };

let columnname = 'column2';
obj.tabl1.columns = obj.tabl1.columns.filter(f=> f.columnname != columnname);
console.log(obj);

Answer 3

一种方法是遍历对象数组，通过其属性标识符找到要删除的对象，然后将它们从数组中删除：

arr = [
{"tablename":"table1","tablecolumns":"yes"},
{"tablename":"table1","columnname":"col1","datatype":"Alphabetic"},
{"tablename":"table2","tablecolumns":"yes"},
{"tablename":"table2","columnname":"tabl2_colu","datatype":null},
{"tablename":"table2","columnname":"tab2_col2","datatype":"Numeric"}
];

function remove_object_by_colname( colname )
  arr.forEach(function( arrayItem, index ) {
    if ( arrayItem.columnname == colname ) {
      arr.splice( index, 1 );
    }
  };
} );

remove_object_by_colname( 'tabl2_colu' ); // Will remove the 4th object from arr.

从对象删除数据

3 个答案: