从对象中删除数据

Posted 2023-03-07

【中文标题】从对象中删除数据

【英文标题】：Delete data from object

【发布时间】：2020-05-03 07:50:29

【问题描述】：

我有下面的对象，需要从结果中删除列，我将动态获取列名。你能帮我如何根据列名删除列及其对应的对象"columnname":"couln2", "datatype":null 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));

【问题讨论】：

分享你需要的结果数组

您的意思是删除属性column 并删除tabl2_colu？ tabl2_colu在哪里？

@Ahsan ,@nopole，请查看更新后的数组详情

【参考方案1】：

如果你想删除数组中的某些列，那么你可以使用Object.entries 而map 你的数组进入另一个数组：

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：

您可以根据columnname 过滤您的数组：

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);

【讨论】：

谢谢你的回答，但我遇到了错误，'ObjectConstructor'.ts 类型上不存在属性'Fromentries'

@user2319726 看起来您需要将"lib": ["es2017"] 添加到您的设置中。 Here yo can read more :)

我需要删除完整的列名对象，我认为现在它只删除列名键值， "tabl1": "tablename":"tabl1","tablecolumns":"yes"," patternCheckStatus":true, "columns": ["columnname":"column1","datatype":"Numeric","patternregex":"jjj",

@user 你能写出想要的输出吗？

请看下面，删除前： "tabl1": "tablename":"tabl1","tablecolumns":"yes","patternCheckStatus":true, "columns": [ "columnname":"column1","datatype":"Numeric","patternregex":"jjj","columnname":"column2","datatype":"UpperCase","patternregex":"hkl;; "], 删除后预期 "tabl1": "tablename":"tabl1","tablecolumns":"yes","patternCheckStatus":true, "columns": ["columnname":"column1" ,"datatype":"Numeric","patternregex":"jjj"],

【参考方案2】：

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，然后从数组中删除该对象：

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.

【参考方案3】：

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

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

然后就可以运行了：

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

【讨论】：

感谢您的回答..请查看更新后的数组