how to remove json object key and value

Question

I have a json object as shown below  where i want to delete the  otherIndustry  entry and its value by using below code which doesn t worked   var updatedjsonobj   delete myjsonobj  otherIndustry      How to remove Json object specific key and its value  Below is my example json object where i want to remove  otherIndustry  key and its value   var myjsonobj             employeeid    160915848          firstName    tet          lastName    test          email    test email com          country    Brasil          currentIndustry    aaaaaaaaaaaaa          otherIndustry    aaaaaaaaaaaaa          currentOrganization    test          salary    1234567         delete myjsonobj   otherIndustry    console log myjsonobj     where the log still prints the same object without removing  otherIndustry  entry from the object

User · Answer

Follow this  it can be like what you are looking    x000D   x000D  var obj     x000D      Objone   one   x000D      Objtwo   two  x000D     x000D   x000D  var key    Objone   x000D  delete obj key   x000D  console log obj      prints    objtwo   two  x000D   x000D   x000D

User · Answer

I had issues with trying to delete a returned JSON object and found that it was actually a string  If you JSON parse   before deleting you can be sure your key will get deleted   let obj  console log this getBody          AED  3 6729  AZN  1 69805  BRL  4 0851  obj   this getBody    delete obj  BRL    console log obj       AED  3 6729  AZN  1 69805  BRL  4 0851  obj   JSON parse this getBody     delete obj  BRL    console log obj       AED  3 6729  AZN  1 69805

User · Answer

Here is one more example   check the reference    x000D   x000D  const myObject     x000D     employeeid    160915848   x000D     firstName    tet   x000D     lastName    test   x000D     email    test email com   x000D     country    Brasil   x000D     currentIndustry    aaaaaaaaaaaaa   x000D     otherIndustry    aaaaaaaaaaaaa   x000D     currentOrganization    test   x000D     salary    1234567  x000D     x000D  const  otherIndustry     otherIndustry2    myObject  x000D  console log otherIndustry2   x000D   as-console-wrapper   x000D    max-height  100   important  x000D    top  0  x000D    x000D   x000D   x000D

User · Answer

There are several ways to do this  lets see them one by one    delete method  The most common way    x000D   x000D  const myObject     x000D       employeeid    160915848   x000D       firstName    tet   x000D       lastName    test   x000D       email    test email com   x000D       country    Brasil   x000D       currentIndustry    aaaaaaaaaaaaa   x000D       otherIndustry    aaaaaaaaaaaaa   x000D       currentOrganization    test   x000D       salary    1234567  x000D     x000D   x000D  delete myObject  currentIndustry    x000D     OR delete myObject currentIndustry  x000D     x000D  console log myObject   x000D   x000D   x000D     By making key value undefined  Alternate  amp  a faster way     x000D   x000D  let myObject     x000D       employeeid    160915848   x000D       firstName    tet   x000D       lastName    test   x000D       email    test email com   x000D       country    Brasil   x000D       currentIndustry    aaaaaaaaaaaaa   x000D       otherIndustry    aaaaaaaaaaaaa   x000D       currentOrganization    test   x000D       salary    1234567  x000D       x000D   x000D  myObject currentIndustry   undefined  x000D  myObject   JSON parse JSON stringify myObject    x000D   x000D  console log myObject   x000D   x000D   x000D     With es6 spread Operator     x000D   x000D  const myObject     x000D       employeeid    160915848   x000D       firstName    tet   x000D       lastName    test   x000D       email    test email com   x000D       country    Brasil   x000D       currentIndustry    aaaaaaaaaaaaa   x000D       otherIndustry    aaaaaaaaaaaaa   x000D       currentOrganization    test   x000D       salary    1234567  x000D     x000D   x000D   x000D  const  currentIndustry     filteredObject    myObject  x000D  console log filteredObject   x000D   x000D   x000D    Or if you can use omit   of underscore js library   const filteredObject     omit currentIndustry   myObject    console log filteredObject     When to use what    If you don t wanna create a new filtered object  simply go for either option 1 or 2  Make sure you define your object with let while going with the second option as we are overriding the values  Or else you can use any of them   hope this helps

User · Answer

delete operator is used to remove an object property  delete operator does not returns the new object  only returns a boolean   true or false  In the other hand  after interpreter executes var updatedjsonobj   delete myjsonobj  otherIndustry      updatedjsonobj variable will store a boolean value   How to remove Json object specific key and its value    You just need to know the property name in order to delete it from the object s properties  delete myjsonobj  otherIndustry      x000D   x000D  let myjsonobj        employeeid    160915848      firstName    tet      lastName    test      email    test email com      country    Brasil      currentIndustry    aaaaaaaaaaaaa      otherIndustry    aaaaaaaaaaaaa      currentOrganization    test      salary    1234567    delete myjsonobj  otherIndustry    console log myjsonobj   x000D   x000D   x000D   If you want to remove a key when you know the value you can use Object keys function which returns an array of a given object s own enumerable properties   x000D   x000D  let value  test   let myjsonobj            employeeid    160915848          firstName    tet          lastName    test          email    test email com          country    Brasil          currentIndustry    aaaaaaaaaaaaa          otherIndustry    aaaaaaaaaaaaa          currentOrganization    test          salary    1234567    Object keys myjsonobj  forEach function key     if  myjsonobj key      value        delete myjsonobj key           console log myjsonobj   x000D   x000D   x000D

User · Answer

function omit obj  key        const   key  ignore     rest    obj      return rest     You can use ES6 spread operators like this  And to remove your key simply call const newJson   omit myjsonobj   quot otherIndustry quot     Its always better if you maintain pure function when you deal with type object in javascript

[javascript] how to remove json object key and value.?

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