Round off decimal value in Nearest 10 or Zero [duplicate]

This question already has an answer here:

C# - Math.Round

5 answers

Following values i would like to convert to round off figure. likes:

60.72 --> 60.70

170.76 --> 170.80

Currently, I'm converted to round off value using below method:

 getFee.ServiceRequestFee.ToString("N")

I'm not sure which Match.Round method suitable to my requirement.

Edit:

protected string Getroundoffdecimalvalue(string servicerequestsfee_val)

    {

        servicerequestsfee_val = Math.Round(Convert.ToDecimal(servicerequestsfee_val), 2).ToString();

        return servicerequestsfee_val;

    }

I used this function even after i return 60.72 only and my expectation should 60.70.

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

marked as duplicate by Community♦ Dec 31 '18 at 4:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So, Bankers’ Rounding?

– Jimi
Dec 31 '18 at 4:11

why not overload math.round to make it suitable for your requirement?

– Common Man
Dec 31 '18 at 4:15

1

See the examples in the link I posted. If you need a string representation, you can use "N2" as format.

– Jimi
Dec 31 '18 at 4:25

@CommonMan - you might be correct. I found the solution through Math.Round(yourNumber, 1). Thanks

– Jignesh Rajput
Dec 31 '18 at 4:35

add a comment |

This question already has an answer here:

C# - Math.Round

5 answers

Following values i would like to convert to round off figure. likes:

60.72 --> 60.70

170.76 --> 170.80

Currently, I'm converted to round off value using below method:

 getFee.ServiceRequestFee.ToString("N")

I'm not sure which Match.Round method suitable to my requirement.

Edit:

protected string Getroundoffdecimalvalue(string servicerequestsfee_val)

    {

        servicerequestsfee_val = Math.Round(Convert.ToDecimal(servicerequestsfee_val), 2).ToString();

        return servicerequestsfee_val;

    }

I used this function even after i return 60.72 only and my expectation should 60.70.

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

marked as duplicate by Community♦ Dec 31 '18 at 4:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So, Bankers’ Rounding?

– Jimi
Dec 31 '18 at 4:11

why not overload math.round to make it suitable for your requirement?

– Common Man
Dec 31 '18 at 4:15

1

See the examples in the link I posted. If you need a string representation, you can use "N2" as format.

– Jimi
Dec 31 '18 at 4:25

@CommonMan - you might be correct. I found the solution through Math.Round(yourNumber, 1). Thanks

– Jignesh Rajput
Dec 31 '18 at 4:35

add a comment |

This question already has an answer here:

C# - Math.Round

5 answers

Following values i would like to convert to round off figure. likes:

60.72 --> 60.70

170.76 --> 170.80

Currently, I'm converted to round off value using below method:

 getFee.ServiceRequestFee.ToString("N")

I'm not sure which Match.Round method suitable to my requirement.

Edit:

protected string Getroundoffdecimalvalue(string servicerequestsfee_val)

    {

        servicerequestsfee_val = Math.Round(Convert.ToDecimal(servicerequestsfee_val), 2).ToString();

        return servicerequestsfee_val;

    }

I used this function even after i return 60.72 only and my expectation should 60.70.

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

This question already has an answer here:

C# - Math.Round

5 answers

Following values i would like to convert to round off figure. likes:

60.72 --> 60.70

170.76 --> 170.80

Currently, I'm converted to round off value using below method:

 getFee.ServiceRequestFee.ToString("N")

I'm not sure which Match.Round method suitable to my requirement.

Edit:

protected string Getroundoffdecimalvalue(string servicerequestsfee_val)

    {

        servicerequestsfee_val = Math.Round(Convert.ToDecimal(servicerequestsfee_val), 2).ToString();

        return servicerequestsfee_val;

    }

I used this function even after i return 60.72 only and my expectation should 60.70.

This question already has an answer here:

C# - Math.Round

5 answers

c# double rounding

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

edited Dec 31 '18 at 4:18

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

asked Dec 31 '18 at 4:07

Jignesh Rajput

3,2852244

marked as duplicate by Community♦ Dec 31 '18 at 4:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Community♦ Dec 31 '18 at 4:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

So, Bankers’ Rounding?

– Jimi
Dec 31 '18 at 4:11

why not overload math.round to make it suitable for your requirement?

– Common Man
Dec 31 '18 at 4:15

1

See the examples in the link I posted. If you need a string representation, you can use "N2" as format.

– Jimi
Dec 31 '18 at 4:25

@CommonMan - you might be correct. I found the solution through Math.Round(yourNumber, 1). Thanks

– Jignesh Rajput
Dec 31 '18 at 4:35

add a comment |

So, Bankers’ Rounding?

– Jimi
Dec 31 '18 at 4:11

why not overload math.round to make it suitable for your requirement?

– Common Man
Dec 31 '18 at 4:15

1

See the examples in the link I posted. If you need a string representation, you can use "N2" as format.

– Jimi
Dec 31 '18 at 4:25

@CommonMan - you might be correct. I found the solution through Math.Round(yourNumber, 1). Thanks

– Jignesh Rajput
Dec 31 '18 at 4:35

So, Bankers’ Rounding?

– Jimi
Dec 31 '18 at 4:11

why not overload math.round to make it suitable for your requirement?

– Common Man
Dec 31 '18 at 4:15

See the examples in the link I posted. If you need a string representation, you can use "N2" as format.

– Jimi
Dec 31 '18 at 4:25

@CommonMan - you might be correct. I found the solution through Math.Round(yourNumber, 1). Thanks

– Jignesh Rajput
Dec 31 '18 at 4:35

add a comment |

2 Answers
2

active

oldest

votes

Math.Round(yourNumber, 1)

The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.

You need to overloadMath.round that takes the decimals parameter of
your choice and convenience.

answered Dec 31 '18 at 4:14

Common Man

1,27221226

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

add a comment |

Use Math.round and if needed convert the same to string.

  Math.Round(var_name,2)

answered Dec 31 '18 at 4:16

Gaurav

57029

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Math.Round(yourNumber, 1)

The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.

You need to overloadMath.round that takes the decimals parameter of
your choice and convenience.

answered Dec 31 '18 at 4:14

Common Man

1,27221226

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

add a comment |

Math.Round(yourNumber, 1)

The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.

You need to overloadMath.round that takes the decimals parameter of
your choice and convenience.

answered Dec 31 '18 at 4:14

Common Man

1,27221226

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

add a comment |

Math.Round(yourNumber, 1)

The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.

You need to overloadMath.round that takes the decimals parameter of
your choice and convenience.

answered Dec 31 '18 at 4:14

Common Man

1,27221226

Math.Round(yourNumber, 1)

The second parameter is number of decimal places to round to. In your case you want 1 decimal place as an end result.

You need to overloadMath.round that takes the decimals parameter of
your choice and convenience.

answered Dec 31 '18 at 4:14

Common Man

1,27221226

answered Dec 31 '18 at 4:14

Common Man

1,27221226

answered Dec 31 '18 at 4:14

Common Man

1,27221226

answered Dec 31 '18 at 4:14

Common Man

1,27221226

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

add a comment |

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

I updated my question. I already tried your suggestion.

– Jignesh Rajput
Dec 31 '18 at 4:19

add a comment |

Use Math.round and if needed convert the same to string.

  Math.Round(var_name,2)

answered Dec 31 '18 at 4:16

Gaurav

57029

add a comment |

Use Math.round and if needed convert the same to string.

  Math.Round(var_name,2)

answered Dec 31 '18 at 4:16

Gaurav

57029

add a comment |

Use Math.round and if needed convert the same to string.

  Math.Round(var_name,2)

answered Dec 31 '18 at 4:16

Gaurav

57029

Use Math.round and if needed convert the same to string.

  Math.Round(var_name,2)

answered Dec 31 '18 at 4:16

Gaurav

57029

answered Dec 31 '18 at 4:16

Gaurav

57029

answered Dec 31 '18 at 4:16

Gaurav

57029

answered Dec 31 '18 at 4:16

Gaurav

57029

add a comment |

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