X + Y gives HTMLObject (Select/option)

I'm trying to add 2 sums, that are getting fetched by an select option.

All I want is to calculate X + Y, but i get a weird error.

Code:

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(this + parseInt(id1) + parseInt(id2));    

});

https://jsfiddle.net/xpvt214o/957709/

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

2

this on your final line of code is casing the issue - remove it and you should be okay?

– Christheoreo
Nov 22 '18 at 12:13

To do: Stop trying to “add” numbers onto this …

– misorude
Nov 22 '18 at 12:13

Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add this to smth?

– Konstantin Kudelko
Nov 22 '18 at 12:14

add a comment |

I'm trying to add 2 sums, that are getting fetched by an select option.

All I want is to calculate X + Y, but i get a weird error.

Code:

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(this + parseInt(id1) + parseInt(id2));    

});

https://jsfiddle.net/xpvt214o/957709/

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

2

this on your final line of code is casing the issue - remove it and you should be okay?

– Christheoreo
Nov 22 '18 at 12:13

To do: Stop trying to “add” numbers onto this …

– misorude
Nov 22 '18 at 12:13

Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add this to smth?

– Konstantin Kudelko
Nov 22 '18 at 12:14

add a comment |

I'm trying to add 2 sums, that are getting fetched by an select option.

All I want is to calculate X + Y, but i get a weird error.

Code:

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(this + parseInt(id1) + parseInt(id2));    

});

https://jsfiddle.net/xpvt214o/957709/

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

I'm trying to add 2 sums, that are getting fetched by an select option.

All I want is to calculate X + Y, but i get a weird error.

Code:

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(this + parseInt(id1) + parseInt(id2));    

});

https://jsfiddle.net/xpvt214o/957709/

javascript jquery html html5

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

edited Nov 22 '18 at 12:18

Peter B

13.3k52044

asked Nov 22 '18 at 12:11

BGH

132

asked Nov 22 '18 at 12:11

BGH

132

asked Nov 22 '18 at 12:11

BGH

132

2

this on your final line of code is casing the issue - remove it and you should be okay?

– Christheoreo
Nov 22 '18 at 12:13

To do: Stop trying to “add” numbers onto this …

– misorude
Nov 22 '18 at 12:13

Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add this to smth?

– Konstantin Kudelko
Nov 22 '18 at 12:14

add a comment |

2

this on your final line of code is casing the issue - remove it and you should be okay?

– Christheoreo
Nov 22 '18 at 12:13

To do: Stop trying to “add” numbers onto this …

– misorude
Nov 22 '18 at 12:13

Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add this to smth?

– Konstantin Kudelko
Nov 22 '18 at 12:14

this on your final line of code is casing the issue - remove it and you should be okay?

– Christheoreo
Nov 22 '18 at 12:13

To do: Stop trying to “add” numbers onto this …

– misorude
Nov 22 '18 at 12:13

Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add this to smth?

– Konstantin Kudelko
Nov 22 '18 at 12:14

add a comment |

5 Answers
5

active

oldest

votes

You have this in your #final_value element. I've also added a change event to your code so it will update when one of the options change:

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

add a comment |

There's a this in the last statement, replace the last line with:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Now it should work.

answered Nov 22 '18 at 12:14

François Lanzeray

8318

add a comment |

Remove this value:

From:

$("#final_value").val(this + parseInt(id1) + parseInt(id2));

To:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Cause if you will use this your function returns a string as:

[object HTMLDocument]11

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

add a comment |

I have altered Gary Thomas Code little bit:

Instead of parseInt(), you can use like this:

var id1 = +$("#one").val();

var id2 = +$("#two").val();

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

add a comment |

Just remove this keyword from your summation as so :

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(parseInt(id1) + parseInt(id2));    

});

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

You have this in your #final_value element. I've also added a change event to your code so it will update when one of the options change:

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

add a comment |

You have this in your #final_value element. I've also added a change event to your code so it will update when one of the options change:

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

add a comment |

You have this in your #final_value element. I've also added a change event to your code so it will update when one of the options change:

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

You have this in your #final_value element. I've also added a change event to your code so it will update when one of the options change:

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = $("#one").val();

        var id2 = $("#two").val();



        $("#final_value").val(parseInt(id1) + parseInt(id2));

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

answered Nov 22 '18 at 12:16

Gary Thomas

1,5151415

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

add a comment |

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

Thank's! That was working! Thank you so much Gary. I wish you the best :-)

– BGH
Nov 23 '18 at 8:11

add a comment |

There's a this in the last statement, replace the last line with:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Now it should work.

answered Nov 22 '18 at 12:14

François Lanzeray

8318

add a comment |

There's a this in the last statement, replace the last line with:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Now it should work.

answered Nov 22 '18 at 12:14

François Lanzeray

8318

add a comment |

There's a this in the last statement, replace the last line with:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Now it should work.

answered Nov 22 '18 at 12:14

François Lanzeray

8318

There's a this in the last statement, replace the last line with:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Now it should work.

answered Nov 22 '18 at 12:14

François Lanzeray

8318

answered Nov 22 '18 at 12:14

François Lanzeray

8318

answered Nov 22 '18 at 12:14

François Lanzeray

8318

answered Nov 22 '18 at 12:14

François Lanzeray

8318

add a comment |

Remove this value:

From:

$("#final_value").val(this + parseInt(id1) + parseInt(id2));

To:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Cause if you will use this your function returns a string as:

[object HTMLDocument]11

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

add a comment |

Remove this value:

From:

$("#final_value").val(this + parseInt(id1) + parseInt(id2));

To:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Cause if you will use this your function returns a string as:

[object HTMLDocument]11

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

add a comment |

Remove this value:

From:

$("#final_value").val(this + parseInt(id1) + parseInt(id2));

To:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Cause if you will use this your function returns a string as:

[object HTMLDocument]11

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

Remove this value:

From:

$("#final_value").val(this + parseInt(id1) + parseInt(id2));

To:

$("#final_value").val(parseInt(id1) + parseInt(id2));

Cause if you will use this your function returns a string as:

[object HTMLDocument]11

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

answered Nov 22 '18 at 12:15

Konstantin Kudelko

1278

add a comment |

I have altered Gary Thomas Code little bit:

Instead of parseInt(), you can use like this:

var id1 = +$("#one").val();

var id2 = +$("#two").val();

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

add a comment |

I have altered Gary Thomas Code little bit:

Instead of parseInt(), you can use like this:

var id1 = +$("#one").val();

var id2 = +$("#two").val();

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

add a comment |

I have altered Gary Thomas Code little bit:

Instead of parseInt(), you can use like this:

var id1 = +$("#one").val();

var id2 = +$("#two").val();

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

I have altered Gary Thomas Code little bit:

Instead of parseInt(), you can use like this:

var id1 = +$("#one").val();

var id2 = +$("#two").val();

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

	$(document).ready(function() {

    $("#one, #two").on("change", () => {

        var id1 = +$("#one").val();

        var id2 = +$("#two").val();

        var total = id1+id2;

        $("#final_value").val(total);

    });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>



<td>

    <select class="full-width select2" name="" id="one" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<td>

    <select class="full-width select2" name="" id="two" data-placeholder="" >

        <option value="1">1</option>

        <option value="2">2</option>

        <option value="3">3</option>

        <option value="4">4</option>

    </select>

</td>

<label for="final_value">Result:</label>

<input type="text" value="" id="final_value">

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

answered Nov 22 '18 at 12:24

Aravind Bhat K

2741214

add a comment |

Just remove this keyword from your summation as so :

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(parseInt(id1) + parseInt(id2));    

});

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

add a comment |

Just remove this keyword from your summation as so :

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(parseInt(id1) + parseInt(id2));    

});

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

add a comment |

Just remove this keyword from your summation as so :

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(parseInt(id1) + parseInt(id2));    

});

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

Just remove this keyword from your summation as so :

$( document ).ready(function() {

    var id1 = $("#one").val();

    $("#final_value").val(id1);



    var id2 = $("#two").val();

    $("#final_value").val(id2);



    $("#final_value").val(parseInt(id1) + parseInt(id2));    

});

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

answered Nov 22 '18 at 12:44

Voice Of The Rain

123211

add a comment |

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