JS: How to remove duplicates in a table?

JS: How to remove duplicates in a table?

I guess there's a simple solution...

Hey!

Maybe something like this:

Array.prototype.unique = function (){
    var r = new Array();
    o:for(var i = 0, n = this.length; i < n; i++){
        for(var x = 0, y = r.length; x < y; x++){
            if(r[x]==this[i]) continue o;}
        r[r.length] = this[i];}
    return r;
}

Use:

var myArray = ["a","b","c","c","a","d","b","b"];
alert(myArray.unique());

Hope that helps.

--

tomaxxi

http://indisnip.WordPress.com/

Tags: InDesign

Similar Questions

  • How to remove duplicates from a Table using PL/SQL

    I created the following tables with data:

    * Table 1: *.



    Department of CREATE TABLE

    (DEPARTMENT_ID 4 NUMBER PRIMARY KEY,

    DEPARTMENT_NAME VARCHAR2 (20).

    ADDRESS VARCHAR2 (20));







    INSERT INTO Department VALUES (10, 'ACCOUNTING', 'NEW YORK');

    INSERT INTO Department VALUES (20, "SEARCH", "DALLAS");

    INSERT INTO Department VALUES (30, 'SALES', 'CHICAGO');

    INSERT INTO Department VALUES (40, 'OPERATIONS', 'BOSTON');

    COMMIT;



    * 2nd table: *.

    CREATE TABLE employee

    (EMPLOYEE_ID NUMBER 4 PRIMARY KEY,

    EMPLOYEE_NAME VARCHAR2 (20).

    USE VARCHAR2 (20).

    MANAGER_ID NUMBER 4,

    HIRE_DATE DATE,

    NUMBER OF SALARY (9, 2)

    COMMISSION (9, 2) NUMBER,

    Department_id number 4 REFERENCES department (DEPARTMENT_ID));






    INSERT INTO employee

    VALUES (7839, 'KING', 'PRESIDENT', NULL, NOVEMBER 17, 81 ', 5000, NULL, 10);

    INSERT INTO employee

    VALUES (7698, "BLAKE", "MANAGER", 7839, MAY 1, 81 ', 2850, NULL, 30);

    INSERT INTO employee

    VALUES (7782, "CLARK", "MANAGER", 7839, JUNE 9, 81 ', 2450, NULL, 10);

    INSERT INTO employee

    VALUES (7566, "JONES", "MANAGER", 7839, APRIL 2, 81 ', 2975, NULL, 20);

    INSERT INTO employee

    VALUES (7654, 'MARTIN', 'SELLER', 7698, 28-SEP-81', 1250, 1400, 30);

    INSERT INTO employee

    VALUES (7499, 'ALLEN', 'SELLER', 7698, FEBRUARY 20, 81 ', 1600, 300, 30);

    INSERT INTO employee

    VALUES (7844, 'TURNER', 'SELLER', 7698, ' 08-SEP-81', 1500, NULL, 30);

    INSERT INTO employee

    VALUES (7900, 'JAMES', 'CLERKS', 7698, 3 DECEMBER 81', 950, NULL, 30);

    INSERT INTO employee

    VALUES (7521, 'WARD', 'SELLER', 7698, FEBRUARY 22, 81 ', 1 250, 500, 30);

    INSERT INTO employee

    VALUES (7902, "FORD", "ANALYST", 7566, DECEMBER 3, 81 ', 3000, NULL, 20);

    INSERT INTO employee

    VALUES (7369, 'SMITH', 'CLERKS', 7902, DECEMBER 17, 81 ', 800, NULL, 20);

    INSERT INTO employee

    VALUES (7788, 'SCOTT', 'ANALYST', 7566, DECEMBER 9, 82 ', 3000, NULL, 20);

    INSERT INTO employee

    VALUES (7876, ' ADAMS, "CLERK", 7788, 12 JANUARY, 83', 1100, NULL, 20);

    INSERT INTO employee

    VALUES (7934, 'MILLER', 'CLERKS', 7782, JANUARY 22, 82 ', 1300, NULL, 10);

    COMMIT;



    I need to print all Department names, employee and salary. I did this, but how can display the name of each Department only once?





    Here is my code:



    DECLARE

    v_department_name department.department_name%TYPE;

    v_employee_name employee.employee_name%TYPE;

    v_salary employee.salary%TYPE;

    CURSOR dept_cur

    IS

    SELECT employee_name, department_name, Department OF employee salary

    WHERE department.department_id = employee.department_id (+)

    ORDER BY department_name, employee_name;

    BEGIN

    OPEN dept_cur.

    EXTRACT the dept_cur IN v_department_name, v_employee_name, v_salary;

    DBMS_OUTPUT. PUT_LINE ('DEPARTMENT_NAME EMPLOYEE_NAME WAGES');

    DBMS_OUTPUT. PUT_LINE('---------------------------------------------------');

    While dept_cur % FOUND

    LOOP

    DBMS_OUTPUT. Put_line (RPAD (v_department_name, 10) |) ' ' || RPAD (v_employee_name, 10). ' ' || v_salary);

    EXTRACT the dept_cur IN v_department_name, v_employee_name, v_salary;

    END LOOP;

    CLOSE Dept_cur;

    END;

    I would appreciate if you could help me with this. I'm sure it's a quick and easy answer, but I can't do it. I spent so much time already.

    Published by: 831522 on January 26, 2011 11:44

    It is a problem of display/reports rather than a problem of duplication of query data. The answer depends on what tools you use to query and display the data. You can use the clause "break it" I suggested earlier if you use SQL Plus.

    Otherwise, you can try this based on the version of Oracle you are using. Someone else may be able to get the data in a single pass.

    SELECT decode(emp_id, 1, department_name) department_name, employee_name, salary
    FROM(SELECT department_name, employee_name, salary,
           ROW_NUMBER() OVER (PARTITION BY department_name order by department_name, employee_name) AS emp_id
         FROM department
         LEFT OUTER JOIN employee ON (department.department_id = employee.department_id)
    ORDER BY department_name, employee_name, emp_id);
    
  • How to remove duplicate pictures

    I imported all my photos in the photo library. How to remove duplicates of Photos (version 1.3) on OS X El Capitan without 3rd party software?

    Photos don't have a tool to detect the duplicates library. It relies on detecting when you import photos.

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    To search for duplicates, you need third-party software. These three are safe to use with Photos:

  • How to remove duplicate files/songs in my music library without having to click on each of them?

    original title: removal of duplicates

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    Hi I have lines like this in a table emp

    ENO ename age
    1 25
    26 2
    3 25
    4 26

    I don't have repititve enames... (here in this example: A).

    If I find something like that I should keep the first line and delete all other lines.

    How can I do this?

    as unique empowerment of ename... (except for this)

    Hi Renon,
    There are different ways could remove duplicate rows, but Franks solution seems good.
    With the help of ID you can get good results in the largest data sets. If you want Nwe minutes remain in the database and all others to delete then make use of eno and not rowid.

    DELETE FROM emp y
    WHERE       eno != (SELECT   Min(eno)
                        FROM     emp x
                        WHERE    x.NAME = y.NAME
                        GROUP BY NAME)
    

    HTH
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  • Photo library - iCloud how to remove duplicates

    All, someone knows something about OSX/IOS 'Photos' app will detect and remove duplicates photo9?  My library 'Photos' 63 299 photos & videos of 2135 from may 2016 and at least 10,000 + of these photos are duplicates (based on an analysis of Photosweeper). I have checked the results of the analysis of Photosweeper by doing a manual visual comparison and check images original in iphoto library (using the function location of show) and confirmed that these 10 000 + photos are images duplicate identical with names of different files in different parts of the photo library database.

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    A few other comments for those who have a double problem:

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    Hi all

    I have an obligation to remove duplicates of string separated by commas:

    Some examples of code (real data can be in millions)

    {code}

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    insert into test_dup_del values (2589, '1, 2, 1, 2, 1, 2');
    commit;

    (code}

    my output would be:

    1234, 1, 2, 3, 4'

    4567, '1, 2, 4'

    7894, '1, 2, 3'

    1478, 1, 2, 3, 4'

    2589, 1, 2'

    Thank you

    Rachid

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    SQL > ed
    A written file afiedt.buf

    1 with test_dup_del (num, dup_val) as)
    2 Select 1234, 1, 2, 3, 4, 3, 1' Union double all the
    3 select 4567, 1, 2, 2, 4, 4, 1' Union double all the
    4 Select 7894, 1, 2, 1, 3, 3, 1' Union double all the
    5 Select 1478, 1, 2, 4, 4, 3, 2' of all the double union
    6 select 2589, 1, 2, 1, 2, 1, 2' of the double
    7      )
    8  --
    9. end of test data
    10-
    11, split as)
    12. Select num
    13, trim (regexp_substr(dup_val,'[^,]+',1,level)) as val
    14 from test_dup_del
    15 connect by num = num prior
    16 and level<=>
    17 and prior sys_guid() is not null
    18                )
    19, dist in (select distinct)
    20                      num
    21                      ,val
    22 split
    23              )
    24 select num
    25, listagg (val, ',') within the Group (order by val) as vals
    dist 26
    27 * group by num
    SQL > /.
    NUM OF VALS
    ---------- ------------------------------
    1234 1,2,3,4
    1478 1,2,3,4
    2589 1.2
    4567 1,2,4
    7894 1,2,3

    5 selected lines.

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    My problem is I want to remove duplicates without losing the edited versions. only those that I have meged in.  The file has a lot of images which is not duplicated, so I can't start the process again?

    You can remove duplicates of Lightroom is removing recognizes them in the grid view.

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    Let us know if that helps.
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  • How to eliminate duplicates in this table

    SELECT sti.item,
    MI: ITEM_DESC,
    Count (STI. Item) item_count,.
    d.dept_name,
    d.group_no,
    d.BUD_INT,
    STI.standard_unit_retail,
    ISC.unit_cost,
    margin (sti.standard_unit_retail - isc.unit_cost).
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    County (sti.item) * isc.unit_cost Tot_Sales,.
    STH. TRAN_DATETIME
    Of
    sa_tran_item sti,
    item_supp_country Sai,
    IM item_master,
    STARTING d,
    sth sa_tran_head
    WHERE
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    and sti.item = im.item
    and d.DEPT = sti. DEPT
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    When I combine this STH. TRAN_DATETIME I have some duplicate values in STH. TRAN_DATETIME repeated dates... How to avoid duplicates and at the same time the group the STH. TRAN_DATETIME...
    Please help... urgent

    Hello

    If you mean that you must get rid of time, then you can put TRUNC around STH. TRAN_DATETIME:

     SELECT sti.item,
    im.ITEM_DESC,
    COUNT(sti.item) item_count,
    d.dept_name,
    d.group_no,
    d.BUD_INT,
    sti.standard_unit_retail,
    isc.unit_cost,
    ( sti.standard_unit_retail-isc.unit_cost ) margin,
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    count(sti.item)*isc.unit_cost Tot_Sales,
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    FROM
    sa_tran_item sti,
    item_supp_country isc,
    item_master im,
    deps d,
    sa_tran_head sth
    WHERE
    sti.item = isc.item
    and sti.item=im.item
    and d.DEPT=sti.DEPT
    and sth.TRAN_SEQ_NO = sti.TRAN_SEQ_NO
    and sth.TRAN_DATETIME between :p1 and :p2
    group by sti.item, im.ITEM_DESC, d.dept_name, d.group_no, d.BUD_INT, sti.standard_unit_retail, isc.unit_cost, (
    sti.standard_unit_retail-isc.unit_cost ), trunc(STH.TRAN_DATETIME)
    

    Herald tiomela
    http://htendam.WordPress.com

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    Hi gurus,

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    1               A1             14-FEB-09 08.18.47 AM          11.00
    2               A2             14-FEB-09 08.18.56 AM       2839.00
    3               A1             15-FEB-09 08.00.02 AM      1299.00
    4               A1             15-FEB-09 07.00.00 AM        789.00
    5               A3             14-FEB-09 08.18.56 AM        312.00
    6               A4             14-FEB-09 08.19.02 AM        233.00
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    2               A2             14-FEB-09 08.18.56 AM       2839.00
    3               A1             15-FEB-09 08.00.02 AM      1299.00
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    Could you please help?

    Hello

    Like this?:

    Connected to Oracle Database 10g Express Edition Release 10.2.0.1.0
    Connected as hr
    
    SQL>
    SQL> with data as(
      2  select 1 as SR_NO, 'A1' as Auth_Code, to_date('14-FEB-09 08.18.47', 'dd-mon-yy hh24:mi:ss') as change_date,    11.00 as cost from dual union all
      3  select 2 as SR_NO, 'A2' as Auth_Code, to_date('14-FEB-09 08.18.56', 'dd-mon-yy hh24:mi:ss') as change_date,  2839.00 as cost from dual union all
      4  select 3 as SR_NO, 'A1' as Auth_Code, to_date('15-FEB-09 08.00.02', 'dd-mon-yy hh24:mi:ss') as change_date,  1299.00 as cost from dual union all
      5  select 4 as SR_NO, 'A1' as Auth_Code, to_date('15-FEB-09 07.00.00', 'dd-mon-yy hh24:mi:ss') as change_date,   789.00 as cost from dual union all
      6  select 5 as SR_NO, 'A3' as Auth_Code, to_date('14-FEB-09 08.18.56', 'dd-mon-yy hh24:mi:ss') as change_date,   312.00 as cost from dual union all
      7  select 6 as SR_NO, 'A4' as Auth_Code, to_date('14-FEB-09 08.19.02', 'dd-mon-yy hh24:mi:ss') as change_date,   233.00 as cost from dual)
      8  select * from data d where change_date = (select max(change_date) from data d2 where d.auth_code = d2.auth_code);
    
         SR_NO AUTH_CODE CHANGE_DATE       COST
    ---------- --------- ----------- ----------
             2 A2        14/02/2009        2839
             3 A1        15/02/2009        1299
             5 A3        14/02/2009         312
             6 A4        14/02/2009         233
    
    SQL>
    

    Kind regards

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    var clipcopies:Array = new Array();

    for (var: lines: uint = 0; lines < 5; lines ++)

    {

    for (var cols: uint = 0; < 2; passes passes ++)

    {

    for (var: Cops: uint = 0; < 2; cops cops ++)

    {

    var character: clips = new clips();

    persona.scaleX =. 5;

    persona.scaleY =. 5;

    persona.gotoAndStop(cols+1+rows*2);

    Persona.x = (COL * (persona.width + 20) + 10) +(cops*300);

    Persona.y = (lines * (persona.height + 20) + 10);

    persona.addEventListener (MouseEvent.CLICK, clickPersona);

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    }

    }

    function clickPersona(e:MouseEvent):void

    {

    var thisclip:uint = clipcopies.indexOf (e.target);

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    removeChild (clipcopies [thisclip + 1]);

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    use:

    var clipcopies:Array = new Array();

    for (var: lines: uint = 0; rows)<5; rows++)="">

    for (var cols: uint = 0; cols)<2; cols++)="">

    for (var: Cops: uint = 0; cops)<2; cops++)="">

    var character: clips = new clips();

    persona.scaleX =. 5;

    persona.scaleY =. 5;

    persona.gotoAndStop(cols+1+rows*2);

    Persona.x = (COL * (persona.width + 20) + 10) +(cops*300);

    Persona.y = (lines * (persona.height + 20) + 10);

    Persona.Name = line + "," + COL + "," + cops;

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    clipcopies.push (Persona);

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    }

    }

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    removeChild (MovieClip (e.currentTarget));

    removeChild (getChildByName (e.currentTarget.name.split (",") [0] + ',' + (1 - int (e.currentTarge t.name.split (",") [1])));)

    }

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     execute immediate 'drop table movies';
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    Note: I use the LabVIEW database connectivity toolkit

    Database: Ms access

    Sorry for my grammatical error

    arun575 wrote:

    Hi friends,

    [...] Now, I want to delete the 3 rows using by selecting the line (i.e., click anywhere in this line and press the button) that day that the rank will be remove from the database. Then remaining will again store in the database.

    Instead of storing data in the database, simply remove the record that you have removed from your FP indicator and recharge the light on the database.  (The first column in my database is named ID - autoindexed, the primary key)

    EDIT - only sub blue-vi is https://lavag.org/files/file/60-array-point-to-rowcol/

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