How to Remove Duplicates Values from table

How to Remove Duplicates Values from table

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How to Remove Duplicates Values from table
Tag : php , By : francisco santos
Date : November 27 2020, 12:01 PM

hope this fix your issue Since your screenshot is of phpMyAdmin, I'm going to assume that we're talking about MySQL here (when asking questions, that's helpful information as various SQL dialects have different tools that can be used).
If you don't need id in your results, you can just use DISTINCT:
  MIN(id) AS id,
  id NOT IN(SELECT MIN(id) FROM YourTableName GROUP BY Username,Video)

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MYSQL Query to make the values as duplicates and remove duplicates from table

Tag : mysql , By : Jack
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have a table with 692256 number of rows ,which has data similar to below , Add an AutoIncrement Key "ID"
from table
where table.ID NOT IN (
            select MAX(ID)
            from table
            group by customer_id, your_function(data) )

How to remove duplicates values and create a table view section in Swift?

Tag : ios , By : Sigfrieg
Date : March 29 2020, 07:55 AM
will help you I would create another dictionary with a string key (which will be your type) and an array of KempfFiber as the values. Then when you get a type you retrieve the array from the dictionary. If it is nil then create a new array, add the kempf object to it, otherwise just add the kempf object to the existing array. You can then get the keys as an array for your sections.
Something like this (where kempfData and sectionKeys are properties of your current class -

for innerDict in plistArray as Array {
    let kempf: KempfFiber = KempfFiber() // Model
    kempf.name = innerDict["Name"] as String
    kempf.detail = innerDict["Detail"] as String
    kempf.imageName = innerDict["ImageName"] as String
    kempf.specGrav = innerDict["SpecGrav"] as String
    kempf.toughness = innerDict["Toughness"] as String
    kempf.pdfFile = innerDict["PDF"] as String

    var theType = innerDict["Type"] as String     

    var kempfArray=self.kempfData[theType]

    if (kempfArray ==nil) {
        kempfArray =[Kempf]()     

    self.kempfData[theType]= kempfArray


 let sectionData=self.kempfData[self.sectionKeys[indexPath.section]]

 if (sectionData != nil) {
     return sectionData!.count
 else {
     return 0
let sectionData=self.kempfData[self.sectionKeys[indexPath.section]]

if (sectionData != nil) {
     let kempf=sectionData[indexPath.row]
     // Configure your cell...
else {
    // Something bad happened

How to remove duplicates from comma separated list by regex in Oracle but I don't want duplicates values?

Tag : sql , By : Ted Leung
Date : March 29 2020, 07:55 AM
This might help you Try this, as per article http://www.dba-oracle.com/t_extract_comma_delimited_strings_oracle_sql.html:
select distinct str from
(select regexp_substr ('ABCD1234, XYZ, ABCD1234, ABCD1234C, ABCD1234, abc, abcX, 1234U, 1234', '[^, ]+',1, rownum) str 
from dual 
connect by level <= regexp_count ('ABCD1234, XYZ, ABCD1234, ABCD1234C, ABCD1234, abc, abcX, 1234U, 1234', '[^, ]+')) v;

MySQL - Select values and remove duplicates by table name

Tag : mysql , By : LinnheCreative
Date : March 29 2020, 07:55 AM
may help you . If I understand correctly, you want to always retain all records from table_saved, plus records from table_default having IDs not appearing in table_saved. One approach is to use a left join to find the unique records from table_default. Then union that with all records from table_saved.
FROM table_default t1
LEFT JOIN table_saved t2
    ON t1.ID = t2.ID
FROM table_saved;

Remove SOME duplicates in data.table column X, but retain duplicates if column Y matches condition

Tag : r , By : user122937
Date : March 29 2020, 07:55 AM
wish of those help We could create an ordered 'STATUS' based on the levels 'A', 'B', 'C', grouped by 'ID', extract the first present levels (after applying droplevels), do a comparison with 'STATUS' to return logical vector, convert it to a numeric index (.I) also have a condition to check if the number of unique elements in the column is 1, return only the first row
test[test[, .I[if(uniqueN(STATUS) == 1) 1 else 
   STATUS == first(levels(droplevels(ordered(STATUS, 
          levels = c("A", "B", "C")))))], ID]$V1]
#1:  1      B
#2:  2      A
#3:  2      A
#4:  2      A
#5:  2      A
#6:  2      A
#7:  3      C
test <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L), STATUS = c("B", "C", "C", "C", "A", "A", "A", 
"A", "A", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C", 
"C", "C", "C", "C", "C", "C", "C", "C", "C", "C")), class = c("data.table", 
"data.frame"), row.names = c(NA, -30L))
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