MYSQL workbench error : You are using safe update mode and you tried to update a table without a WHERE that uses a KEY column

 SET SQL_SAFE_UPDATES = 0;

how to find the table name using column name in mysql

 SELECT DISTINCT TABLE_NAME 

    FROM INFORMATION_SCHEMA.COLUMNS
    WHERE COLUMN_NAME IN ('columnA','ColumnB')
        AND TABLE_SCHEMA='YourDatabase';

linq error At least one object must implement IComparable

 Well, you're trying to use SortedSet<>... which means you care about the ordering. But by the sounds of it your Player type doesn't implement IComparable<Player>. So what sort order would you expect to see?

Basically, you need to tell your Player code how to compare one player with another. Alternatively, you could implement IComparer<Player> somewhere else, and pass that comparison into the constructor of SortedSet<> to indicate what order you want the players in. For example, you could have:

public class PlayerNameComparer : IComparer<Player>
{
    public int Compare(Player x, Player y)
    {
        // TODO: Handle x or y being null, or them not having names
        return x.Name.CompareTo(y.Name);
    }
}

Then:

// Note name change to follow conventions, and also to remove the
// implication that it's a list when it's actually a set...
SortedSet<Player> players = new SortedSet<Player>(new PlayerNameComparer());

linq orderby collection property

 IQueryable<Parent> data = context.Parents.OrderBy(p=>p.Children.OrderBy(chi => chi.Name).FirstOrDefault());

How to execute sql queries in workbench mysql ?

 If you are from MS SQL background and first time working with MySql , you may be struggling for finding the shortcut key to execute the sql query in workbench. 

Here is the short cut to execute the SQL query in workbench MySql. 

Default key mapping

1. Execute (All or Selection) -> Ctrl + Shift + Enter.
2. Execute Current Statement -> Ctrl + Enter.

How to assign property value while initializing class in typescript

 While creating new object you want to set the value of few fields of class , then you can use following feature of typescript to do that : 

class Person {
    public name: string = "default"
    public address: string = "default"
    public age: number = 0;

    public constructor(init?:Partial<Person>) {
        Object.assign(this, init);
    }
}

let persons = [
    new Person(),
    new Person({}),
    new Person({name:"John"}),
    new Person({address:"Earth"}),    
    new Person({age:20, address:"Earth", name:"John"}),
];

How to make sure application always run with admin access

 Most of the time it happen with programmer , that they forget to run the application (Visual Studio/VS Code or any other IDE) with admin access. Because of that in result they face weird errors. So instead of every time facing same issue again and again, we can set the application permission in such way that its always run with admin access. 

Here are the steps which you can use to make sure your application always run with admin access. 

how to make sure application always run with admin access

1. Open Start.
2. Search for the app that you want to run elevated.
3. Right-click the top result, and select Open file location. ...
4. Right-click the app shortcut and select Properties.
5. Click on the Shortcut tab.
6. Click the Advanced button.
7. Check the Run as administrator option.