Table To Dictionary <String,Object> using Lambda Expression

Table To Dictionary <String,Object> using Lambda Expression

By : phass
Date : November 22 2020, 02:59 PM
may help you . You CAN do this with a Dictionary.Add, but you cannot with ToDictionary. The reason for this is that ToDictionary does let you change specify a key selector function 'row => row["USERNAME"].ToString() + "|" + row["SQL_FULLTEXT"].ToString()' in your case, but it does not allow you to change the underlying enumerable type.
You can only create a dictionary of type Dictionary (assuming table is of type DataTable) using ToDictionary();
code :

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Select boolean expression inside dictionary using lambda expression

Select boolean expression inside dictionary using lambda expression

By : past5
Date : March 29 2020, 07:55 AM
I wish did fix the issue. It sounds like you're completely ignoring the fact that you're using a dictionary. It sounds like you might want:
code :
public bool CheckMap(int? key, int? checkValue)
    List<int?> values;
    if (!pegMap.TryGetValue(key, out values))
        return false;
    return values.Contains(checkValue);
Getting Object Property to String Using Lambda Expression in C#

Getting Object Property to String Using Lambda Expression in C#

By : zjxte
Date : March 29 2020, 07:55 AM
I hope this helps . Take a look at the answer to this: Workaround for lack of 'nameof' operator in C# for type-safe databinding?
In your case, if you implement this Generic Class:
code :
public class Nameof<T>
    public static string Property<TProp>(Expression<Func<T, TProp>> expression)
        var body = expression.Body as MemberExpression;

        if(body == null)
            throw new ArgumentException("'expression' should be a member expression");

        return body.Member.Name;
void BindToDataSource(IEnumerable<MyClass> list) 
    myComboBox.DataSource = list;
    myComboBox.DisplayMember = Nameof<MyClass>.Property(e => e.MyProperty);
Java Lambda Expression to UpperCase string value within an Object

Java Lambda Expression to UpperCase string value within an Object

By : Karan Choudhary
Date : March 29 2020, 07:55 AM
Hope this helps I have a lambda expression in which I use a UnaryOperator to convert a string value into uppercase, if not null: , You could make convertToUpper a generic method:
code :
public static final <T> void convertToUpper(T object1, Function<? super T, String> getter,
        BiConsumer<? super T, String> setter) {
    setter.accept(object1, toUpperCase.apply(getter.apply(object1)).toString());
Casting error adding Guid value to Dictionary<string, object> inside an IQueryable.Select() lambda expression sele

Casting error adding Guid value to Dictionary<string, object> inside an IQueryable.Select() lambda expression sele

By : Deepak Sharma
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further The problem happens for all value types. Value types are not objects, but C# is supposed to automatically box value types, but where you are building an expression yourself, you might be bypassing that.
The following change converts everything to object, thus boxing value types.
code :
var eKey = Expression.Constant(selectedProperty);
var eValue = Expression.Property(parameter, selectedProperty);
Expression boxed = Expression.Convert(eValue, typeof(object));
var eAdd = Expression.ElementInit(addMethod, eKey, boxed);
Bad return type in lambda expression String cannot be converted to Object

Bad return type in lambda expression String cannot be converted to Object

By : David Riedl
Date : March 29 2020, 07:55 AM
hop of those help? First things first, (a, b) -> a + ", " + b) will not compile simply because the returned value must be a Subject rather than a String, I believe you have eventually figured this out given your latest edit states the code is now working without duplicates when using the (a, b) -> a merge function.
The reason why you're not getting the expected result given your latest update is that you're using the merge function (a, b) -> a meaning "if two given Subjects have the same Student then keep the first student and discard the second".
code :
Map<Student, List<Subject>> result = markList
                        v -> new ArrayList<>(singletonList(v.getSubject())),
                        (a, b) -> {a.addAll(b); return a;}));
Map<Student, List<Subject>> resultSet = markList.stream()
                        mapping(Mark::getSubject, toList())));
resultSet.forEach((student, subjects) ->
                System.out.printf("%s %s : %s ", student.getFirstName(), student.getLastName(), 
                        subjects.stream().map(Subject::getName).collect(joining(", "))));
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