Help with Listview:Urgent

Hello friends,

I m vijay from india. I am making a project in vb6 but at
this movement i am having a problem in list view. The problem is.......


I am using list view.the listview's check box property is true ,i want
to delete all record,those are checked or selected in listview but
records are not deleted from database.i m using Access 2000.so anyone
can help me. plzzzzzzz


so plzzzz give ce code for deleting selected record from database as
well as listview.



Thankx
(Vijay)

Welcome to the Forums.

Moved :)

Quote:

---

Originally Posted by vijay2006

Hello friends,

I m vijay from india. I am making a project in vb6 but at
this movement i am having a problem in list view. The problem is.......


I am using list view.the listview's check box property is true ,i want
to delete all record,those are checked or selected in listview but
records are not deleted from database.i m using Access 2000.so anyone
can help me. plzzzzzzz


so plzzzz give ce code for deleting selected record from database as
well as listview.



Thankx
(Vijay)

---

Please show the code that you are using to delete the records.

A long time ago I had came across this problem, the quick and dirty solution was

1. copy the list in to a Array
2. Clear the list
3. remove the the Item of array like array(4) = ""
4. if array(i) <> "" then Add item

the proper solution is long.. so i am now bored of typeing

You must be getting Item/Index already deleted kind of error

What happens when you deletes

Dim i as integer
For i=0 to List1.Items.Count
If List1.Items(i).Selected then List1.Items(i).Remove
next

I Don't remember the properties of List and proper sytax of vb6 as i am switched to .NET

See When you removes the item the Indexes of all the below item changes and but var i keep on going so if you have deleted 6th item the item next to it will become the 6th one.

Now try solving the problem yourself. A Hint is there below this line if you still confused then tell me i will give you complete code select all to see the hint

Hint: A simple GoTo will do the trick

============================
Plz see this code. I m using this
============================

VB Code:

If lvList.ListItems.count > 0 Then
                Dim ANS As Integer
                ANS = MsgBox("Are you sure that you want to clear all the checked records in the list?" & vbCrLf & vbCrLf & "NOTE: This will clear the all the checked records in the list from the due checks.", vbQuestion + vbYesNo)
                'Me.MousePointer = vbHourglass
                If ANS = vbYes Then
                    Dim i As Integer
                    For i = 1 To lvList.ListItems.count
                        If isRecordExist("tblPDC_Manager", "CheckNo", CLng(LeftSplitUF(lvList.ListItems(i).Tag))) = True Then
                            If lvList.ListItems(i).Checked = True Then
                               ChangeValue strCon, "tblPDC_Manager", "Cleared", "Y", , "WHERE CheckNo=" & CLng(LeftSplitUF(lvList.ListItems(i).Tag))
                            End If
                        End If
                    Next i
                    i = 0
                    
                    MsgBox "Record has been successfully cleared", vbInformation, "Confirm"
                End If
                ANS = 0
                Me.MousePointer = vbDefault
            Else
                MsgBox "No record to clear.", vbExclamation
            End If

Quote:

---

Originally Posted by Vishalgiri

A long time ago I had came across this problem, the quick and dirty solution was

1. copy the list in to a Array
2. Clear the list
3. remove the the Item of array like array(4) = ""
4. if array(i) <> "" then Add item

the proper solution is long.. so i am now bored of typeing

You must be getting Item/Index already deleted kind of error

What happens when you deletes

Dim i as integer
For i=0 to List1.Items.Count
If List1.Items(i).Selected then List1.Items(i).Remove
next

I Don't remember the properties of List and proper sytax of vb6 as i am switched to .NET

See When you removes the item the Indexes of all the below item changes and but var i keep on going so if you have deleted 6th item the item next to it will become the 6th one.

Now try solving the problem yourself. A Hint is there below this line if you still confused then tell me i will give you complete code select all to see the hint

Hint: A simple GoTo will do the trick

---

Ya this is right but i want to delete record from database which is chacked in the listview if u have a proper solution then plzzzz help

What does the ChangeValue sub/function look like? I guess I'll also need to know what data is in the Tags.

BTW I edited your post and added VBCode tags to make it easier to read.

Quote:

---

Originally Posted by MartinLiss

What does the ChangeValue sub/function look like? I guess I'll also need to know what data is in the Tags.

BTW I edited your post and added VBCode tags to make it easier to read.

---

VB Code:

============================================
There are some Procedure i m using ,so try it andhelp me 
if any one done it then plzz upload full source code 
============================================
'Procedure that will change the value at once
Public Sub ChangeValue(ByRef srcCN As Connection, ByVal srcTable As String, ByVal srcField As String, ByVal srcValue As String, Optional isNumber As Boolean, Optional srcCondition As String)
    If srcCondition <> vbNullString Then srcCondition = " " & srcCondition
    If isNumber = True Then
        srcCN.Execute "UPDATE " & srcTable & " SET " & srcField & " =" & srcValue & " " & srcCondition
    Else
        srcCN.Execute "UPDATE " & srcTable & " SET " & srcField & " ='" & srcValue & "'" & " " & srcCondition
    End If
End Sub
 
'--------------------------------------------------------------
'Function used to check if the record exit or not.
Public Function isRecordExist(ByVal sTable As String, ByVal sField As String, ByVal sStr As String, Optional isString As Boolean) As Boolean
    Dim rs As New Recordset
 
    rs.CursorLocation = adUseClient
    If isString = False Then
        rs.Open "Select * From " & sTable & " Where " & sField & " = " & sStr, strCon, adOpenStatic, adLockOptimistic
    Else
        rs.Open "Select * From " & sTable & " Where " & sField & " = '" & sStr & "'", strCon, adOpenStatic, adLockOptimistic
    End If
    If rs.RecordCount < 1 Then
        isRecordExist = False
    Else
        isRecordExist = True
    End If
    Set rs = Nothing
End Function
'------------------------------------------------------------------
Public Function LeftSplitUF(ByVal srcUF As String) As String
    If srcUF = "*~~~~~*" Then LeftSplitUF = "": Exit Function
    Dim i As Integer
    Dim t As String
    For i = 1 To Len(srcUF)
        If Mid$(srcUF, i, 7) = "*~~~~~*" Then
            Exit For
        Else
            t = t & Mid$(srcUF, i, 1)
        End If
    Next i
    LeftSplitUF = t
    i = 0
    t = ""
End Function

'-------------------------------------------------------------------------
Plzzz check this code and help me. I am making a project Health club management system ,If will complete this project then i will upload it here and every one go through and find the error in project then i will correct the error and make a perfact project.

thankx Friends for supporting me.

'-------------------------------------------------------------------------

You can make your code easier to read if you surround the code with VBCode tags as in this example.

[vbcode]
'Your code
[/vbcode]

which will cause it to show up like this.

VB Code:

' Your code

PLZZZZZ help

Put a breakpoint on the ChangeValue strCon, "tblPDC_Manager"... line and trace the code (line by line) by pressing F8 multiple times. At some point you will see the code flow diverge from where you expect it to go. You can also examine the values of the variables in each line you come to.

if it possible then plzzzz give me full working code plzzzzzz

I doubt if anyone could tell what is wrong by just looking at the code. In case you don't know how to debug, what you do is to put the mouse in the margin to the left of the line that I mentioned and click the mouse. That will place a red dot there which is a breakpoint. Run the code and it will stop there since I assume that that line isn't being skipped. Then press F8 one line at a time and see where the code goes.

VB Code:

For i = 1 To lvList.ListItems.count
                        If isRecordExist("tblPDC_Manager", "CheckNo", CLng(LeftSplitUF(lvList.ListItems(i).Tag))) = True Then
                            If lvList.ListItems(i).Checked = True Then
                               ChangeValue strCon, "tblPDC_Manager", "Cleared", "Y", , "WHERE CheckNo=" & CLng(LeftSplitUF(lvList.ListItems(i).Tag))
                            End If
                        End If
                    Next i

ithink there are some internal problem becose the code is give not any error but in this code the program is not come in to the for loop so problem in loop .If any one solve this problem then write the solution here.

I have two questions for you:

1) Why are you using two different member names?

2) Do you understand what I was trying to explain concerning breakpoints?

If you can answer yes to the second question why not put a breakpoint someplace where you know the code goes and step throught (F8) the code from there. That the ONLY way this problem will be solved.

This time I am not Agree with my MartinLiss (he is like my Elder Brother).

I have too much respect for MartinLiss but If Vijay Would know how to debug and do some of the trick you mentioned then why he would be here to ask questions?

Quote:

---

Originally Posted by vijay2006

Ya this is right but i want to delete record from database which is chacked in the listview if u have a proper solution then plzzzz help

---

See when you removes the selected records from listbox at that time Fire some query to delete that specific record from database.

I hope you know SQL.


And one more thing, You should not use two diffrent member names, if your password is lost then you can get it mailed.

Also Can't you understand MartinLiss have told you many times that use vbcode tag.

This is very disapointing yaar, As an Indian we are the Gurus of Programming, and you are bluring the image of the Great Indian Programmers.

So be carefull next time we are always here to help you but No one will do anything for you, ya we can help you. But Keep in mind "there are no free lunches in the world"