Search Results

Search found 60556 results on 2423 pages for 'access list'.

Page 211/2423 | < Previous Page | 207 208 209 210 211 212 213 214 215 216 217 218  | Next Page >

  • Can't set up Usermin correctly to allow users to login outside of local network, what am I missing?

    - by thecraic
    I'm fairly new at creating a server, but the biggest problem I am currently having at the moment is getting Usermin set up to be accessible from outside the LAN. I talked to other people that use it and was told that all I need to do is type the url:20000 to access the login screen, but that doesn't work. I have also tried the ip:20000 and that doesn't lead to anything. Instead I get the error message: Error - Bad Request This web server is running in SSL mode. Try the URL https://hostname:10000/ instead. (where hostname is my server's hostname) I know it must be a configuration issue, but I have checked all my settings and as far as I can tell I don't have the ports blocked anywhere. I have the correct ports forwarded on my router and my server firewall doesn't have the port block either. Is there anything I am missing? Any help would be appreciated and I will add more information upon request. Thank You.

    Read the article

  • Is it possible to impersonate another WAP by intercepting communication with other client?

    - by OSX NINJA
    There is a well known WAP that lots of people use. Someone comes in with a laptop equipped with a sniffer. The laptop sniffs people trying to log on to the WAP. It intercepts the connection, and when people try to log on to the WAP, they unknowingly log on through that person's laptop instead. All communication between the WAP and people's laptops go through that person's laptop. That person's laptop is able to block access to certain websites that the WAP would normally allow.

    Read the article

  • Problem accessing the remote working space on my new SBS 2008 box

    - by Dabblernl
    This supposedly easy to install OS is starting to drive me nuts... SYMPTOMS: When trying to connect to the remote workplace I get (and ignore) the security warning because I am currently testing with the self issued certificate. After loggin in the remote workplace's main screen displays but the images on it do not load. When I try to click the email link I am thrown back to the login screen. If I try the login to exchange directly by typing in the remote.mydomain.com/owa address I get a 403 error that I am denied access. The problem occurs on both a vista and a win 7 machine. It seems that some security setting is playing tricks with me. How can I troubleshoot this?

    Read the article

  • How to connect to internet 2 or more pcs with an usb wifi adapter

    - by bhaoahd
    In one pc I have an USB Wifi adapter (Realtek RTL8187L Chipset), from there a 10 meters cable to an outdoor antenna. With that I connect to a public Access Point. Now I want to be able to connect more computers to internet using that same connection that comes from the outdoor antenna, what options do I have to do this? I do have another usb wifi adapter with a small omni antenna, and a router encore ENHWI-G/A. This router can be used as a Wifi AP, but it needs a modem that works with the RJ45 connector. Is there a way I could take internet from the outdoor antenna, and create another AP indoor using that router to "repeat" that wifi that comes from the outdoor antenna? If creating this second AP is not possible, should I use some sort of local network between computers, connect my main PC with the outdoor antenna, and share the connection through the lan? (I would really prefer to have a second AP indoor so I can connect other devices like a Palm, but I am not sure how could this be possible with this router and the usb wifi adapter)

    Read the article

  • SQL update table from another table

    - by LtDan
    Using SQL in Access, trying to "Update" a table, with the user name, from another table. The 3rd line below (SQLnm2...) says error-2465 cant find field '|'. I've tried changing the expression many ways but no success. Any assistance would be greatly appreciated. Dim SQLnm As String Dim SQLnm2 As String SQLnm2 = SQLnm2 & "', '" & [Employees]![NBK] & "');" SQLnm = " Update tbl_DateTracking SET NBK = " SQLnm = SQLnm & "'" & SQLnm2 & "' WHERE " SQLnm = SQLnm & "CaseId = '" & CaseId & "' AND OCC_Scenario = '" & OCC_Scenario & "';" DoCmd.RunSQL SQLnm

    Read the article

  • Moving to the next line to populate an excel file from VBA

    - by edmon
    I have the below code that takes certain fields from my MS Access (A small Hotel Reservation Database)form and populates defined cells in the said Excel file. Dim objXLApp As Object Dim objXLBook As Object Set objXLApp = CreateObject("Excel.Application") Set objXLBook = objXLApp.Workbooks.Open("Y:\123files\File\Hotel Reservation.xls") objXLApp.Application.Visible = True objXLBook.ActiveSheet.Range("B2") = Me.GuestFirstName & " " & GuestLastName objXLBook.ActiveSheet.Range("C2") = Me.PhoneNumber objXLBook.ActiveSheet.Range("E2") = Me.cboCheckInDate objXLBook.ActiveSheet.Range("F2") = Me.cboCheckOutDate objXLBook.ActiveSheet.Range("H2") = Me.RoomType objXLBook.ActiveSheet.Range("I2") = Me.RoomNumber End Sub How can I keep populating a new Guest to the same Excel file just on the next row?

    Read the article

  • Prevent acccess to the C drive

    - by Jenko
    Is it possible to prevent regular users from accessing the C drive via Windows Explorer? they should be allowed to execute certain programs. This is to ensure that employees cannot steal or copy out proprietary software even though they should be able to execute it. One way would be to change the option in windows Group Policy and set the "shell" to something other than "explorer.exe". I'm looking for a similar windows setting that just hides the C drive or otherwise prevents trivial access. This is for Windows XP/7.

    Read the article

  • Remote connect to mysql server?

    - by LF4
    I've been trying to figure out why I keep getting this error when I try to connect to the MySQL server with the following commands. $~ mysql -u username -h SQLserver -p Enter password: ERROR 1045 (28000): Access denied for user 'username'@'myIP' (using password: YES) I've done the following: Port is open in the firewall other wise I wouldn't get the error it'd just timeout. MySQL server not running with skip-networking or bind-address username has host as '%' and I can connect locally so the password is correct. GRANT USAGE ON *.* TO username@% IDENTIFIED BY 'password'; FLUSH PRIVILEGES; I wanted to know if anyone had ideas or ran into this issue before and solved it? mysql> select user, host from mysql.user where user='username'; +----------+------+ | user | host | +----------+------+ | username | % | +----------+------+ 1 row in set (0.00 sec) mysql> show grants for 'username'; +----------------------------------------------------------------------------------------------------------------+ | Grants for username@% | +----------------------------------------------------------------------------------------------------------------+ | GRANT ALL PRIVILEGES ON *.* TO 'username'@'%' IDENTIFIED BY PASSWORD '*F42AD03PASSWORDHASHADF4021C86B' | | GRANT ALL PRIVILEGES ON `DB2`.* TO 'username'@'%' | +----------------------------------------------------------------------------------------------------------------+ 2 rows in set (0.00 sec)

    Read the article

  • How to make your apache application accessible within network

    - by guest
    I have a Windows XP machine where I have installed WAMP and made a PHP based web application. I can access the web application from within this machine by using the browser and pointing to: http://localhost/myApp/ --- and the page loads fine. Now I want this site (http://localhost/myApp) to be accessible to all machines within the network (and may be later, to the general public as well). I am quite new to this, how do I make my site accessible to all machines within the network and to the general public in the internet? I tried modifying the httpd.conf file in Apache (WAMP) by changing Listen 80 to Listen 10.10.10.10:80 (where I replaced 10.10.10.10 with the actual IP of this windows xp machine). I also tried "Put Online" feature in WAMP. None seem to work though. How do I make it accessible?

    Read the article

  • Temporarily block other users from network printer

    - by TecBrat
    I found where someone else asked this question here, but they did not get a working answer. We have a printer that is shared. It has it's own network card, so we all have equal access to it. (none of our computers owns it) One of our users needs to print on specialty paper and we need to be sure not to print when that paper is in the printer. Our current method is "Hey, don't print anything right now!" Obviously this method is not preferred because it does not enforce itself. :-) I think all our PCs are running Win7 Home. The printer in question is an HP Laserjet 2200. Is there a way that we can make this happen?

    Read the article

  • opening a GUID usb stick in windows 7

    - by altomic
    so I have a mac which I have files on. I put in a GUID formatted usb stick and dropped some files on to it. took the stick and plugged in to my netbook with win7. In Devices & Printers it shows up. It also appears in "safely remove hardware". no actual letter or device when i search for it in other ways. Question - how do I access the files of my GUID usb stick on my windows 7 netbook? thanks

    Read the article

  • Format text to 5 chars from a number

    - by Wheelersg
    In Access, I used a query to sum some numbers and appended the answer to another table(table2). Now I need to export the number as a text with 5 positions but can't seem to get it to hard code all 5 positions. I have it formatted as text, field length 5, custom foramt "00000" (also tried @@@@@). Example: 3 + 3 + 1 = 7. THen append the 7 to table2. It always shows as 7. I need it to shows as 00007.

    Read the article

  • Jquery Livesearch with quicksilver plugin to include not just the <li>s

    - by 133794m3r
    Ok, what i'm trying to do here is to make the exact code found here here ordered list and make it so that it doesn't just not work when i try to add additional elements into the list. Also i'm planning on using the more effecient one linked to at the end but i cannot put it here so you'll have to find that link on your own sadly. Since i'm trying to use this for a knowledge base page i want to allow people to be able to search through the items on the page and go to the proper part(ie loading it into the viewing area) but it's not letting me even include a simple anchor in there. if there is anyway to do something similar or to edit it so that it'll include everything within the <li> that'd be great. I don't know if anyone out there has done something like this before, but if they have and wouldn't mind sharing the code with me i'd be extremely happy. If no one has but does know how to make it include everything within the <li> including text also a great thing to have. I imagine that i won't be the only person out there in this world with my exact query.

    Read the article

  • va_arg with pointers

    - by Yktula
    I want to initialize a linked list with pointer arguments like so: /* * Initialize a linked list using variadic arguments * Returns the number of structures initialized */ int init_structures(struct structure *first, ...) { struct structure *s; unsigned int count = 0; va_list va; va_start(va, first); for (s = first; s != NULL; s = va_arg(va, (struct structure *))) { if ((s = malloc(sizeof(struct structure))) == NULL) { perror("malloc"); exit(EXIT_FAILURE); } count++; } va_end(va); return count; } The problem is that clang errors type name requires a specifier or qualifier at va_arg(va, (struct structure *)), and says that the type specifier defaults to int. It also notes instantiated form at (struct structure *) and struct structure *. This, what seems to be getting assigned to s is int (struct structure *). It compiles fine when parentheses are removed from (struct structure *), but the structures that are supposed to be initialized are inaccessible. Why is int assumed when parentheses are around the type argument passed to va_arg? How can I fix this?

    Read the article

  • How to keep form items selected after post request?

    - by Ole Jak
    I have a simple html form. On php page. A simple list is placed on form. I submit this form (selected list items) to this page so it gives me page refresh. I want items which were POSTED to be selected after form was submited. How to do such thing? For my form I use such code: <form action="FormPage.php" method="post"> <select id="Streams" class="multiselect ui-widget-content ui-corner-all" multiple="multiple" name="Streams[]"> <?php $query = " SELECT s.streamId, s.userId, u.username FROM streams AS s JOIN user AS u ON s.userId = u.id LIMIT 0 , 30 "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); while ($row = mysql_fetch_array($streams_set)){ echo '<option value="' , $row['streamId'] , '"> ' , $row['username'] , ' (' , $row['streamId'] ,')' ,'</option> '; } ?> </select> <br/> <input type="submit" class="ui-state-default ui-corner-all" name="submitForm" id="submitForm" value="Play Stream from selected URL's!"/> </fieldset> </form>

    Read the article

  • Initialize listitem with blanks?

    - by VBartilucci
    Say I have a list made up of a listitem which contains three strings. I add a new listitem, and try to assign the values of said strings from an outside source. If one of those items is unassigned, the value in the listitem remains as null (unassigned). As a result I get an error if I try to assign that value to a field on my page. I can do a check on isNullOrEmpty for each field on the page, but that seems inefficient. I'd rather initialize the values to "" (Empty string) in the codebehind and send valid data. I can do it manually: ClaimPwk emptyNode = new ClaimPwk(); emptyNode.cdeAttachmentControl = ""; emptyNode.cdeRptTransmission = ""; emptyNode.cdeRptType = ""; headerKeys.Add(emptyNode); But I have some BIG list items, and writing that for those will get tedious. So is there a command, or just plain an easier way to initialize a listitem to empty string as opposed to null? Or has anyone got a better idea?

    Read the article

  • Disable ARC with Xcode 5

    - by user2187565
    First, sorry for my bad english, I'm french and had 15years old but StackOverFlow is for me the best forum for developers. So, in the previous versions of Xcode, we can disable ARC (Automatic Reference Counting) in the project settings when we create the project. Not now with Xcode 5 and ARC to pose me a problem: with an property list file, for the reading step, Xcode send me an error: "implicit conversion of 'int' to 'id' is disallowed with ARC". I had not the problem with the same code with Xcode 4. In my property list file, The keys are numbers and also in my viewController.m . NIKOS M.: No problem, but I don't see how I can add compiler flag with the 5th version of Xcode. The code (with french string...): NSString *error; NSString *rootPath = [NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES) objectAtIndex:0]; NSString *plistPath = [rootPath stringByAppendingPathComponent:@"Save.plist"]; NSArray *keys = [NSArray arrayWithObjects:@"valeurCompteur1", @"valeurCompteur2", @"valeurCompteur3", @"valeurCompteur4", @"valeurCompteur5", @"nomCompteur1", @"nomCompteur2", @"nomCompteur3", @"nomCompteur4", @"nomCompteur5", nil]; NSArray *objs = [NSArray arrayWithObjects: compteur1, compteur2, compteur3, compteur4, compteur5, nameC1, nameC2, nameC3, nameC4, nameC5, nil]; REVIEW: When I disallow ARC for the target, an warning persist. How I can resolve that please ? Thank you very much.

    Read the article

  • Replacing words in string

    - by abkai
    Okay, so I have the following little function: def swap(inp): inp = inp.split() out = "" for item in inp: ind = inp.index(item) item = item.replace("i am", "you are") item = item.replace("you are", "I am") item = item.replace("i'm", "you're") item = item.replace("you're", "I'm") item = item.replace("my", "your") item = item.replace("your", "my") item = item.replace("you", "I") item = item.replace("my", "your") item = item.replace("i", "you") inp[ind] = item for item in inp: ind = inp.index(item) item = item + " " inp[ind] = item return out.join(inp) Which, while it's not particularly efficient gets the job done for shorter sentences. Basically, all it does is swaps pronoun etc. perspectives. This is fine when I throw a string like "I love you" at it, it returns "you love me" but when I throw something like: you love your version of my couch because I love you, and you're a couch-lover. I get: I love your versyouon of your couch because I love I, and I'm a couch-lover. I'm confused as to why this is happening. I explicitly split the string into a list to avoid this. Why would it be able to detect it as being a part of a list item, rather than just an exact match? Also, slightly deviating to avoid having to post another question so similar; if a solution to this breaks this function, what will happen to commas, full stops, other punctuation? It made some very surprising mistakes. My expected output is: I love my version of your couch because you love I, and I'm a couch-lover. The reason I formatted it like this, is because I eventually hope to be able to replace the item.replace(x, y) variables with words in a database.

    Read the article

  • Extended slice that goes to beginning of sequence with negative stride

    - by recursive
    Bear with me while I explain my question. Skip down to the bold heading if you already understand extended slice list indexing. In python, you can index lists using slice notation. Here's an example: >>> A = list(range(10)) >>> A[0:5] [0, 1, 2, 3, 4] You can also include a stride, which acts like a "step": >>> A[0:5:2] [0, 2, 4] The stride is also allowed to be negative, meaning the elements are retrieved in reverse order: >>> A[5:0:-1] [5, 4, 3, 2, 1] But wait! I wanted to see [4, 3, 2, 1, 0]. Oh, I see, I need to decrement the start and end indices: >>> A[4:-1:-1] [] What happened? It's interpreting -1 as being at the end of the array, not the beginning. I know you can achieve this as follows: >>> A[4::-1] [4, 3, 2, 1, 0] But you can't use this in all cases. For example, in a method that's been passed indices. My question is: Is there any good pythonic way of using extended slices with negative strides and explicit start and end indices that include the first element of a sequence? This is what I've come up with so far, but it seems unsatisfying. >>> A[0:5][::-1] [4, 3, 2, 1, 0]

    Read the article

  • Removing specific tags in a KML file

    - by Legion
    I have a KML file which is a list of places around the world with coordinates and some other attributes. It looks like this for one place: <Placemark> <name>Albania - Durrës</name> <open>0</open> <visibility>1</visibility> <description>(Spot ID: 275801) show <![CDATA[<a href="http://www.windguru.cz/int/index.php?go=1&vs=1&sc=275801">forecast</a>]]></description> <styleUrl>#wgStyle001</styleUrl><Point> <coordinates>19.489747,41.277806,0</coordinates> </Point> <LookAt><range>200000</range><longitude>19.489747</longitude><latitude>41.277806</latitude></LookAt> </Placemark> I would like to remove everything except the name of the place. So in this case that would mean I would like to remove everything except <name>Albania - Durrës</name> The problem is, this KML file includes more than 1000 of these places. Doing this manually obviously isn't an option, so then how can I remove all tags except for the name tags for all of the items in the list? Can I use some kind of program for that?

    Read the article

  • How to match words as if in a dictionary, based on len-1 or len+1? Python

    - by pearbear
    If I have a word 'raqd', how would I use python to have a spellcheck, so to speak, to find the word 'rad' as an option in 'spellcheck'? What I've been trying to do is this: def isbettermatch(keysplit, searchword): i = 0 trues = 0 falses = 0 lensearchwords = len(searchword) keysplits = copy.deepcopy(keysplit) searchwords = copy.deepcopy(searchword) #print keysplit, searchwords if len(keysplits) == len(searchwords)-1: i = 0 while i < len(keysplits): j = 0 while j < lensearchwords: if keysplits[i] == searchwords[j]: trues +=1 searchwords.pop(j) lensearchwords = len(searchwords) elif keysplits[i] != searchwords[j]: falses +=1 j +=1 i +=1 if trues >= len(searchwords)-1: #print "-------------------------------------------------------", keysplits return True keysplit is a list like ['s', 'p', 'o', 'i', 'l'] for example, and the searchword would be a list ['r', 'a', 'q', 'd']. If the function returns True, then it would print the keyword that matches. Ex. 'rad', for the searchword 'raqd'. I need to find all possible matches for the searchword with a single letter addition or deletion. so ex. 'raqd' would have an option to be 'rad', and 'poted' could be 'posted' or 'potted'. Above is what I have tried, but it is not working well at all. Help much appreciated!

    Read the article

  • Set List View Size Android

    - by Sandeep
    Hello , I am using List View in my project where i have used a xml file which is used to create the list item.Then i have used it programmatically in my class which is extended by ListActivity. But the problem is i have to add a button in the bottom of screen which is not related to list view but List view covers all the screen. So,is there any way to add button in bottom with list view in android. My Code is :- import android.app.ListActivity; import android.content.Intent; import android.os.Bundle; import android.widget.AdapterView; import android.widget.ArrayAdapter; import android.widget.ImageView; import android.widget.ListView; import android.widget.TextView; import android.widget.Toast; import android.widget.AdapterView.OnItemClickListener; import android.view.LayoutInflater; import android.view.View; import android.view.ViewGroup; import android.view.Window; public class Options extends ListActivity { String[] items; @Override public void onCreate(Bundle icicle) { super.onCreate(icicle); requestWindowFeature(Window.FEATURE_RIGHT_ICON); items = getResources().getStringArray(R.array.chantOption_array); setListAdapter(new IconicAdapter()); ListView lv = getListView(); lv.setTextFilterEnabled(true); lv.setBackgroundResource(R.drawable.ichant_logo); setFeatureDrawableResource(Window.FEATURE_RIGHT_ICON, R.drawable.icon_t); lv.setOnItemClickListener(new OnItemClickListener() { public void onItemClick(AdapterView<?> parent, View view, int position, long id) { // When clicked, show a toast with the TextView text Toast.makeText(getApplicationContext(), items[position], Toast.LENGTH_SHORT).show(); if ("Gayatri Mantra".equals(items[position].toString())) { int[] timeintervals = { 23900, 24000 }; // startChantActivity(TotalMala_loop,Total_Bead_Loop,BacgroundImage,Icon,Title,BeadsTotalTimeIntervals+totalTimeDurationOfAudio) startChantActivity(2, 108, R.drawable.gayatri, R.raw.gayatri, R.drawable.icon_gayatri, "Gayatri Mantra", timeintervals); } if ("Om Mani Padme Hum".equals(items[position].toString())) { int[] timeintervals = { 5500, 8200, 11100, 13900, 34100, 36700, 39500, 42300, 59300, 62000, 64800, 67600, 124600 }; // startChantActivity(TotalMala_loop,Total_Bead_Loop,BacgroundImage,Icon,Title,BeadsTotalTimeIntervals+totalTimeDurationOfAudio) startChantActivity(2, 108, R.drawable.ommanipadmehum, R.raw.om_mani, R.drawable.icon_padme, "Om Mani Padme Hum", timeintervals); } if ("Sai Ram".equals(items[position].toString())) { // Audio time interval for bead+total time duration of audio int[] timeintervals = { 4800, 7500, 10400, 12600, 15800, 18600, 21600, 24400, 25000 }; // startChantActivity(TotalMala_loop,Total_Bead_Loop,BacgroundImage,Icon,Title,BeadsTotalTimeIntervals+totalTimeDurationOfAudio) startChantActivity(2, 108, R.drawable.sairam, R.raw.sairam, R.drawable.icon_sairam, "Sai Ram", timeintervals); } if ("Aum".equals(items[position].toString())) { // Audio time interval for bead+total time duration of audio int[] timeintervals = { 12850, 13000 }; // startChantActivity(TotalMala_loop,Total_Bead_Loop,BacgroundImage,Icon,Title,BeadsTotalTimeIntervals+totalTimeDurationOfAudio) startChantActivity(2, 108, R.drawable.aum, R.raw.aum, R.drawable.ico_aum, "Aum", timeintervals); } } }); } class IconicAdapter extends ArrayAdapter { IconicAdapter() { super(Options.this, R.layout.list_item, items); } public View getView(int position, View convertView, ViewGroup parent) { LayoutInflater inflater = getLayoutInflater(); View row = inflater.inflate(R.layout.list_item, parent, false); TextView label = (TextView) row.findViewById(R.id.label); label.setText(" "+items[position]); ImageView icon = (ImageView) row.findViewById(R.id.icon); if (items[position].equals("Gayatri Mantra")) { icon.setImageResource(R.drawable.icon_gayatri); } if (items[position].equals("Om Mani Padme Hum")) { icon.setImageResource(R.drawable.icon_padme); } if (items[position].equals("Sai Ram")) { icon.setImageResource(R.drawable.icon_sairam); } if (items[position].equals("Aum")) { icon.setImageResource(R.drawable.ico_aum); } return (row); } } protected void startChantActivity(int mala_loop, int beads_loop, int background, int media, int titleIcon, String title, int[] timeintervals) { Bundle bundle = new Bundle(); bundle.putInt("mala_loop", mala_loop); bundle.putInt("beads_loop", beads_loop); bundle.putInt("background", background); bundle.putInt("media", media); bundle.putInt("titleIcon", titleIcon); bundle.putString("title", title); bundle.putIntArray("intervals", timeintervals); Intent intent = new Intent(this, ChantBliss.class); intent.putExtras(bundle); startActivityForResult(intent, this.getSelectedItemPosition()); } } And Corresponding xml file is: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content" android:orientation="horizontal" > <ImageView android:id="@+id/icon" android:layout_width="wrap_content" android:paddingLeft="2px" android:paddingRight="2px" android:paddingTop="2px" android:layout_height="wrap_content" /> <TextView android:id="@+id/label" android:layout_width="wrap_content" android:layout_height="wrap_content" android:textSize="22sp" android:textColor="#ff000000" /> </LinearLayout> Thanks in Advance: Sandeep

    Read the article

  • architecture mismatch between the Driver and Application?

    - by shane87
    I am using JDBC to connect to my microsoft access database. I get the following exception when I try to connect to the database: java.sql.SQLException: [Microsoft][ODBC Driver Manager] The specified DSN contains an architecture mismatch between the Driver and Application I am using 64bit windows7, and I am using eclipse which is also a 64bit version My database is a microsoft access database and it seems that the driver is a 32bit driver which is causing the problem. I read somewhere that microsoft has not released a 64bit driver for microsoft access! Any help on how to solve this problem would be greatly appreciated.

    Read the article

  • Call a protected method from outside a class in PHP

    - by Chad Johnson
    I have a very special case in which I need to call a protected method from outside a class. I am very conscious about what I do programmingwise, but I would not be entirely opposed to doing so in this one special case I have. In all other cases, I need to continue disallowing access to the internal method, and so I would like to keep the method protected. What are some elegant ways to access a protected method outside of a class? So far, I've found this. I suppose it may be possible create some kind of double-agent instance of the target class that would sneakily provide access to the internals...

    Read the article

  • General ODBC Error in VBA

    - by raam
    Hi am populating the data from MS Access By Using VBA i am using below mentioned code.if i am run the same code in MS 2007 then It run properly but if i am run the same code in MS 2003 it gives the "General ODBC Error" how to solve this problem Any help would be appreciated!! Thanks in advance Sub Button2_Click() Dim varConnection As String Dim varSQL As String Dim cal, cal1, x varConnection = "ODBC; DSN=MS Access Database;DBQ=D:\Box\Generate.mdb;Driver={Driver do Microsoft Access (*.mdb)}" ' varSQL = "SELECT * FROM Empdata" With ActiveSheet.QueryTables.Add(Connection:=varConnection, Destination:=ActiveSheet.Range("C7")) .CommandText = varSQL .Name = "Query-39008" .Refresh BackgroundQuery = False End With End Sub

    Read the article

< Previous Page | 207 208 209 210 211 212 213 214 215 216 217 218  | Next Page >