You can try the following to get ADB working: First search for adb.exe in processes of task manager and end it and try to run application. If the above doesn’t work or you can’t see adb.exe in task manger go for the next way Open command prompt, give path of platform-tools and write following commands … Read more
the operator ‘.’ is prior than new No it isn’t. The operator . is not prior to new: they are both Primary, so they have equal precedence, so they evaluate left to right. why this can be executed right? The real question is why can it be compiled, and I’ve answered that. It hasn’t tell … Read more
to take a year from current date: import datetime year = datetime.datetime.today().year 1 solved How to exclude year from date? [closed]
Finally, I discovered that the problem was in the code. Just small mistake using substr func. So, the correct code is: $buffer = file_get_contents(“image.png”); $length = filesize(“image.png”); if (!$buffer || !$length) { die(“Reading error\n”); } $_buffer=””; for ($i = 0; $i < $length; $i++) { $_buffer .= sprintf(“%08b”, ord($buffer[$i])); } echo $_buffer.”<br>”; $nb = “”; … Read more
If the line you mentioned, System.Windows.Forms.Application.Run(new Form1()); is in a file in a different namespace to xpartygo, you’ll need to edit it to System.Windows.Forms.Application.Run(new xpartygo.Form1()); to sort the namespaces. Alternatively, place the file with the above line into the same namespace. In future, pasting the exact error message as output from the compiler, along with … Read more
Try Select instead of Where to get a list of all can_main_key values: var candidate = db_can_records.CandidateMains.Select(m => m.can_main_key).ToList(); To get all the values for a single row of CandidateMains where you know the key try: var candidate = db_can_records.CandidateMains.FirstOrDefault(m => m.can_main_key == variableContainingRequiredId); solved Lambda expressions in c#
JSONObject jObject = new JSONObject(response); JSONSONArray p = jObject.getJSONArray(“SizeOptions”); for(int i=0;i<p.length();i++) { JSONObject jObjectValue=p.getJSONObject(i); String name = jObjectValue.getString(“Name”); } 2 solved Parsing Json from server
I don’t know what do you want to do with this code but there are many errors in your code. check below your code without errors <input type=”text” id=’name’ onmouseover=”return show(this,’namemessage’)” onmouseout=”return hide(‘namemessage’)” /> <span id=’namemessage’></span> <br/> <input type=”text” id=’email’ onmouseover=”return show(this,’emailmessage’)” /> <span id=’emailmessage’></span> <script> function show(inp,spanId){ var msg=document.getElementById(spanId); msg.innerHTML='<font color=”red” size=”6″>’+inp.id+'</font>’; } function … Read more
I suggest to implement the toString() method for each pojo and then simply use System.out.println(yourObject) for example for the Pessoa class you could write something like this: @Override public String toString() { return String.format(“Pessoa [idade=%s, nome=%s, altura=%s, peso=%s, matricula=%s]”, idade, nome, altura, peso, matricula); } 2 solved How to print values in Java?
I can see why you’re having trouble with this pointer variable: it isn’t one. It is a reference. You cannot have a non-const reference to a const thing. That would violate the const, and allow the n = 11 line to succeed, thus again violating the original const. const means “I cannot and will not … Read more
Why don’t you make mNsdUtils to be a member field if it should be accessed from the outside of the function? The root cause saying mNsdUtils is null from the MainActivity.onClickDiscover() implementation. Implement in the method itself, as the context is different throwing a null pointer exception. A small tip for you: read the stack … Read more
You should certainly read more on regex. A first hint is that when you want to capture a pattern you need to enclose it in parentheses. e.g. (\d+). For this example though, the code you need is: match = re.match(r’C(\d+)([F|G])(\d+)\.PNG’, s) first_id = match.group(1) fg_class = match.group(2) second_id = match.group(3) 1 solved How to extract … Read more
You don’t have to use auto if you don’t want to, but you can of course. The type returned by std::vector<int>::begin() and std::vector<int>::end() is a std:vector<int>::iterator (or std:vector<int>::const_iterator, depending on context), it is not an int. You could as well have written this instead: for(vector<int>::iterator it=vect1.begin(); it<vect1.end();it++){ cout<<*it<<endl; } or for(vector<int>::const_iterator it=vect1.begin(); it<vect1.end();it++){ cout<<*it<<endl; } … Read more
why output is 0 not “lol” ? because you are printing an integer and not an instance of that Main class you can do the following public class Main { @Override public String toString(){ return “lol”; } public static void main(String[] args) { // int aaa=0; Main myMain = new Main(); System.out.println(myMain); } } note … Read more
My guess is that: does not have the recent changes is because the data has not been posted. And cannot be from what I can see. You’re trying to insert and update tuples from two tables, but you fetched only a foreign key of a detail table. Imagine, that you’d like to update this resultset … Read more