[ad_1] You can use Bundle class and put_extra data to retrieve and put. Example : put_extra Intent i = new Intent(); i.putExtra(“name_of_extra”, myParcelableObject); to read data Bundle b = new Bundle(); b = getIntent().getExtras(); Object myParcelableObject = b.getString(“name_of_extra”); [ad_2] solved How can I share values between a Service and Fragment in Android? [duplicate]
[ad_1] The endings are different 7i18704 5i18704 4i18704 Following your comment I have updated and they GROUP as expected. What do you get when you try this? CREATE TABLE #Advertisements ( ID INT IDENTITY(1,1), Url VARCHAR(200) ) INSERT INTO #Advertisements VALUES (‘http://example.com’) INSERT INTO #Advertisements VALUES (‘http://example.com’) INSERT INTO #Advertisements VALUES (‘http://example.com’) SELECT COUNT(*) c, … Read more
[ad_1] var keyVal = []; for (var i = 0; i < itemname.length; i++) { keyVal.push({ itemname: itemname[i], itemqty: itemqty[i] }); } [ad_2] solved Combining two arrays into one javascript json object
[ad_1] To group the sublists in a general way you can: Code: def linking_sublists(lists): index = {} sets = [] for l in lists: found = None for i in l: if i in index: # this list has an element we have already seen if found: # combine two sets to_remove = index[i] if … Read more
[ad_1] Not much unless you’re root. If you are root, services that require files with certain permissions will stop working. sudo is one of those. 3 [ad_2] solved Chmod recursively for /
[ad_1] I don’t understand the confusion. The split() function return a list of all subparts of your string by removing all occurences of the given argument. For example, if you have the following string : “Hello world!” and split this one by split(“o”) then your output will be : [“Hell”, ” w”, “rld!”] With a … Read more
[ad_1] It looks like a bug in Server did in fact introduce a duplicate signing identity. I reported it as rdar://21080937, if you’d like to dupe it. In order to fix it, I had to learn about how Xcode Server stores signing identities (thanks entirely to an extremely helpful answer to an unrelated question). Xcode … Read more
[ad_1] I don’t know which problems did you have writing this in C#, but the code is very simple: var y = 20; var p = 50000; var r = 0.05; var c = 6521; var result = p * Math.Pow(1 + r, y) + c * ((Math.Pow(1 + r, y + 1) – (1 … Read more
[ad_1] Create random date from today Have while loop, Check generated already exist in exclude dates array (continue loop until you find date which is not in dates array) const randomDateFromToday = (exclude_dates, max_days = 365) => { const randomDate = () => { const rand = Math.floor(Math.random() * max_days) * 1000 * 60 * … Read more
[ad_1] Indeed push_back may trigger reallocation, which is a slow process. It will not do so on every single push_back though, instead it will reserve exponentially more memory each time, so explicit reserve only makes sense if you have a good approximation of resulting vector size beforehand. In other words, std::vector already takes care of … Read more
[ad_1] Alright I get what your asking for now. The javascript below will make a call from your page to an external script (ajax/myscripthere.php) which will handle the SQL. In the php script, we simply handle our SQL like we normally would if this was not an AJAX call. Please accept answer if this is … Read more
[ad_1] Instead of: Promise.all(stuffArray).then((result) => { console.log(‘result: ‘, result); }); Use: Promise.all(stuffArray.map(func => func())).then((result) => { console.log(‘result: ‘, result); }); [ad_2] solved Promise.all invoking array variable full of functions [duplicate]
[ad_1] Instead of going to hell, as suggested in your comment above, I’ll provide you an approach. Create a server-side component. Your client sends it some protocol (such as XEP-0050) and the service sends your client an invite to the correct room. If you want existing clients to be able to join, also accept a … Read more
[ad_1] Pretty simple: In [8]: course_name Out[8]: ‘Post Graduate Certificate Programme in Retail Management (PGCPRM) (Online)’ In [9]: print re.sub(‘\([A-Z]+\)\s*’, ”, course_name) Post Graduate Certificate Programme in Retail Management (Online) In [17]: print re.search(‘\(([A-Z]+)\)\s*’, course_name).groups()[0] PGCPRM 0 [ad_2] solved Extract only first match using python regular expression [duplicate]
[ad_1] Do you want something like this? genOutList= ( from i in genOutList where !genAccList.Any(x=>x.Contract==i.Contract) select i ).ToList(); Or genOutList.RemoveAll(x=>genAccList.Any(i=>i.Contract==x.Contract)); 0 [ad_2] solved How can I merge this two bits of code [closed]