repeat

how to repeat alarm after 1 day in android

那年仲夏 提交于 2019-12-04 17:04:40
I m using the alarm manager in my app and i want to repeat alarm after one day. the alarm should be invoked after one day when invoked once by time.Please help. Thanks in advance. if(str_freqSchedule.equals(checkForDaily)){ Calendar calendar = Calendar.getInstance(); //calendar.set(Calendar.DAY_OF_YEAR, calendar.get(Calendar.DAY_OF_YEAR)); calendar.set(Calendar.HOUR_OF_DAY, hr); calendar.set(Calendar.MINUTE, min); calendar.set(Calendar.SECOND,0); Intent intent = new Intent(this, AlarmReceiverActivity.class); PendingIntent pendingIntent = PendingIntent.getBroadcast( this.getApplicationContext()

how to create a repeating list of links inside a li tag using apache wicket?

隐身守侯 提交于 2019-12-04 15:12:42
In Apache Wicket I would like to create a repeating list of links from code. I am not sure what the template should be to get to an html result that looks like this: <li><a href="whatevenrlink1">link1</a></li> <li><a href="whatevenrlink2">link2</a></li> <li><a href="whatevenrlink3">link3</a></li> so after much testing this worked for me. HTML should look like: <ul> <ui wicket:id="LinkList"><a wicket:id="Link"><span wicket:id="Text"/></a></ui> </ul> and then the repeating view code will be: RepeatingView view = new RepeatingView("LinkList"); add(view); WebMarkupContainer list = new

Django - Tips to avoid repeating code in views

不打扰是莪最后的温柔 提交于 2019-12-04 13:05:35
问题 I'm moving from a PHP background into Django development via python, mostly for the sake of tackling a MVC (or MVT) that I feel makes the most sense, although in this pattern I've started to notice a lot of repeated code in my views. For example, when logged in I have information regarding the user that I would like to appear on every page, although when using render_to_response and in every view this is required I have to grab the information and pass it to the render_to_response function. I

Generate multiple serial graphs/scatterplots from data in two dataframes

﹥>﹥吖頭↗ 提交于 2019-12-04 12:55:30
I have 2 dataframes, Tg and Pf, each of 127 columns. All columns have at least one row and can have up to thousands of them. All the values are between 0 and 1 and there are some missing values (empty cells). Here is a little subset: Tg Tg1 Tg2 Tg3 ... Tg127 0.9 0.5 0.4 0 0.9 0.3 0.6 0 0.4 0.6 0.6 0.3 0.1 0.7 0.6 0.4 0.1 0.8 0.3 0.9 0.9 0.6 0.1 Pf Pf1 Pf2 Pf3 ...Pf127 0.9 0.5 0.4 1 0.9 0.3 0.6 0.8 0.6 0.6 0.6 0.7 0.4 0.7 0.6 0.5 0.1 0.6 0.5 0.3 0.3 0.3 Note that some cell are empty and the vector lengths for the same subset (i.e. 1 to 127) can be of very different length and are rarely the

How to idiomatically call C++ functions based on variable value?

微笑、不失礼 提交于 2019-12-04 07:27:36
Suppose I have a data type enum TreeTypes { TallTree, ShortTree, MediumTree } . And I have to initialize some data based on one particular tree type. Currently I have written this code: int initialize(enum TreeTypes tree_type) { if (tree_type == TallTree) { init_tall_tree(); } else if (tree_type == ShortTree) { init_short_tree(); } else if (tree_type == MediumTree) { init_medium_tree(); } return OK; } But this is some kind of stupid code repetition. I am not using any of the powerful C++ capabilities like templates. How could I write this code better? Thanks, Boda Cydo. Your code is OK for two

Python - Repeating code with a while loop

老子叫甜甜 提交于 2019-12-04 05:51:15
问题 So here is my question. I have a chunk of input code that I need to repeat in case the input is wrong. So far this is what I have (note that this is just an example code, the actual values I have in print and input are different: input_var_1 = input("select input (1, 2 or 3)") if input_var_1 == ("1"): print ("you selected 1") elif input_var_1 == ("2") print ("you selected 2") elif input_var_1 == ("3") print (you selected 3") else: print ("please choose valid option") What do I add after the

UILocalNotification fires after reinstalling the app

喜你入骨 提交于 2019-12-04 05:33:48
My app has an alarm function using UILocalNotification, and it works great. However, if the user uninstalls the app, then later REINSTALLS it, he would receive all the "in between" notifications at once. I have tried to call: [[UIApplication sharedApplication] cancelAllLocalNotifications]; if it's the first time the app is launched, but it doesn't help, because the notification is received even before application:didFinishLaunchingWithOptions: is called. This was worse in 4.0 when the alarm was repeated even if the user has deleted the app, but at least that bug was fixed by Apple in later

css start repeating background from defined position

这一生的挚爱 提交于 2019-12-04 04:51:17
#container{ background:url(images/bg-main.png) repeat-y; width: 903px; background-position: 0px 687px; background-position: bottom; height: 1200px; margin-left: auto; margin-right: auto; } #content{ background:url(images/bg-wood.png) repeat-y; width: 903px; height: 678px; margin-left: auto; margin-right: auto; } #content div is inside #container div. I want #container 's background to start repeating at 687px from top. Is it possible? EDIT: Is it possible that first x pixels of div (from top) have emtpy space and after x pixels backgrund starts? background : url('image path') 0px 287px repeat

Combinations with repetition

核能气质少年 提交于 2019-12-04 04:17:45
I'm using Mathematica 7 and with a combinatorica package function I can get all combinations of a certain number from a list of elements where the order doesn't matter and there is no repetition.e.g: in: KSubsets[{a, b, c, d}, 3] out: {{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}} I cannot find a function that will give me all combinations of a certain number from a list of elements where the order doesn't matter and there is repetition. i.e. the above example would include elements like {a,a,b},{a,a,a},{b,b,b}...etc in the output. It may require a custom function. If I can come up with one I

Python Repeat Elements in One List Based on Elements from Another

*爱你&永不变心* 提交于 2019-12-04 04:14:34
问题 Given the following lists: a = [0, 5, 1] b = [1, 2, 1] I'd like to repeat each element of [a] by the number of its corresponding position in [b] to produce this: [0, 5, 5, 1] i.e. 0 occurs 1 time, 5 occurs 2 times, and 1 occurs 1 time. 回答1: In [7]: a = [0, 5, 1] In [8]: b = [1, 2, 1] In [9]: list(itertools.chain(*(itertools.repeat(elem, n) for elem, n in zip(a, b)))) Out[9]: [0, 5, 5, 1] In [10]: b = [2, 3, 4] In [11]: list(itertools.chain(*(itertools.repeat(elem, n) for elem, n in zip(a, b))