concatenation

问题 I am trying to concatenate two lists, one with just one element, by doing this: print([6].append([1,1,0,0,0])) However, Python returns None . What am I doing wrong? 回答1: Use the + operator >>> [6] + [1,1,0,0,0] [6, 1, 1, 0, 0, 0] What you were attempting to do, is append a list onto another list, which would result in >>> [6].append([1,1,0,0,0]) [6, [1,1,0,0,0]] Why you are seeing None returned, is because .append is destructive, modifying the original list, and returning None . It does not

问题 I have this weird problem. Why do these two implementations return different results? $db = DbBase::getInstance(); $stmt = $db->prepare('SELECT round(round(9.50 * :amount, 2) * 23 * 0.01, 2)'); $stmt->execute(array(':amount' => 1)); echo $stmt->fetchColumn(); Result: 2.18 $db = DbBase::getInstance(); $stmt = $db->prepare('SELECT round(round(9.50 * 1, 2) * 23 * 0.01, 2)'); $stmt->execute(); echo $stmt->fetchColumn(); Result: 2.19 When I bind the amount it gives me different result. I'd rather

问题 I'm trying to concatenate a Sass variable with a string. How do I do it? Below there is a code example. @mixin setIcons($pathImages:$pathImages, $extensionIcon:$extensionIcon { @each $class-icon in $class-icons-30 { .snw-tui-icon-#{$class-icon} { @include icon($class-icon, $pathImages, $extensionIcon, 30px); } .snw-tui-icon-**#{$class-icon} + "selected"** { @include icon(**$class-icon + "selected"**, $pathImages, $extensionIcon, 30px); } } 回答1: You want to interpolate. $sidebar-width: 20px;

问题 Hi I've 2 byteBuffers and I want to concatenate them together to a single byteBuffer. I found a similar question here but none of the suggestions there worked for me. 回答1: You can do it like this ByteBuffer b3 = ByteBuffer.allocate(b1.limit() + b2.limit()); b3.put(b1); b3.put(b2); 来源： https://stackoverflow.com/questions/30734268/concatenate-two-bytebuffers-to-a-single

问题 I've seen plenty of discussion on using paste and paste0 to concatenate two strings in r. However, this does not appear to be working for two string variables. I have a data frame that looks like the following. series_id year period value footnote_codes 1 LASBS260000000000003 1983 M01 15.1 2 LASBS260000000000003 1983 M02 15.0 3 LASBS260000000000003 1983 M03 14.8 4 LASBS260000000000003 1983 M04 14.6 I wish to combine the year variable to the period variable to generate a new variable in the

问题 I'm trying to be a good erlanger and avoid "++". I need to add a tuple to the end of a list without creating a nested list (and hopefully without having to build it backwards and reverse it). Given tuple T and lists L0 and L1: When I use [T|L0] I get [tuple,list0] . But when I use [L0|T] , I get nested list [[list0]|tuple] . Similarly, [L0|L1] returns [[list0]|list1] . Removing the outside list brackets L0|[T] produces a syntax error. Why is "|" not symmetric? Is there a way to do what I want

问题 im trying to concatenate 3 columns in a select query to show in one column in the results. the column is called DelPostalName and for some reason always shows a '0' when i run the select query. like its trying to add up the strings but theres not actual numbers to add. ive been googling string concatenation and this seems to be the correct syntax. any ideas? isc_orders.ordShipFirstName + ' ' + isc_orders.ordshiplastname + isc_orders.ordshipcompany as DelPostalName, 回答1: The result appears as

问题 I am trying to use moviepy to generate video with texts. First, I want to show one messages and then another one. In my case I want to show "Dog" for one second and than "Cat Cat". For that I use the following code: from moviepy.editor import * def my_func(messeges): clips = {} count = 0 for messege in messeges: count += 1 clips[count] = TextClip(messege, fontsize=270, color='green') clips[count] = clips[count].set_pos('center').set_duration(1) clips[count].write_videofile(str(count) + '.avi'

问题 I would like to join results returned in the set in MySQL with a comma as a separator string. For example, set returned contains: COLUMN_X john jerry maria joseph gugla I would like to receive the result as: COLUMN_X-concat john,jerry,maria,joseph,gugla is that possible? thanks. SELECT CONCAT(rooms.ID,",") FROM rooms AS rooms LEFT JOIN inter AS i ON rooms.ID=i.value WHERE xxx=999 doesn't work as I would like it to as it returns separate results. 回答1: SELECT GROUP_CONCAT(COLUMN_X SEPARATOR ','

问题 I'm using SonarLint that shows me an issue in the following line. LOGGER.debug("Comparing objects: " + object1 + " and " + object2); Side-note: The method that contains this line might get called quite often. The description for this issue is "Preconditions" and logging arguments should not require evaluation (squid:S2629) Passing message arguments that require further evaluation into a Guava com.google.common.base.Preconditions check can result in a performance penalty. That's because