i have an AutoNumber field which created in format IN -{0000}, this is my format.here i can insert upto 10k recoeds{0000}.what will happend if limit exceed more than 10k..?
I know that if you create a custom object, with an Auto-number Name field with a format, e.g. MYC-{0000} you will be able to create records from MYC-0000 to MYC-9999.  The moment you create 10,000 records Salesforce happily extends the range and you can then create MYC-10000 to MYC-99999.  My question is, what is the limit to this - for obvious reasons its a difficult one to test. I know instances that have exceeded 999,999 e.g. MYC-1000000 but am interested to know how far this extends to.  EDIT: When the field has a data type of Text the maximum length is 80. Is it safe to assume that the same is true for Autonumber (though that would be a very big number)?
How do I have a camera follow my object in Unity? I have an object that automatically moves by itself and I want the main camera to automatically follow it.(Like in games such as geometry dash and jetpack joyride)  This is the code for the automatic moving object in case it is needed:  using UnityEngine; using System.Collections; public class automove : MonoBehaviour { public static int movespeed = 5; public Vector3 userDirection = Vector3.right;    public void Update() {     transform.Translate(userDirection * movespeed * Time.deltaTime);  } }   So does anyone know any good scripts I could add to the main camera to follow this object that automatically moves? Thanks in advance! There are many red and black spots in my photos,what are those? pip: AttributeError: 'module' object has no attribute 'SSL_ST_INIT' $$\lim_{x\to 0} \frac{\ln(1+x)}x$$  The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)   $$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$  $$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$  $$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$  At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula. I want to do:  cat update_via_sed.sh | sed 's/old_name/new_name/' &gt; new_update_via_sed.sh   in my program.  But I want to use variables, e.g.  old_run='old_name_952' new_run='old_name_953'   I have tried using them but the substitution doesn't happen (no error). I have tried:  cat update_via_sed.sh | sed 's/old_run/new_run/' cat update_via_sed.sh | sed 's/$old_run/$new_run/' cat update_via_sed.sh | sed 's/${old_run}/${new_run}/' Asking the position of a person in a sequence How to ask the position(?) of a president (like 10th) of a country, correctly? The situation is somewhat like-  var someVar = some_other_function(); someObj.addEventListener("click", function(){     some_function(someVar); }, false);   The problem is that the value of someVar is not visible inside the listener function of the addEventListener, where it is probably being treated as a new variable. After cruising through a lot of material online, and answers over here, my understanding of contravariant and covariant vectors are, in a finite-dimensional vector space, suppose we have a vector, which is contravariant to its basis. However, using the metric tensor, we can map this vector to a one form in the dual space, which acts as a vector space for this example. This one form, or covactor, varies similarly to the basis in our original vector space. However, the covectors have a different basis, which is the dual basis, and are expanded in terms of those. Here's where my confusion arises : Many diagrams over the internet give a geometric picture of the scenario, by claiming that contravariant components of an 'arrow' are found by drawing parallel lines along the axes, and checking where they intersect. Co-variant components could be found by sketching the perpendicular lines on these axes. So, these materials are treating covariant and contravariant as different representations of the same arrow, while I'm inclined to believe they are completely different. However, if we let go of rigour, in exchange for a 'more' geometric understanding, I think we are allowed to do this. The diagrams are usually of the form :  However, even if we assume that contravariant and covariant are different representations of the same object, this particular diagram still seems wrong to me. Particularly the locations of $x_1,x_2$. In this diagram, they are located on the span of the original basis. Shouldn't covector components be located on the span of the dual basis ? This diagram suggests to me, that co-variant components have the same basis, as the contravariant ones. Shouldn't the diagram look more like this ?  This second diagram seems to fit the concept better according to me. This is because the dual vector must be 'contravariant' with respect to the 'dual basis'. This means, their components must be found out by sketching parallel lines along the span of the dual basis. These lines intersect the original axis at a right angle, which is to be expected, as dual bases are orthogonal to the original bases. Moreover, this second picture can also show in a much better way, how scaling up of the original basis, scales down the 'contravariant components' and the 'dual basis', which in turn scales up the 'covariant' components.This shows that the components of the dual vector are covariant to the original basis. This is something that is not readily visible in the first diagram. So, am I correct in assuming this second 'geometrical' representation is correct ? I know this doesn't make much sense, because as the mathematicians said, that vectors are completely different from one forms, and one forms should be represented using the number of hyperplanes intercepted by our arrow. However, I've seen most course material refer to it in this way, and frankly it is easier to visualize. However, most of this material used the first picture. Can anyone point out my mistakes, if any, and tell me if the first picture is correct, or the second one ? Or in this case 'less-flawed'. Problem from Herstein on group theory The problem is:  If $G$ is a finite group with order not divisible by 3, and $(ab)^3=a^3b^3$ for all $a,b\in G$, then show that $G$ is abelian.  I have been trying this for a long time but not been able to make any progress. The only thing that I can think of is: $$ab\cdot ab\cdot ab=aaa\cdot bbb\implies(ba)^2=a^2b^2=aabb=(\text{TPT})abba.$$ Now, how can I prove the last equality? If I write $aabb=abb^{-1}abb$, then in order for the hypothesis to be correct, $b^{-1}abb=ba\implies ab^2=b^2a$. Where am I going wrong? What should I do? We roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by 6? So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?  Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six? Euclidean distance vs Squared reconfiguration issues with dpkg I ran into some problems installing a package with apt. The terminal hung up during an install and I saw no other choice but to close the terminal. The problem, however, is that I can't seem to solve the issues that this has caused.   As is typical, I get:  E: Could not get lock /var/lib/dpkg/lock - open (11: Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/), is another process using it?   This gets me nowhere and I'm locked out until I run sudo dpkg --configure -a. Makes sense. However, this process spits the following out and then hangs on the last line:  Setting up virtualbox-dkms (5.2.10-dfsg-6) ... Removing old virtualbox-5.2.10 DKMS files...  -------- Uninstall Beginning -------- Module:  virtualbox Version: 5.2.10 Kernel:  4.15.0-23-generic (x86_64) -------------------------------------  Status: This module version was INACTIVE for this kernel. depmod...  DKMS: uninstall completed.  ------------------------------ Deleting module version: 5.2.10 completely from the DKMS tree. ------------------------------ Done. Loading new virtualbox-5.2.10 DKMS files... Building for 4.15.0-23-generic Building initial module for 4.15.0-23-generic   A typical solution I've read is that you need to fix the broken packages. When I try to fix the broken packages (sudo apt install -f), the system also hangs on the same line, although the output is different:  Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. 3 not fully installed or removed. After this operation, 0 B of additional disk space will be used. Setting up virtualbox-dkms (5.2.10-dfsg-6) ... Removing old virtualbox-5.2.1  0 DKMS files...  ------------------------------ Deleting module version: 5.2.10 completely from the DKMS tree. ------------------------------ Done. Loading new virtualbox-5.2.10 DKMS files... Building for 4.15.0-23-generic Building initial module for 4.15.0-23-generic   I am unable to get out of it unless I kill the process. This I do by checking what is currently running related to apt (ps aux |grep apt), and then I kill the relevant processes with sudo kill -9 process_number_here.  If I try to run things after that, it tells me I still need to reconfigure dpkg.  I tried looking around for more solutions to dpkg and all I really see is the nuclear option:  sudo rm /var/lib/dpkg/lock sudo rm /var/cache/apt/archives/lock sudo rm /var/lib/apt/lists/lock   I also delete the files in /var/lib/dpkg/updates because when I did some digging one of the files inside had the exact same text as the above long output... but I fully admit to being out of my league here and I don't know what to do.   Taking a look at the error messages, I recognize that it hangs on building something related to the kernel. But I have no idea where to even start solving that. Any help is appreciated. My system is ubuntu 18.04 Set active image node with python There's any way to change it directly by Python, or there's any way to set by Python what node image is the active to display in viewport (Textured) and bake? What does “pull sb. out of the hat” mean? Vertically align horizontal inline-block elements What should be created first in a video game? Is there a way to open a raster that is stored in ESRI V10.0 Geodatabase in QGIS V2.4?    I can open the vector layer, but there does not seems to be a way to open a raster. I have gotten as far as to create the bootable usb, booting into that bootable usb and getting to the install options however i quit because I had some concerns.  I currently have a Samsung 850 evo 1tb that has win 10ninstalled. I want to dual boot with Ubuntu 18.04 by installing it on a brand new unformatted uninitialized Samsung 860bevo 250gb. Should i format the brand new ssd as a new volume in win 10 and then proceed with the Ubuntu install on this volume. What else should i do? Im a first time Ubuntu user and installer.  Any and all help is appreciated. Friendly_Id and Reserved Words -- How can i replace the reserved word? A rubber viewfinder cover is connected to my camera strap.  What's it for, and when do I use it?  What happens to a photo if I don't use it?