I've got an warning message with my site saying "session_start(): open(/home/content/55/5307955/tmp/sess_qphb8rav2kmt5kavptqh1mik06, O_RDWR) failed: No such file or directory (2) ". I've been reading basics about PHP session and seesion_start(), so far I understand that when session_start it either starts a new session or resume the existing session. What does it have to do with opening a file?
Are $_SESSION variables stored on the client or the server?
How do you auto resize cells in Excel? Take as an example the well-known . In the original paper, the regression model was used to model the amount people were willing to pay for improved air quality.  But the 'people' whose behaviour was being modelled was never specified. Presumably, it was not limited to 'residents of Greater Boston'. Only implausibly would it be 'humans anywhere on the planet'. Nor, 'all US residents'.   So what inference are we invited by the authors to draw about the scope they intend for their theory? Perhaps 'all residents of large US metro areas'. But - what about other First World metro areas? The first two sections of the Boston paper scrupulously avoid any reference which might suggest that they had a particular 'population' to which their 'sample' was to apply.  There is also the dimension of time - the data came from the 70s, whereas the theory is has no time limitations spelt out.  Plus - going back to first principles - how could the Boston data be treated as a sample of a population including, say, the Chicago metro area when no data from the Chicago area was considered?   Besides, lots of the covariates in the Boston regression weren't samples at all, but totals - so, for example, the proportion of blacks was supplied from census stats.  My question: how can we conceive the Boston data as supporting the geographically unlimited propositions in the paper? Is there a mathematical way of showing equivalence or approximation of the Boston data analysis to other cities? Or is it just assumed as a matter of common sense - such that the paper's authors would be amazed to think anyone could have any doubts on the matter? Force new users to go through animated tutorial before they post for the first time Do not allow people to post the minute they created an account.  Let them ask the question but not allow them to post.  Let them stew for 30 minutes and force them through a nice animated tutorial on what they should do before they post and enforcing the fact that bad questions will not get an answer.  If they had a Need to know! basic rules introduction forced on them asking them at the end to check a box I understand! some might ask better questions and help the community instead of flooding the site with bad questions.  The presentation should be on a timer without the ability to skip and highlight the similar questions asked before that might give them the answer.  Also for the first next 9 posts they should have to check a box next to the first 5 of each similar questions before posting.  I do not mean to be harsh, but the truth is I see more negative votes on questions then positive. It's like 90% of the time people give -1 to first questions. Limit connection rights on SQL Server I have an app to deploy in production that uses 'honor system' security. That is, all users connect to the DB using a SQL user/passwd credential and the app manages permissions itself. The latter part doesn't bother me as much as the fact that the connection object contains embedded credentials and can be copied around freely. I'm try to find some way to limit connections to a more limited set of clients. I can create firewall rules to limit by IP, of course. Is there any way to 'prequalify' SQL logins either by Machine account or domain membership? Source shell script automatically in terminal How can I automatically source a particular shell script when I open a terminal window by right clicking somewhere and choosing "open in terminal"?  For example, every time I open a terminal I need to run the following command line:  source myscript   How can I make it so that I don't have to type this automatically?  My script is written in tsch:  #!/bin/tcsh setenv DISPLAY 127.0.0.1:10.0 cd /ast/dcm/data   I'm using CentOS 7. How can I `alias sudo !!`? I'm trying to set an alias for sudo !! in Bash. I tried alias sbb='sudo !!', but it interprets that as a literal  !! and prints   sudo: !!: command not found   If I use double quotes, it substitutes the double bang in the string itself, so that doesn't work.  Is there any way to make this work? Or an alternative alias? ` How do I report a bug related to Stack Exchange? Does Stack Exchange contain any edit suggestion or bug submission system, or such type of discussion forum to inform SO or its administrators about suggestions and bug reports that relate to SO itself? According to the , you can harden a MySQL server by adding passwords, or removing the anonymous accounts.      If you want to prevent clients from connecting as anonymous users   without a password, you should either assign a password to each   anonymous account or else remove the accounts.   Before hardening, my users table looked like this.   mysql&gt; select user,host,password from mysql.user; +------------------+-----------+-------------------------------------------+ | user             | host      | password                                  | +------------------+-----------+-------------------------------------------+ | root             | localhost | *F3A2A51A9B0F2BE246XXXXXXXXXXXXXXXXXXXXXX | | root             | gitlab    |                                           | | root             | 127.0.0.1 |                                           | | root             | ::1       |                                           | |                  | localhost |                                           | |                  | gitlab    |                                           | | debian-sys-maint | localhost | *95C1BF709B26A5BAXXXXXXXXXXXXXXXXXXXXXXXX | | myuser           | localhost | *6C8989366EAF75BB6XXXXXXXXXXXXXXXXXXXXXXX | +------------------+-----------+-------------------------------------------+   I've remove all anonymous accounts, so that the user table now looks like this.  (I'm using puppet to manage the users, but puppet effectively performs a DROP USER command).   mysql&gt; select user,host,password from mysql.user; +------------------+-----------+-------------------------------------------+ | user             | host      | password                                  | +------------------+-----------+-------------------------------------------+ | root             | localhost | *F3A2A51A9B0F2BE246XXXXXXXXXXXXXXXXXXXXXX | | debian-sys-maint | localhost | *95C1BF709B26A5BAXXXXXXXXXXXXXXXXXXXXXXXX | | myuser           | localhost | *6C8989366EAF75BB6XXXXXXXXXXXXXXXXXXXXXXX | +------------------+-----------+-------------------------------------------+   Why is it that I am still able to login to my test system without a username or a password? What do I need to do to prevent any unwanted users from logging in?   root@gitlab:~# mysql Welcome to the MySQL monitor.  Commands end with ; or \g. Your MySQL connection id is 98 Server version: 5.5.32-0ubuntu0.12.04.1 (Ubuntu) .... mysql&gt;    Update:  I've also just discovered that I can login as root, without entering a password.  Update2: I found  which has some good information, but does not solve the issue.   There are no anonymous users.  mysql&gt; select user,host,password from mysql.user where user=''; Empty set (0.00 sec)   I login as root@localhost.  mysql&gt; select USER(),CURRENT_USER(); +----------------+----------------+ | USER()         | CURRENT_USER() | +----------------+----------------+ | root@localhost | root@localhost | +----------------+----------------+    I do not have a default password, or skip-grant-tables defined in my.cnf  root@gitlab:~# cat /etc/mysql/my.cnf |grep -i 'skip-grant-tables'|wc -l 0 root@gitlab:~# cat /etc/mysql/my.cnf |grep -i 'pass'|wc -l 0   Update3:  I have tried performing these steps with puppet, (which should perform a flush privileges automatically). I have also manually flushed privileges and also tried restarting mysql.   Update4: I've also tried changing the mysql root password and flushed privileges. No luck, I can still log in as any user without a password. Change style of labels of Tikz nodes of a given style Problem with >> inside of \tt I have an inline code example with a nested Java generic declaration, using the typewriter typeface. However, the closing &gt;&gt; is rendering as a black box. A cursory Google search suggested $&gt;&gt;$, but this removes the typewriter font. Using \textgreater in place of the actual &gt; also encounters the same problem. Here is the offending snippet:  ...declaration {\tt List&lt;Map&lt;Integer, String&gt;&gt;} is...   Any ideas? IMO 1987 - function such that $f(f(n))=n+1987$ Show that there is no function $f: \mathbb{N} \to \mathbb{N}$ such that $$f(f(n))=n+1987, \ \forall n \in \mathbb{N}$$  This is problem 4 from IMO 1987 - see, for example, . How can I parse dates and convert time zones in Perl? Where can I find 32-bit version of Ubuntu? Why is it that Frisian is considered the closest related language to English? It is commonly asked "What is the closest language to English?" and the equally common answer is         Except that there is rarely a reason given for this connection; the most that is given are baldly stated 'facts' (e.g. "English and Dutch are in the same 'Low German' group")  Can anyone give any substantive reasons? For example, an comparative overlap of vocabulary, a distance metric of phonology, privately shared syntactic rules among all the rules in West German languages. How to make the taskbar blink my application like Messenger does when a new message arrive? Is there an API call in .NET or a native DLL that I can use to create similar behaviour as Windows Live Messenger when a response comes from someone I chat with? Does ClamAV Scan Windows virus? I'm new to Ubuntu.  I'm now setting up an OwnCloud server which supports ClamAV file scan.  Since most of the users connected to the server uses Windows OS, can ClamAV scan for Windows virus?  I also did some research that such as BitDefender. It clams that it scans real time, if someone upload files to the server, will BitDefender scan the file even OwnCloud doesn't support?  By the way, I heard that if I update the system frequently, I don't have to install any antivirus that scan for Linux virus. Is that right? Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ Prove by induction the summation of $\frac1{2^n}$ is greater than or equal to $1+\frac{n}2$.  We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$$ for all positive integers.   I have resolved that the following attempt to prove this inequality is false, but I will leave it here to show you my progress. In my proof, I need to define P(n), work out the base case for n=1, and then follow through with the induction step. Strong mathematical induction may be used.  This is equivalent to $$\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n}2\;.$$  Let $P(n)$ be summation shown above.   Base case for $n=1$, the first positive integer,   $$\sum_{k=0}^1\frac1{2^k}=\frac1{2^0}+\frac1{2^1}=1+\frac12=\frac32\ge 1+\frac12=\frac32\;,$$  so base case is true.  Induction step: Assume $P(n)$ is true and implies $P(n+1)$. Thus   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge\frac1{2^{n+1}}+\sum_{k=0}^n\frac1{2^k}\ge 1+\frac{n+1}2\;.$$  This can be written as   $$\sum_{k=0}^{n+1}\frac1{2^k}\ge \frac1{2^{n+1}}+1+\frac{n}2\ge 1+\frac{n+1}2\;.$$  I work the math out but I get stuck contradicting my statement. Please show your steps hereafter so I can correct my mistakes. Let A and B be sets. Show that A is a subset of B if and only if for any set C, one has A union C is a subset of B union C. Can you verify my proof if it is right?  Let $A$ and $B$ be sets.  (a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$. (b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.  Edited Solution :  (a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.      Solution 1 :(&lt;=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.      Solution 2 :(&lt;=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.  (b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.  (&lt;=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B. Applying for a Schengen Visa without itinerary I am currently residing in Ireland with work authorization permit. In the next few months I am planning for a trip to few schengen countries. I have not yet decided my itinerary. Is there any Schengen country embassy which can process a tourist Schengen visa for 90 days? How to define binary operation on arbitrary set in order to create a group structure.