I have an sql table like this: TABLE: Info  ID      |   JOB    |  Value --------|----------|------------------ 1       |   leader |  3 2       |   Host   |  212 1       |   User   | 4 2       |   leader |  5 2       |   Host   |  4  I'm trying to get the rows where a duplicate ID must contain 3 and 4 from the Value column. I'm currently using this  query:   select * from info where value = 3 and 4    group by id having count(*) &gt; 1;  my result using this query ID     |   JOB    |  Value --------|----------|------------------ null    |   null   |  null  whereas my expected result should be: ID      |   JOB    |  Value --------|----------|------------------ 1       |   leader |  3 1       |   User   | 4  Please help as I'm not sure what I'm doing wrong
This is a very basic query I can't figure out....  Let's say I have a two column table like this:  userid  |  roleid --------|--------    1    |    1    1    |    2    1    |    3    2    |    1   I want to get all distinct userids that have roleids 1, 2 AND 3.  Using the above example, the only result I want returned is userid 1.  How do I do this?
I recently got a lenovo Y40 laptop and after i installed ubuntu 14.04 (also in the live usb) Wireless doesn't work. It says that it is disabled by hardware switch, however the hardware switch doesnt work (its intention is to put on airplane mode but does nothing when pressed in ubuntu) wireless works in windows and the BIOS has it enabled. I tried resetting the BIOS but it still doesn't work. Here are some command outputs  lspci:  03:00.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller (rev 10) 04:00.0 Network controller: Intel Corporation Wireless 3160 (rev 93)   rfkill list all:   0: ideapad_wlan: Wireless LAN     Soft blocked: no     Hard blocked: yes 1: ideapad_bluetooth: Bluetooth     Soft blocked: no     Hard blocked: yes 2: hci0: Bluetooth     Soft blocked: no     Hard blocked: no 3: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no   lswh:     *-network         description: Ethernet interface         product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller         vendor: Realtek Semiconductor Co., Ltd.         physical id: 0         bus info: pci@0000:03:00.0         logical name: eth0         version: 10         serial: f8:a9:63:31:2e:b6         size: 1Gbit/s         capacity: 1Gbit/s         width: 64 bits         clock: 33MHz         capabilities: bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation         configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=full firmware=rtl8168g-3_0.0.1 04/23/13 ip=192.168.1.4 latency=0 link=yes multicast=yes port=MII speed=1Gbit/s         resources: irq:64 ioport:4000(size=256) memory:c0604000-c0604fff memory:c0600000-c0603fff *-pci:3      description: PCI bridge      product: Lynx Point-LP PCI Express Root Port 4      vendor: Intel Corporation      physical id: 1c.3      bus info: pci@0000:00:1c.3      version: e4      width: 32 bits      clock: 33MHz      capabilities: pci normal_decode bus_master cap_list      configuration: driver=pcieport      resources: irq:59 memory:c0500000-c05fffff    *-network DISABLED         description: Wireless interface         product: Wireless 3160         vendor: Intel Corporation         physical id: 0         bus info: pci@0000:04:00.0         logical name: wlan0         version: 93         serial: a0:88:69:90:0a:38         width: 64 bits         clock: 33MHz         capabilities: bus_master cap_list ethernet physical wireless         configuration: broadcast=yes driver=iwlwifi driverversion=3.13.0-32-generic firmware=22.24.8.0 latency=0 link=no multicast=yes wireless=IEEE 802.11abgn         resources: irq:66 memory:c0500000-c0501fff The motivation for this question is in my answer to this post :       (1) If I take $A$, $B$ and $C$ three groups such that $A\times B$ is isomorphic to $A\times C$ then do we have $B$ is isomorphic to $C$ ?   In the post a counter example is given when $A$ is an infinitely generated group. However I cannot get a counter-example when $A$, $B$ and $C$ are assumed to be finite groups. Note that there is no reason a priori if $\phi$ is an isomorphism from $A\times B$ to $A\times C$ that $\phi(A\times\{1_B\})=A\times\{1_C\}$.   In the case of finite groups it is true because of the Krull-Schmidt theorem : .  A slightly more general question is the following :     (2) If $B$ and $C$ are two finitely generated groups, can I always find a finitely generated group $A$ such that $A\times B$ is isomorphic to $A\times C$?   If $A$ is not assumed to verify any property but to be a group we can take :  $$A=B^{\mathbb{N}}\times C^{\mathbb{N}} $$  Furthermore replacing :$ \prod_{\mathbb{N}}$ by $\bigoplus_{\mathbb{N}}$ (or more likely the restricted product for groups but I am not sure of a well known notation, it is the subgroup of the product with all but a finite number components being trivial) will give you the same result for countably infinite groups.   $\underline{\text{Edit}}$ : To be clear about what I am asking. (1) is kown to be true for finite groups whereas (2) is known to be true when we are considering groups without restriction (even just for countably infinite groups). Let us pretend we are working on the smaller class of groups : the finitely generated groups. Is there any counter example to $(1)$? Is there any counter example to $(2)$?   $\underline{\text{Edit 2}}$: Here is a partial answer, from the classification of finitely generated abelian groups $(1)$ is always true if all groups are assumed to be abelian finitely generated groups. Now assume that $A\times B=G=A\times C$. Then you have (derived groups) :  $$D(G)=D(A)\times D(B)=D(A)\times D(C) $$  So the abelianizations verify :  $$A^{ab}\times B^{ab}=A^{ab}\times C^{ab}$$  Because we are now in the realm of finitely generated abelian groups we get :  $$B^{ab}=C^{ab} $$  So $(2)$ cannot be true in general for instance you can take $B:=\mathbb{Z}$ and $C=\mathbb{F}_5$, those two groups have different abelianizations.   The new question can be found here : I have a problem with energy conservation in case of interfering waves.  Imagine two harmonic waves with amplitudes $A$. They both carry energy that is proportional to $A^2$, so the total energy is proportional to $2A^2$. When they interfere, the amplitude raises to $2A$, so energy is now proportional to $4A^2$ and bigger than before.  The equivalent question is what happens to the energy with the superposition of two waves that interfere destructively.  Also, if someone could comment on the statement about this problem in my physics book (Bykow, Butikow, Kondratiew):      The sources of the waves work with increased power during the interference because they feel the wave from the other source. Can you Dash twice on the same turn? The Dash action gives you extra movement for the current turn (equal to your speed). If I have an effect (like the Expeditious Retreat spell or the monk's Step of the Wind feature) which allows me to dash as a bonus action, can I also dash as my regular action, for triple movement? Can I use my laptop as a second monitor? Is it possible to make my laptop's screen act as a second monitor for my desktop computer? I have a quick question. Looking at the next GIF that I made, what is the quickier way you can come up with to delete the excess material that goes beyond the wall? It is basically the attachment of the handrail to the wall what is giving me this "problem".    My solution would be using the boolean modifier with a cube or something, so obviously I know how to do it. But I'm looking for the best solution, i.e., does Blender let me select the excess material in any way? For example, is there a quick way to intersect those planes? I read about intersecting planes in Blender and seems like a difficult task (not worth it in this case). So I discarded that solution.  Thanks. I changed my username and now I cannot play online I don't understand this and I'm now worried as I have donated to a lot of servers and I don't want to lose this account. On the launcher it says my new username but when I log on it says "Account not Authenticated with minecraft.net" and "Non-Valid Username".  How can I regain the ability to use my account online? Consider a string under tension, for example, a string on a guitar. When a guitar string is plucked, it vibrates at a certain frequency. When the tension on the string is increased by twisting the tuning peg at the end of the neck, the frequency of the string increases. According to , increasing the tension on a string also increases the speed of the wave moving through it. According to  the speed of a wave is higher in a more dense medium.  Does this mean that increasing the tension also increases the density?  This does not seem very logical to me, since density is the "compactness" of a substance, or how close together the individual molecules are. By twisting the tuning peg on a guitar, you rotate the cylinder that the string is fixed to, wrapping it around more tightly. This pulls more of the string around that cylinder, similarly to when you reel in a fishing rod. Since more of the string's molecules are now wrapped around the peg, there should be fewer molecules over the same distance between the bridge and the nut (the two ends of the vibrating portion of the string), therefore bringing the density down. Show that a norm on an inner product space satisfies parallelogram law. Hence use the parallelogram law to show that the space of continuous real functions defined on the interval $[a,b]$ is not a Hilbert space. Here I did the first part. Let $X$ be an inner product space and $x,y\in X$ consider $\Vert x+y\Vert ^2$ and $\Vert x-y\Vert ^2$ Then on adding I get $\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2 +\Vert y\Vert ^2)$ For the second part I don’t know how can a parallelogram law prove $C[a,b]$ is not a Hilbert space, help me please. What is the difference between logistic regression and Fractional response regression? Why can I no longer easily navigate inside of this mesh? PHP: Right way to declare variable before use in loop I have a variable that is built in loop. Something like:  $str = ""; for($i = 0; $i &lt; 10; $i++) $str .= "something";   If $str = "" is ommitted, I get undefined variable notice, but I thought php auto-declare a variable the first time it sees undeclared one?  How do I do this right? As you are all most likely aware already all domain registrars ask for your personal information.   Such personal information includes:   Forename and surname Address information Email address Telephone number   Personal information via WHOIS  Depending on the domain type such as  or  this information may become immediately available after the registration of the domain via a simple whois. If you do not want that (and in most cases you certainly don't) they offer a service where you can remove this information from the whois service but of course hiding your personal information adds yearly costs to the domain  purchase.  What happens if I use fake details when registering a domain? I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$. I’m a big fan of ampersands (possibly too much …) and I’d like to replace all mentions of “et al.” (included in citations) by \&amp; al. – both in the main text (I use natbib citations of the form [Author1 et al., 1999]), and in the actual bibliography?  How can I do that, considering that BibTeX is inserting the words for me, and I have no control over the text? Preferably, I’d like to do this without creating my own bibliography style sheet (since I want to make this work regardless of the actual BibTeX style that is used), but if anyone can tell me how I need to edit the style sheet file, that would be fine as well (makebst doesn’t seem to offer an option for this …). My professor was lecturing today and he made this statement which I was unable to verify.  (I worded it nicer)  There is no map which is both smooth and onto from $S^1$ to $S^1$$\times$ $S^1$.  When he said this he included the original "one can clearly see" statement which is usually when the most people seem to be confused in his class.  Thus, I was hesitant to ask why.  So, please no one lecture me on the fact that I should have asked anyway.  If it is really that obvious, I would rather look stupid here than in class :) Can I send in for repair a lens that has a shattered front element?     UPDATE: The front element was not shattered, it was the UV filter. Factory diagnosis: Lens assembly had been impacted; cleaned and checked.  Helpful notes by Crowley, David Richerby (x2), Michael Clark (x2), and   mattdm (on element price vs UV filter).      Canon cleaned and checked my lens back to like new quality for $179 labor, no parts, not including shipping.      Jim Garrison answered my question, but Crazy Dino has the best info all around for a lens element shatter.   Original Question:  I recently opened my camera bag to find a shattered front element on a Canon EF 16-35mm f/2.8 L II USM lens.   There are a lot of micro glass shards on the second element.  Does anyone have experience with this sort of thing?  Some of the following are questions I have:   Can I get it repaired? What will the cost be roughly? Will the lens ever work the same and have a good image quality? Should I ship it as is, or attempt cleaning it with air? Where is the best place to cry a little?   Thanks, any good info on this would be appreciated.  I will never rest a lens on another lens, I nearly lost a 24-70mm L that shattered a UV filter also.   Good reason to have a UV always, never thought it could happen to me.  Update: Looking at it more it looks like maybe just the front ring came undone and the UV filter broke?  I can't remember what it looked like exactly before.  I threw away the glass shards (polished sand) in the Nevada desert.      UPDATE: Primary reason for thinking more than just the UV was broken,   UV filter split into two rings, depositing the locking collar into the   lens barrel. How do I prove this form of mean value theorem for integral? I only knew the standard mean value theorem for integrals. (i.e. $\int_a^b f(x)dx= f(c)(b-a)$ for some $c$ between $[a,b]$ where $f$ is continuous. This is directly derived by applying mean value theorem and Fundamental theorem of calculus) I'm taking numerical analysis this year and there is one theorem stated without a proof in my text. That is:  Let $f,g$ be continuous real-valued functions on $[a,b]$ such that $g$ is nonnegative. Then there exists some $c$ between $a,b$ such that $\int_a^b fg(x) dx = f(c) \int_a^b g(x) dx$  How do I prove this? Difference between "did + verb" and just "verb+ed" This is something I always get confuse with (native language German). As far as I know "did + verb" is something done and finished in the past. But isn't it the same with "verb + ed"? Slipstream Windows 7 + Service Pack 1