How to address second-order SQL injection without using PDO or mysqli binding in PHP application. I did many research but all point to PDO which I do not have. mysql_real_escape_string PHP escape is just for protection against the quotes. Right now we are having legacy PHP which using mysql_connect. We cannot upgrade our system to use PDO or MySQLi since it would require dependencies installation. To over come injection we using escape. But this does not gives protection against second order SQL injection. $name = $_POST['name']; $user = &quot;Select user from user where name = $name&quot; now use $id to fetch user location. Note the current code does not do join as this is only example. $loc = &quot;Select * from location where user = $user&quot; currently we are using escape for $name and $user to prevent injection. But the issue is if query from first one contain SQL query then we have issue. So we need a way to escape the query that way second query will be safe
If user input is inserted without modification into an SQL query, then the application becomes vulnerable to , like in the following example:  $unsafe_variable = $_POST['user_input'];   mysql_query("INSERT INTO `table` (`column`) VALUES ('$unsafe_variable')");   That's because the user can input something like value'); DROP TABLE table;--, and the query becomes:  INSERT INTO `table` (`column`) VALUES('value'); DROP TABLE table;--')   What can be done to prevent this from happening?
Problem: Let $f$ be defined for all real $x$, and suppose that   $$|f(x)-f(y)|\le (x-y)^2$$  for all real $x$ and $y$. Prove $f$ is constant.  Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 1. login and payoff are nouns. But can they be used as verbs? I know that words like login and payoff are properly nouns but I increasingly notice many (not at all uneducated) people use them as verbs:     Will you payoff your credit card this month?   and     Please check this when you login.   To me (a non-native English user), that seems incorrect because separating the core noun from a preposition is what I was taught: Please check this when you log in. (My emphasis)  Is this use proper or not? What is the plural of "cottage tariff"? What software works well for viewing massive TIFF images on Windows 7? $M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Rounded caps on a Bezier Curve pipe This may be a simple question.  Essentially I am procedurally generating random 3D figures using a bezier curve and bezier circle bevel object, however, one of the only things I can't figure out how to do is how to round the end caps of the object.   I have tried using a taper object like suggested  but have had only limited success.  This is about as close as I've gotten:   I should note that all of this is done through Python and I'm not modeling any of this by hand. So any suggestions for how I can make the rounded caps programmatically would be greatly appreciated. I'm in trouble with time dilation: Suppose that there's two people on the Earth (A,B), they are twins and each other has a clock. (So they are at the same reference frame). B travels in a spaceship and is orbiting around the Earth, as B's speed has increased there's some time dilation. So it's supposed that if A "looks" to B, B's clock will run slower than A's, and viceversa. And if one day B decides come back to the Earth, he will be younger than his twin and the hour of his clock would be different (earlier than A's clock) because of time dilation.  My first question is, why each person see the other's clock as the slow one? (In fact, I found some information that may solved it but I'm not sure if it's the right answer, and if it is, I don't understand it. -> It's because everyone thinks about himself as the reference frame. In other words, that for B, he is the stationary and who moves is A, so he sees A's clock run slower as B is "stopped" and the same for A. B is who is moving and A the stationary.)  The other question I have is that if it's true that each one sees the other as the slow one there's something that i missed out. We all agree that B (after going to the space and came back to the Earth) he is younger than his brother. So, why B don't see A moving and aging very fast if A see his brother and his things going very slow? Let's imagine they were looking to each other all the time, if both of them see the other going slowly once they meet and "discover" B is the young one... how is that possible?  [I think that in one episode of Cosmos of Carl Sagan (though may be I'm mistaken) he said that a neutrino "borned" in the Big Bang could have seen the creation of the Universe until today in few seconds due to its high speed, here starts my doubts: or i misunderstood something or there's contradictory information]  (I don't know anything about physics, only what is taught at high school as I'm a teenager so it'd be better to answer with no kind of calculation as I'd not understand it.) Regular Expressions: Is there an AND operator? Obviously, you can use the | (pipe?) to represent OR, but is there a way to represent AND as well?  Specifically, I'd like to match paragraphs of text that contain ALL of a certain phrase, but in no particular order. Why are objects at rest in motion through spacetime at the speed of light? $1 \leq z \overline {z} \leq 4$ and $|\Im(z)|&lt;\Re (z)$  How can I solve this system of inequalities? ($\Im$ is the imaginary part and $\Re$ is the real part of a complex number).  I have tried so far: $z=x+yi$ and then I get $1 \leq x^2+y^2 \leq 4$ and $ |y|&lt;x$. How can I solve this? An inequality for a sum of sine functions I want to show for $x\in\mathbb{R}$, the following inequality holds. $$ \left|\sin x\right|+\left|\sin (x+1)\right|+\left|\sin (x+2)\right|&gt;8/5. $$ This trigonometric inequality has been verified by Mathematica using the Plot commend. However, using the sum formula for the sine function seems doesn't work, and I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks! I am using Snow Leopard and Windows 7 with bootcamp. I would like to be able to save some files to my OSX partition from Windows. I would also like to have files (say, photos, source code, etc) that I can manage with both partitions. Is it possible to get the two partitions to "play nice"? concurrentModificationException Simple question about the definition of Brownian motion I have a question concerning the definiton of Brownian motion. Usually (e.g. on Wikipdia) one demands a brownian motion $\lbrace B_t\rbrace_{t\in[0,\infty)}$ to satisfy the following condition:  $\forall 0\leq t_0&lt;t_1&lt;...&lt;t_m: B_{t_1}-B_{t_0},...,B_{t_m}-B_{t_{m-1}}$ are independent random variables.   But today I come across with formally another definition of Brownian motion! Instead of the above condition the book demands the following property:  $\forall t,h\geq 0$, $B_{t+h}-B_t$ is independent of $\lbrace B_u : 0\leq u\leq t \rbrace$.  Now I'm wondering whether both properties are indeed equivalent or not?! Hopefully someone can help me. How to delete (invalid) files with colon in their name under Windows? Why is it possible to conclude everything from a false statement? Possible Duplicates:             I heard a professor of mathematics explaining that the following conslusions are true.   $\textrm{False}\Rightarrow \textrm{True}$ $\textrm{False}\Rightarrow \textrm{False}$   Could someone please explain to me why both conclusions are true? I am especially interested in the first case. I'd like to hear two explanations if possible: One for mathematicians and one for non-mathematicians. :-) Resize multiple layers around their respective anchor points How to quotient $\mathbb Z [\sqrt{-11}] / (1+\sqrt{-11})$? That's it. I want to do this quotient,  $\mathbb Z [\sqrt{-11}] / (1+\sqrt{-11})$. My first idea was to see which elements are in the ideal:  $(a+b\sqrt{-11})(1+\sqrt{-11}) = a+ a\sqrt{-11} + b\sqrt{-11} -11b$  So I want to make and application who send this elements to $0$, and then see the Kernel of the application and apply the Isomorphism Theorem.  Any ideas? When we measure the speed of a moving element we do it with the help of a reference frame. Now if we need to measure the speed of time, is it possible? Does time actually have a speed? LaTeX seems to be having problems handling nested optional arguments. I've included a minimal code example that generates the error:  \documentclass[12pt,oneside]{book} \begin{document}  \newcommand{\termbad}[2][]{#2#1} \newcommand{\termgood}[2][]{#2#1}  \newcommand{\up}[1]{#1} \newcommand{\dd}[1][]{#1}  \termgood{\up{Tart}}{Pop} \termbad[\up{Tart}]{Pop}      \termgood{\dd[Tart]}{Pop} \termbad[{\dd[Tart]}]{Pop} %\termbad[\dd[Tart]]{Pop}   \end{document}   If you run this code, it works. If you remove the % in the second to last line, it fails with:  Runaway argument?  Tart{Pop} \par \par \end {document}  !File ended while scanning use of \dd.   After further tinkering, I have discovered the problem: LaTeX is reading the ] for the inner optional argument as the end of the outer optional argument and then passing a malformed argument in. So it is passing in \dd[Tart as the optional argument to \termbad instead of \dd[Tart].  This still feels like a bug to me. Nesting braces in arguments works, brackets should as well. The following are workarounds:   Enclose the inner optional argument in braces (as in the line before %). Redefine your inner function to require its argument instead of using optional arguments. Redefine your outer function to require its argument instead of using optional arguments.   Any one of the three works around the issue.  (Reposted here; question was originally posted to stackexchange and then closed instead of being migrated.)