conditional

问题 I am trying to come up with a query to report revenue. It will require 2 tables: clicks and offers. Revenue is calculated by the number of conversions * commission for the offer. Conversions are stored in the clicks table in a field called "conversionDate", and the commission for each offer is stored in the offers table. There needs to be a condition in the query to ignore any clicks that did not convert (meaning conversionDate is NULL) when adding up revenue for the offer. What I've got

问题 I am trying to make a bookmarklet that when clicked will check the URL of the current tab/window to see if it contains 'char1' and/or 'char2' (a given character). If both chars are present it redirects to another URL, for the other two it will append the current URL respectively. I believe there must be a more elegant way of stating this than the following (which has so far worked perfectly for me) but I don't have great knowledge of Javascript. My (unwieldy & repetitive) working code

问题 I asked this question, and it turned out that when removing an attribute from an element, checking whether the element exists first using elem.xxx!==undefined makes the runtime faster. Proof. Why is it quicker? There's more code to go through and you'll have to encounter the removeAttribute() method whichever way you go about this. 回答1: Well, first thing you need to know is that elem.xxx is not the same as elem.getAttribute() or any other method relative to the attribute. elem.xxx is a

问题 Site: http://clientfiles.priworks.com/lgmanagedportfolios/test/investment.html I have been working at this web page for a while now trying to get the left side of the first submenu link to line up with the left of the 1st parent/main menu link. I don't think there is a logical way of doing it but after much fudging I am close to getting it just right. The position is consistent for all win browsers and all Mac browsers (with some conditional CSS) but not for Win and Mac browsers together, for

问题 Site: http://clientfiles.priworks.com/lgmanagedportfolios/test/investment.html I have been working at this web page for a while now trying to get the left side of the first submenu link to line up with the left of the 1st parent/main menu link. I don't think there is a logical way of doing it but after much fudging I am close to getting it just right. The position is consistent for all win browsers and all Mac browsers (with some conditional CSS) but not for Win and Mac browsers together, for

问题 I have a data frame that is 200 rows by 6 columns. I am interested in computing the total times that a value in Col A is less than a specific number. The number can be hard coded. I do not know where to begin... 回答1: For a slightly more complex problem, use the "which" to tell the "sum" where to sum: if DF is the data frame: Ozone Solar.R Wind Temp Month Day 1 41 190 7.4 67 5 1 2 97 267 6.3 92 7 8 3 97 272 5.7 92 7 9 Example: sum the values of Solar.R (Column 2) where Column1 or Ozone>30 AND

问题 So I see that this question has already been asked, however the answers were a little vague and unhelpful. Okay, I need to implement a c expression using only "& ^ ~ ! + | >> <<" The expression needs to resemble: a ? b : c So, from what I've been able to tell, the expression needs to look something like: return (a & b) | (~a & c) This works when a = 0, because anding it with b will give zero, and then the or expression will return the right side, (~a & c) which works because ~0 gives all ones

问题 So I see that this question has already been asked, however the answers were a little vague and unhelpful. Okay, I need to implement a c expression using only "& ^ ~ ! + | >> <<" The expression needs to resemble: a ? b : c So, from what I've been able to tell, the expression needs to look something like: return (a & b) | (~a & c) This works when a = 0, because anding it with b will give zero, and then the or expression will return the right side, (~a & c) which works because ~0 gives all ones

问题 Sometimes I see chunks of Scala code, with several nested levels of conditionals and matchings, that would be much clearer using an explicit return to exit from the function. Is there any benefit in avoiding those explicit return statements? 回答1: A return may be implemented by throwing an exception, so it may have a certain overhead over the standard way of declaring the result of a method. (Thanks for Kim Stebel for pointing out this is not always, maybe not even often, the case.) Also, a

问题 This question already has answers here : Dynamically access object property using variable (13 answers) Closed 2 years ago . function removeNumbersLargerThan(num, obj) { for (var key in obj) { if (!isNaN(obj[key]) && obj[key] > num) { delete obj.key; } } return obj; } var obj = { a: 8, b: 2, c: 'montana' } removeNumbersLargerThan(5, obj); console.log(obj); // Should be {b: 2, c: 'montana'} The function should remove any property that meets the 'if' condition inside the 'for' loop, but it