excel - adding wildcard to look up formula? -

-

Look at the function I'm using the following formula as part of a multi-result. Many of these threads have been posted in one row below each other, so if I write Catering on the sheet 1 in cell K22 then my formula Matching word appears in column, produces name in sheet 2 on b and column a . In the formula to duplicate this lookup some times to get other values ​​from other columns. Similar matching results

For example, on my sheet 2 my columns look like: 

  Name of the location Place number Amy Hotel London 1 Dave Hotel Manchester 2 Mike Catering Birmingham 3

I have to use wild cards around my cell K22 in this formula to show that if I describe the type of hotel I Is like; I.e. hotels, hotels ... etc ... get results? ... I know that this index match can be used, but can I use them with the following code? At the moment it is an array formula, but I do not get any results when I try to do it like this. Tanks = IF (ISERROR (Index $ 7 = "*" & Sheet1 $ K $ 22 & "*", Line ($ B $ 1: $ B $ 7) !), Line (1: 1)), 1)), "", index (Sheet2 $! A $ 1: $ D $ 7, small (IF (Sheet2 $ B $ 1: $ B $ 7 = "*" & Amp; Sheet1 $ K $ 22 & "*", Line ($ B $ 1: $ B $ 7), Line (1: 1)), 1)) & amp; "-" and IF (Easter (index (sheet 2! $ A $ 1: $ D $ 7, small (if (sheet 2! $ B $ 1: $ B $ 7 = "*" and Sheet1! $ K $ 22 & amp; "*", line ($ B $ 1: $ B $ 7), line (1: 1)), 3)), "index" (Sheet2 $ A $ 1: $ D $ 7, Small (IF (Sheet2 $ B $ 1: $ B $ 7 = "*" and Sheet1! $ K $ 22 & amp; "*", ROW ($ B $ 1: $ B $ 7)), ROW (1 : 1)), 3)) & amp; "-" and IF (Easter (index (sheet 2! $ A $ 1: $ D $ 7, small (if (sheet 2! $ B $ 1: $ B $ 7 = "*" and Sheet1! $ K $ 22 & amp; "*", line ($ B $ 1: $ B $ 7), line (1: 1)), 4)), "index" (Sheet2 $ A $ 1: $ D $ 7, Small (IF (Sheet2 $ B $ 1: $ B $ 7 = "*" and Sheet1! $ K $ 22 & amp; "*", ROW ($ B $ 1: $ B $ 7)), ROW (1 : 1)), 4 <

P>

function (and in other contexts), you can not use them here. When you have = Using wildcard with direct comparison, i.e.

Sheet2! $ B $ 1: $ B $ 7 = "*" and Sheet1! $ K $ 22 & "*"

* Not a Wild Card , Is interpreted in the form of a literal star

replace all the above examples with

ISNUMBER (SEARCH (sheet1! $ K $ 22, sheet 2! $ B $ 1: $ B $ 7))


Comments

Post a Comment

Popular posts from this blog

c# - SignalR: "Protocol error: Unknown transport." when navigating to hub -

-
Image
I am trying to use the ASP.NET signal for the first time - here's what I did: created a new web Has been added to the project and the following signal R packages: I have created the following category: [Hub-name ("chat")] Public class ChatHub: Hub {Public Zero Halo () {Clients.All.hello ( ); }} and the following startup configuration: public partial class startup {public zero configuration (IAppBuilder app) {configure (app); App.MapSignalR (); }} However, whenever I navigate localhost: (port) / api / signalr / chat I get the following error: Protocol error: Unknown transport. I am running a new browser and trying different people so remove it from the API / URL, and just use the localhost: (potter) / signaller / Chat
Read more

class - Kivy: how to instantiate a dynamic classes in python -

-
Using KV lang on my Python code, I have taken a hard time trying to instantiate a dynamic class, the following Consider the code: My test.kv file looks like this: & lt; MyPopup @ popup: Title: 'Hello' size_hunt: (1, .6) Grid layout: ID: RouteGrid column: 2 padding: ['8dp', '4dp', '8dp', '4dp'] Difference: '8dp' label: text: 'some text here' button: text: 'OK' On_press: do_something () & lt; MyGrid & gt;: Rows: 1 Button: Text: 'Show Popup' on_press: root.pop.show () Then import Kivy.app into my test.py: I get the following error: raise Factory Exception ('Uncount Class' '% name) kivy.factory.FactoryException: Unknown class Can anyone tell me how to do this properly, what do I remember? If you need more information please let me know Thank you. Takes a call to the factory before the KV file is loaded, so you want the class to Are not yet in existence. As long as no clas...
Read more

svn - Cannot merge back into trunk, must be ancestrally related. But they *are* related as far as I understand -

-
Image
So I'm trying to merge back into my trunk, but I'm getting errors saying that I Because they are not related to the ancestors but I am not clear why this is why. Here is the graph generated by TartiSesWN The trunk is at the bottom left (14698) Currently, I am starting with the merger in the trunk (15 9 7 9) from the top right, this is an example which claims that it is not related to the ancestors, some disparities in it That is what a coworker presented, and I'm not sure it could be the reason for this. If you notice, in the right middle, the branch with modification number 15406 actually divides a tag. It seems incredibly wrong to me, but I'm not sure. A few months ago the coworker did this (he does not work here anymore). However, they should be related to all ancestors because they come from all trunk, right? If this is true, then why am I getting this error? Will a tag be divided? The head of the trunk is 16063 (as per the figure), 14698 merge-s...
Read more