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

java - Algorithm negotiation fail SSH in Jenkins -

-
I am trying to do SSH on the local server from Jenkins, but the following error has been thrown: [SSH] Exception: Algorithm negotiation fail com.jcraft.jsch.JSchException: algorithm talk com.jcraft.jsch.Session.receive_kexinit (Session.javaatar20) at com.jcraft.jsch.Session.connect (Sesson at com.jcraft.jsch.Session.connect (Session.java:150) at org.jvnet.hudson.plugins.SSHSite.createSession (SSHSite.java:141) on failed failed org.jvnet.hudson.plugins. java: 286) .SSHS on org.jvnet.hudson.plugins.SSHBuildWrapper.executePreBuildScript (SSHBuildWrapper.java:75) on Org.jvnet.hudson.plugins.SSHBuildWrapper.setUp (SSHBuildWrapper.java59) ite.executeCommand (SSHSite.java151) on hudson.model.Build $ BuildExecution.doRun hudson.model.AbstractBuild $ AbstractBuildExecution.run (AbstractBuild.java:533) to (Build.java:154) hudson.model.Run.execute ( Run.java:1754) in ended .model.FreeStyleBuild.run (FreeStyl hudson.model.ResourceController.execute (ResourceController.java:89) on hudson.mod...
Read more