SearchObject | In many of my projects I needed an object that performs | REST library

Is the folder index being kept just for the reason of ordering the result set by folder path? If that's the case, why not keep full folder paths as a sortable field in the original index? This way you'll be able to order the result set by folder paths, assuming the folder path order you want is alphabetical.

For example:

Doc1: “field1”

Doc2: ”field1”

Doc3: “field1\subfield11\subfield111”

Doc4: ”field2”

The member access operators (. and []) are really high precedence. await is fairly high, but not that high. (In fact, only the grouping operator [()] has higher precedence than member access.) await in your code applies to the result of findOne.

You can see the precedence and associativity of operators on MDN.

Here's my take on the Fetch API,

What you want here, is a so called semi-join in relational algebra which is modeled through the EXISTS predicate in SQL and also supported in JPA. You have to use a correlated subquery with an exists predicate. Something along the lines of the following:

I found the solution to the problem myself. I dropped the options part completely and went with the following approach which works just as intended:

in Kusto there are several helper methods you can use. In your case parse_url can be helpful as it will contain an array of query parameters that can be easily accessed:

There was essentially 3 things wrong with your code

1. displayResult dis not take any parameter yet you were passing the result to display to it - add an argument (res below) and change this to res
2. The loop where you were trying to display the result was not set up correctly. The loop was doing for (i = 0; i = tempList.length; i++) change to for (i = 0; i < tempList.length; i++).
3. You were not clearing the table between searches so I added sResultTable.innerHTML = ""; before displaying results.

There are a lot of other improvements which could be made to your code which I've skipped for now - that will get you going.

In addition, I don't know what the repeated loops were for so I removed them for the sake of brevity to show a working example below:

1. There is nothing wrong with the Java versions.
2. Your application might be having multiple Windows come up before the desired window has come, so he lost his top level component.

Just use switch to window method and pass tile before findElements is called. This should solve the issue.

The reason this is happening is that python runs the regex until it stops matching. Currently, it's actually matching once for every character and keeping it if it's a not a space because the .*? is matching nothing and it moves to the next character. The reason that lazy operator is matching nothing is because there's nothing forcing it to, so it chooses the laziest option it can: doing no work.

For example, here's what the matches look like currently: https://regex101.com/r/AzOrJZ/2

You can get what it looks like after a single match by following these directions:

How can I substitute a regex only once in Python?

In order to improve performance, here's the regex I'd suggest for performance and fewer matches:

\s*(\S+(?:\s+\S+)*)\s*

This essentially matches all leading whitespace characters \s+ then groups all non-whitespace \S+ followed by one or more whitespace with non-whitespace (\s+\S+)*. Once all the non-whitespace characters are matched, then it matches the trailing whitespace.

https://regex101.com/r/AzOrJZ/4