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Here is the command to print the current version of npm
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npm -v
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For example
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npm view redux version
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3 years ago
Use the below code
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async function printFiles () {
const files = await getFilePaths()

for await (const contents of files.map(file => fs.readFile(file, 'utf8'))) {
console.log(contents)
}
}
***
3 years ago
NVM (Node version manager) is a portable solution for managing multiple versions of node
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nvm uninstall v4.1.0
> nvm install v8.1.2
> nvm use v8.1.2
> nvm list
v4.2.0
v5.8.0
v6.11.0
-> v8.1.2
system
***
3 years ago
One of the way to get the solution to the above question is
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numpy.arange(100).reshape(25,4).tolist()
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3 years ago
That's the normal structure for ngClass is:
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[ngClass]="{'classname' : condition}"
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just use this
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***
3 years ago
You could use **url_for(only_path: false)**
3 years ago
Use rand(range)
If you need percentage then just use rand()
If you need an integer value from 1 to x, not including x, call rand(x).
If you need a number in a specific range, call rand(x,y) where x is the lower range and y is upper range
3 years ago
The "use strict" mode instructs the browser to use strict mode which is a reduced safe feature of javascript.
Some features of strict mode are
Disallows global variable
All property name in the object should be unique.
Forbid the **With** keyword
3 years ago
first, check for the latest version
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npm -v
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Now update it to the latest version
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npm install -g npm@latest
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3 years ago
There are only five:

" "
' '
< <
> >
& &

Escaping characters depends on where the special character is used.
For Text

The safe way is to escape all five characters in text. However, the three characters ", ' and > needn't be escaped in text:
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"'>
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Attributes
The safe way is to escape all five characters in attributes. However, the > character needn't be escaped in attributes:
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***
The ' character needn't be escaped in attributes if the quotes are ":


3 years ago
Use awk for these purposes, as it allows you to perform more complex checks.

Lines not containing foo:
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awk '!/foo/'
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Lines containing neither foo nor bar:
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awk '!/foo/ && !/bar/'
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Lines containing neither foo nor bar which contain either foo2 or bar2:
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awk '!/foo/ && !/bar/ && (/foo2/ || /bar2/)'
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3 years ago
If you want the regex test to only fail if the entire string matches, the following will work:
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^(?!hede$).*
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For exact matching use this
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myStr !== 'foo'
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3 years ago