Substring in String test, like Python: if "this" in thisString

[Ross]

Does anyone have a suggestion for a nice compact test for a substring
in a string? I'm porting some stuff from Python where I can just test
easily with

a="here are some words"
if "are" in a:
print "Yes it's There"

Something tells me a regex is a likely approach...

Thx in advance...
Ross.

 

[Dmitry A. Soshnikov]

Does anyone have a suggestion for a nice compact test for a substring
in a string?  I'm porting some stuff from Python where I can just test
easily with

  a="here are some words"
  if "are" in a:
     print "Yes it's There"

Something tells me a regex is a likely approach...

Thx in advance...
Ross.

'here are some words'.indexOf('are') != -1 && alert('Yes it's There');

 

[Stevo]

Does anyone have a suggestion for a nice compact test for a substring
in a string? I'm porting some stuff from Python where I can just test
easily with

a="here are some words"
if "are" in a:
print "Yes it's There"

Something tells me a regex is a likely approach...

Thx in advance...
Ross.

if ( a.indexOf("are") > -1 )
alert("Yes, it's there");

 

[Ross]

Thank you Dmitry and Stevo! That looks great, I can live with that.

Ross.

 

[Thomas 'PointedEars' Lahn]

Does anyone have a suggestion for a nice compact test for a substring
in a string? I'm porting some stuff from Python where I can just test
easily with

a="here are some words"
if "are" in a:
print "Yes it's There"

Something tells me a regex is a likely approach...

You are on the right track:

if (/are/.test(a))
{
console.log("Yes it's There");
}

The presented solution using String.prototype.indexOf() is probably more
efficient with simple expressions but less flexible. For example, given
text that contains only characters that are also in the ASCII character set,
you can search for a word (here: are):

if (/\bare\b/.test(a))
{
console.log("Yes it's There");
}

And you can search case-insensitive:

if (/\bare\b/i.test(a))
{
console.log("Yes it's There");
}

You cannot do this as easily and as efficiently without Regular Expressions.


PointedEars