Create a program or function which takes a list of strings as input, and outputs the longest string that is a substring of all input strings. If there are several substrings of equal length, and no longer substring, output any one of them.

• This may mean outputting the empty string.


• If there are several valid outputs, you may output any one of them. You are not required to give consistent outpput for a given input so long as the output is always valid.


• There will always be at least one string in the input, but there might not be a non-empty string.


• All printable ASCII characters may appear in the input. You may assume those are the only characters that appear.


• You may take input or produce output by any of the default methods.


• 
  Standard loopholes aren't allowed.


• This is code-golf - the fewer bytes of code, the better.

Test cases:

[Inputs] -> [Valid outputs (choose one)]

["hello", "'ello"] -> ["ello"]
["very", "much", "different"] -> [""]
["empty", "", "STRING"] -> [""]
["identical", "identical"] -> ["identical"]
["string", "stRIng"] -> ["st", "ng"]
["this one", "is a substring of this one"] -> ["this one"]
["just one"] -> ["just one"]
["", "", ""] -> [""]
["many outputs", "stuptuo ynam"] -> ["m", "a", "n", "y", " ", "o", "u", "t", "p", "s"]
["many inputs", "any inputs", "ny iii", "yanny"] -> ["ny"]
["%%not&", "ju&#st", "[&]alpha_numeric"] -> ["%"]

Jelly, 12 bytes

Ẇ€œ&/LÐṀḢ¹L?

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Last four bytes are there because of the requirement to only output one answer.

JavaScript (Node.js), 114 105 bytes

a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.search(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)

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Probably still golfable.
y.indexOf must be used in place of y.search if the strings may contain special regex characters, at the cost of 1 more byte.

Brachylog (v2), 3 bytes

sᵛw

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Full program. Input from standard input (as a JSON-style list of strings), output to standard output.

Explanation

sᵛw
s Find a substring
 ᵛ of every element of the input; the same one for each
 w and output it.

Tiebreak order here is set by the s, favouring the longest substring (the secondary tiebreak doesn't matter, but IIRC it's position within the first element of the input).

Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.

Python 3, 174 bytes

def a(b):
 a=();c=list(sum((list(d[f:e]for f in range(e+1))for e in range(len(d)+1)),[])for d in b)
 for i in c[0]:
 if all(i in j for j in c):a+=(i,)
 return max(a,key=len)

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