Brachylog (v2), 3 9 bytes

{sᵛ}ᶠlᵒtw

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

Explanation

{sᵛ}ᶠlᵒtw

 s         Find a substring

  ᵛ          of every element {of the input}; the same one for each

{  }ᶠ      Convert generator to list

     lᵒt   Take list element with maximum length

        w  Output it

Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)

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 2, 82 bytes

f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)

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Takes input splatted. Will time out for inputs where the first string is long.

The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.

Python 2, 94 bytes

lambda l:max(set.intersection(*map(g,l)),key=len)

g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}

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A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.

Jelly, 12 6 bytes

Ẇ€f/ṫ0

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Thanks to @JonathanAllan for saving 6 bytes!

Ruby 2.6, 76 59 54 bytes

->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}

Try it online! - Ruby 2.5 version (56 bytes)

How?

Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.

Thanks Kirill L for -2 bytes and histocrat for another -2

R, 119 116 108 106 bytes

function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]

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Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.

-3 bytes thanks to Kirill L.

-8 bytes using lapply instead of Map

-2 bytes thanks to Kirill L. again, removing braces

05AB1E, 14 9 8 bytes

€Œ.«ÃéθJ

-6 bytes thanks to @Adnan.

Try it online or verify all test cases.

Explanation:

€Œ       # Get the substring of each string in the (implicit) input-list

  .«     # Right-reduce this list of list of strings by:

    Ã    #  Only keep all the strings that are present in both list of strings

     é   # Sort by length

      θ  # And pop and push its last item

         # The substrings exclude empty items, so if after the reduce an empty list remains,

         # the last item will also be an empty list,

       J # which will become an empty string after a join

         # (after which the result is output implicitly)

Zsh, 126 123 117 112 111 bytes

-3 bytes from arithmetic for, -6 bytes from implicit "$@" (thanks roblogic), -5 bytes from removing unneeded { }, -1 byte from short form of for

for l

{a= i=

for ((;i++<$#l**2;))a+=($l[1+i/$#l,1+i%$#l])

b=(${${b-$a}:*a})}

for s ($b)

(($#x<$#s))&&x=$s

<<<$x

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We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.

for l; {                            # implicit "$@"

    a= i=                           # empty a and i

    for (( ; i++ < $#l**2; ))       # compound double loop using div/mod

        a+=( $l[1+i/$#l,1+i%$#l] )  # append to a all possible substrings of the given line

#               1+i/$#l             # 1,1,1...,  1,1,2,2,2,... ...,  n,n

#                       1+i%$#l     # 1,2,3...,n-1,n,1,2,3,... ...,n-1,n

#       a+=( $l[       ,     ] )    # append that substring to the array

    b=( ${${b-$a}:*a} )

#         ${b-$a}                   # if b is unset substitute $a

#       ${       :*a}               # take common elements of ${b-$a} and $a

#   b=(               )             # set b to those elements

}

for s ($b)                          # for every common substring

    (( $#x < $#s )) && x=$s         # if the current word is longer, use it

<<<$x                               # print to stdout

Perl 6, 62 bytes

{~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}

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I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.

Haskell, 80 bytes

import Data.List

f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]

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Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.

TSQL query, 154 bytes

USE master

DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)

INSERT @ values('string'),('stRIng');



SELECT top 1x FROM(SELECT

distinct substring(a,f.number,g.number)x,i

FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D

GROUP BY x ORDER BY-sum(i),-len(x)

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Made case sensitive by declaring the column 'a' with collation containing CS (case sensitive).

Splitting all strings from 2540 starting positions(many identical) but the useful values range between 1 and 2070 and ending 0 to 22 characters after starting position, the end position could be longer by changing the type to 'P' instead of 'L', but would cripple performance.

These distinct strings within each rownumber are counted. The highest count will always be equal to the number of rows in the table variable '@'. Reversing the order on the same count will leave the substring with most matches on top of the results followed by reversed length of the substring will leave longest match with most matches on top. The query only select the top 1 row.

In order to get all answers, change the first part of the query to

SELECT top 1with ties x FROM

C# (Visual C# Interactive Compiler), 320 257 bytes

l=>(string.Join(",",l.Select(s=>new int[s.Length*s.Length*2].Select((i,j)=>string.Concat(s.Skip(j/-~s.Length).Take(j%-~s.Length))))

.Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x =>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));

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Props to @Expired Data and @dana

Japt -h, 8 bytes

(I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)

mã rf iP

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mã rf iP     :Implicit input of array

m            :Map

 ã           :  Substrings

   r         :Reduce by

    f        :  Filter, keeping only elements that appear in both arrays

      i      :Prepend

       P     :  An empty string

             :Implicit output of last element

Japt v2.0a0 -hF, 8 bytes

Îã f@eøX

Thanks to Shaggy for saving 3 bytes

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Îã              //Generate all substrings of the first string

 f@             //Filter; keep the substrings that satisfy the following predicate:

   e            //    If all strings of the input...

    øX          //    Contain this substring, then keep it

-h              //Take last element

-F              //If last element is undefined, default to empty string

Retina 0.8.2, 48 bytes

M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)

O#$^`

$.&

1G`

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M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)

For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.

O#$^`

$.&

Sort the substrings in reverse order of length.

1G`

Keep only the first, i.e. the longest substring.

Python 3, 137 bytes

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

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Python 2, 103 bytes

lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)

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This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.

C# (Visual C# Interactive Compiler), 147 145 bytes

a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}

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Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes

may be golfed more certainly

map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h

TIO

JavaScript (ES6),  98  92 bytes

a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])

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Bash 4+, 295... 178 bytes

Not pretty but at least it works.
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-37 by general cleaning up ;
-52 by plagiarising from the zsh answer ;
-26 by replacing array with a loop ;
-2 thanks to GammaFunction

for l;{ d=${#l}

for((i=0;i<d**2;i++)){ a="${l:i/d:1+i%d}" k=

for n;{ [[ $n =~ $a ]]&&((k++));}

((k-$#))||b+=("$a");};}

for e in "${b[@]}";do((${#e}>${#f}))&&f="$e";done

echo "$f"

Here's the original ungolfed script with comments

Java (JDK), 176 bytes

a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}

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This is a rather naive implementation.

Red, 266 174 bytes

func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u

next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]

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Changed the recursion to iteration and got rid of the sorting.

Perl 5, 87 bytes

my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r

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JavaScript (Node.js), 106 bytes

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

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a=>(                      // Main function

 F=(                      //  Helper function to run through all substrings in a[0]

  l,                      //   Length

  n,                      //   Start position

  w=a[0].substr(n,l)      //   The substring

 )=>

 l?                       //   If l > 0:

  n<0?                    //    If n < 0:

   F(--l,L-l)             //     Check another length

  :a.some(                //    If n >= 0: 

   y=>y.indexOf(w)<0      //     Check whether there is any string not containing the substring

                          //     (indexOf used because of presence of regex special characters)

  )?                      //     If so:

   F(l,n-1)               //      Check another substring

  :w                      //     If not, return this substring and terminate

                          //     (This function checks from the longest substring possible, so

                          //      it is safe to return right here)

 :""                      //   If l <= 0: Return empty string (no common substring)

)(

 L=a[0].length,           //  Starts from length = the whole length of a[0]

 0                        //  And start position = 0

)

Gaia, 15 bytes

eḋ¦&⊢⟨:l¦:⌉=¦⟩∇

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e		| eval as code

 ḋ¦		| find all non-empty substrings

   &⊢		| Reduce by set intersection

              ∇	| and return the first element where

      ⟨:l¦:⌉=¦⟩	| the length is equal to the max length

PowerShell, 165 163 87 bytes

-76 bytes thanks to Nail for the awesome regexp.

"$($args-join'

'|sls "(?<=^.*)(.+)(?=(.*n.*1)*.*$)"-a -ca|% m*|sort Le*|select -l 1)"

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Less golfed:

$multilineArgs = $args-join"`n"

$matches = $multilineArgs|sls "(?<=^.*)(.+)(?=(.*n.*1)*.*$)" -AllMatches -CaseSensitive|% matches

$longestOrNull = $matches|sort Length|select -Last 1

"$longestOrNull"

Charcoal, 30 bytes

≔⊟θη≔⁰ζＦＬη«≔✂ηζ⊕ι¹ε¿⬤θ№κεＰε≦⊕ζ

Try it online! Link is to verbose version of code. This algorithm is more efficient as well as shorter than generating all substrings. Explanation:

≔⊟θη

Pop the last string from the input list into a variable.

≔⁰ζ

Zero out the substring start index.

ＦＬη«

Loop over all possible substring end indices. (Actually this loops from 0 excluding the length, so the value is adjusted later.)

≔✂ηζ⊕ι¹ε

Obtain the current substring.

¿⬤θ№κε

Check whether this substring is contained in all of the other input strings.

Ｐε

If it is then overprint any previously output substring.

≦⊕ζ

Otherwise try incrementing the substring start index.

JavaScript (Node.js), 90 bytes

f=(a,s=e=0,w='')=>a[0][e]?f(a,s+(r=a.some(x=>!x.includes(t),t=a[0].slice(s,++e))),r?w:t):w

Try it online! Test cases shamelessly stolen from @Arnauld. Port of my Charcoal answer.