2018 Stockholm Open – Doubles
# 2018 Stockholm Open – Doubles

Oliver Marach and Mate Pavić were the defending champions, but chose not to defend their title.

Luke Bambridge and Jonny O'Mara won the title, defeating Marcus Daniell and Wesley Koolhof in the final, 7–5, $7–6^{(10–8)}.$

## Seeds

* Marc López / Aisam-ul-Haq Qureshi (quarterfinals)
* Jack Sock / Jackson Withrow (quarterfinals)
* Marcus Daniell / Wesley Koolhof (final)
* Ken Skupski / Neal Skupski (quarterfinals)

## Draw

### Key

* Q = Qualifier
* WC = Wild card
* LL = Lucky loser
* Alt = Alternate
* SE = Special exempt
* PR = Protected ranking
* ITF = ITF entry
* JE = Junior exempt
* w/o = Walkover
* r = Retired
* d = Defaulted
* SR = Special ranking

### Draw

| First round | First round | First round | First round | First round | Quarterfinals | Quarterfinals | Quarterfinals | Quarterfinals | Quarterfinals | Semifinals | Semifinals | Semifinals | Semifinals | Final | Final | Final | Final |
| - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - | - |
| 1 | M López A-u-H Qureshi | 6 | 6 | | | | | | | | | | | | | | |
| | A de Minaur J Millman | 4 | 2 | | 1 | M López A-u-H Qureshi | 5 | 4 | | | | | | | | | |
| | J Cerretani S Tsitsipas | $6^4$ | 6 | [9] | | S Bolelli I Dodig | 7 | 6 | | | | | | | | | |
| | S Bolelli I Dodig | $7^7$ | 4 | [11] | | | | | | | S Bolelli I Dodig | 3 | 4 | | | | |
| 4 | K Skupski N Skupski | 4 | $7^7$ | [10] | | | | | | | L Bambridge J O'Mara | 6 | 6 | | | | |
| | F Nielsen J Salisbury | 6 | $6^2$ | [7] | 4 | K Skupski N Skupski | 4 | $7^9$ | [7] | | | | | | | | |
| | L Bambridge J O'Mara | 6 | $7^7$ | | | L Bambridge J O'Mara | 6 | $6^7$ | [10] | | | | | | | | |
| | T Daniel M Marterer | 4 | $6^3$ | | | | | | | | | | | | L Bambridge J O'Mara | 7 | $7^{10}$ |
| | M Ebden R Lindstedt | 6 | 6 | | | | | | | | | | | 3 | M Daniell W Koolhof | 5 | $6^8$ |
| WC | M Eriksson A Göransson | 2 | 4 | | | M Ebden R Lindstedt | $7^7$ | 2 | [9] | | | | | | | | |
| WC | E Ymer M Ymer | 4 | 6 | [4] | 3 | M Daniell W Koolhof | $6^3$ | 6 | [11] | | | | | | | | |
| 3 | M Daniell W Koolhof | 6 | 4 | [10] | | | | | | 3 | M Daniell W Koolhof | $7^7$ | $7^9$ | | | | |
| | A Behar J Nedunchezhiyan | 0 | 6 | [3] | | | | | | | J Benneteau L Pouille | $6^5$ | $6^7$ | | | | |
| | J Benneteau L Pouille | 6 | 4 | [10] | | J Benneteau L Pouille | 4 | 6 | [12] | | | | | | | | |
| | F Martin F Škugor | $6^{12}$ | $7^8$ | [8] | 2 | J Sock J Withrow | 6 | 2 | [10] | | | | | | | | |
| 2 | J Sock J Withrow | $7^{14}$ | $6^6$ | [10] | | | | | | | | | | | | | |
