diff --git a/.prettierrc b/.prettierrc
index dece4466002ba..90ec8297e71b4 100644
--- a/.prettierrc
+++ b/.prettierrc
@@ -29,7 +29,8 @@
                 "solution/1500-1599/1555.Bank Account Summary/Solution.sql",
                 "solution/1600-1699/1667.Fix Names in a Table/Solution.sql",
                 "solution/1900-1999/1972.First and Last Call On the Same Day/Solution.sql",
-                "solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql"
+                "solution/2600-2699/2686.Immediate Food Delivery III/Solution.sql",
+                "solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql"
             ],
             "options": {
                 "parser": "bigquery"
diff --git a/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README.md b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README.md
index 1a2757ba6178c..b6985dc7f1a63 100644
--- a/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README.md	
+++ b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README.md	
@@ -71,7 +71,29 @@ In total, the highest number of consecutive transactions is 3, achieved by custo
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT
+            customer_id,
+            date_sub(
+                transaction_date,
+                INTERVAL row_number() OVER (
+                    PARTITION BY customer_id
+                    ORDER BY transaction_date
+                ) DAY
+            ) AS transaction_date
+        FROM Transactions
+    ),
+    t AS (
+        SELECT customer_id, transaction_date, count(1) AS cnt
+        FROM s
+        GROUP BY 1, 2
+    )
+SELECT customer_id
+FROM t
+WHERE cnt = (SELECT max(cnt) FROM t)
+ORDER BY customer_id;
 ```
 
 <!-- tabs:end -->
diff --git a/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README_EN.md b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README_EN.md
index cd7b3fec3f4b8..0aa47586d3908 100644
--- a/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README_EN.md	
+++ b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/README_EN.md	
@@ -65,7 +65,29 @@ In total, the highest number of consecutive transactions is 3, achieved by custo
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT
+            customer_id,
+            date_sub(
+                transaction_date,
+                INTERVAL row_number() OVER (
+                    PARTITION BY customer_id
+                    ORDER BY transaction_date
+                ) DAY
+            ) AS transaction_date
+        FROM Transactions
+    ),
+    t AS (
+        SELECT customer_id, transaction_date, count(1) AS cnt
+        FROM s
+        GROUP BY 1, 2
+    )
+SELECT customer_id
+FROM t
+WHERE cnt = (SELECT max(cnt) FROM t)
+ORDER BY customer_id;
 ```
 
 <!-- tabs:end -->
diff --git a/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql
new file mode 100644
index 0000000000000..f213a7dfb1845
--- /dev/null
+++ b/solution/2700-2799/2752.Customers with Maximum Number of Transactions on Consecutive Days/Solution.sql	
@@ -0,0 +1,23 @@
+# Write your MySQL query statement below
+WITH
+    s AS (
+        SELECT
+            customer_id,
+            date_sub(
+                transaction_date,
+                INTERVAL row_number() OVER (
+                    PARTITION BY customer_id
+                    ORDER BY transaction_date
+                ) DAY
+            ) AS transaction_date
+        FROM Transactions
+    ),
+    t AS (
+        SELECT customer_id, transaction_date, count(1) AS cnt
+        FROM s
+        GROUP BY 1, 2
+    )
+SELECT customer_id
+FROM t
+WHERE cnt = (SELECT max(cnt) FROM t)
+ORDER BY customer_id;