Assuming you are using a SQL database and the table name is 'students' and the column with the email ids is 'email', the query would look something like this:

```sql
SELECT COUNT(*) AS student_count
FROM students
WHERE NOT (email LIKE '_%@%.com');
```

Here's the step by step explanation:

1. `SELECT COUNT(*) AS student_count`: This part of the query will count the number of rows returned by the query and give it an alias name 'student_count'.

2. `FROM students`: This part specifies the table we are querying, which is 'students' in this case.

3. `WHERE NOT (email LIKE '_%@%.com')`: This is the condition for the query. It will return the rows where the email does not match the pattern '_%@%.com'. The underscore (_) in the LIKE clause represents exactly one character, the percent (%) represents zero, one, or multiple characters. So this pattern will match any string that has at least one character before the '@', then any characters, then '.com' at the end. The NOT operator negates the condition, so it returns the rows that do not match this pattern, i.e., the rows with invalid email ids.

Question

Assuming you are using a SQL database and the table name is 'students' and the column with the email ids is 'email', the query would look something like this:

```sql
SELECT COUNT(*) AS student_count
FROM students
WHERE NOT (email LIKE '_%@%.com');
```

Here's the step by step explanation:

1. `SELECT COUNT(*) AS student_count`: This part of the query will count the number of rows returned by the query and give it an alias name 'student_count'.

2. `FROM students`: This part specifies the table we are querying, which is 'students' in this case.

3. `WHERE NOT (email LIKE '_%@%.com')`: This is the condition for the query. It will return the rows where the email does not match the pattern '_%@%.com'. The underscore (_) in the LIKE clause represents exactly one character, the percent (%) represents zero, one, or multiple characters. So this pattern will match any string that has at least one character before the '@', then any characters, then '.com' at the end. The NOT operator negates the condition, so it returns the rows that do not match this pattern, i.e., the rows with invalid email ids.

Knowee AI · Accepted Answer

Assuming you are using a SQL database and the table name is 'students' and the column with the email ids is 'email', the query would look something like this:

```sql
SELECT COUNT(*) AS student_count
FROM students
WHERE NOT (email LIKE '_%@%.com');
```

Here's the step by step explanation:

1. `SELECT COUNT(*) AS student_count`: This part of the query will count the number of rows returned by the query and give it an alias name 'student_count'.

2. `FROM students`: This part specifies the table we are querying, which is 'students' in this case.

3. `WHERE NOT (email LIKE '_%@%.com')`: This is the condition for the query. It will return the rows where the email does not match the pattern '_%@%.com'. The underscore (_) in the LIKE clause represents exactly one character, the percent (%) represents zero, one, or multiple characters. So this pattern will match any string that has at least one character before the '@', then any characters, then '.com' at the end. The NOT operator negates the condition, so it returns the rows that do not match this pattern, i.e., the rows with invalid email ids.

Write a query to display the number of students having invalid email id.The valid email ids should have “@” anywhere after the first character and should end with “.com”. Give an alias name for the count as ‘student_count’.

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