![]() Select demo.empcode,demo.name,sq_name_new1,(case when sq_name_new1 like format('%%%s%%',demo.name) THEN count(sq_name_new1) else (when demo.empcode is not null then count(demo.id) End) END) AS indivisual from res_scheduledjobs LEFT JOIN demo on demo.id=res_scheduledjobs.assigned_technician group by res_scheduledjobs.assigned_technician,sq_name_new1,demo.empcode,demo.name (select hr_employee.emp_code,hr_employee.id,concat(resource_resource.name,' ',hr_employee.middle_name,' ',hr_employee.last_name) as name from hr_employee inner join resource_resource on resource_resource.id=hr_employee.resource_id) In the above query demo.name column comes from CTE.so my whole query look like: with demo(empcode,id,name) as In this tutorial, you have learned how to use the PostgreSQL COUNT() function to return the number of rows in a table.I want to write nested case when condition in query to store the value that will come from one case when condition and another case when condition into same new column.to get this kind of result i am writing the query as: (case when sq_name_new1 like format('%%%s%%',demo.name) THEN count(sq_name_new1) else (when demo.empcode is not null then count(demo.id) End) END) AS indivisual from res_scheduledjobs ![]() HAVING COUNT (customer_id) > 40 Code language: SQL (Structured Query Language) ( sql ) For example, the following statement finds customers who have made more than 40 payments: SELECT You can use the COUNT function in a HAVING clause to apply a specific condition to groups. Here is the partial output: PostgreSQL COUNT() with HAVING clause The following query illustrates the idea: SELECTĬustomer_id Code language: SQL (Structured Query Language) ( sql ) To get the number of payments by the customer, you use the GROUP BY clause to group the payments into groups based on customer id, and use the COUNT() function to count the payments for each group. Payment Code language: SQL (Structured Query Language) ( sql ) PostgreSQL COUNT() with GROUP BY clause To get the distinct amounts which customers paid, you use the COUNT(DISTINCT amount) function as shown in the following example: SELECT COUNT ( DISTINCT amount) Here is the output: 2) PostgreSQL COUNT(DISTINCT column) example Payment Code language: SQL (Structured Query Language) ( sql ) The following statement uses the COUNT(*) function to return the number of transactions in the payment table: SELECT COUNT(*) Let’s use the payment table in the sample database for the demonstration. ![]() For example, we can use the COUNT() with the GROUP BY clause to return the number of films in each film category. We often use the COUNT() function with the GROUP BY clause to return the number of items for each group. In this form, the COUNT(DISTINCT column) returns the number of unique non-null values in the column. SELECT COUNT( column)Ĭondition Code language: SQL (Structured Query Language) ( sql ) COUNT(DISTINCT column) However, it does not consider NULL values in the column. Similar to the COUNT(*) function, the COUNT(column) function returns the number of rows returned by a SELECT clause. Because multiple transactions see different states of data at the same time, there is no direct way for COUNT(*) function to count across the whole table, therefore PostgreSQL must scan all rows. This is related to the PostgreSQL MVCC implementation. If you use the COUNT(*) function on a big table, the query will be slow. When you apply the COUNT(*) function to the entire table, PostgreSQL has to scan the whole table sequentially. SELECT COUNT(*)Ĭondition Code language: SQL (Structured Query Language) ( sql ) The COUNT(*) function returns the number of rows returned by a SELECT statement, including NULL and duplicates. The following statement illustrates various ways of using the COUNT() function. The COUNT() function is an aggregate function that allows you to get the number of rows that match a specific condition of a query. Summary: in this tutorial, you will learn how to use the PostgreSQL COUNT() function to count the number of rows in a table.
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