sql-delete | 易学教程

Multiple LEFT JOIN in Access

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I have the following query, which works for MySQL: DELETE `test1`, `test2`, `test3`, `test4` FROM `test1` LEFT JOIN `test2` ON test2.qid = test1.id LEFT JOIN test3 ON test3.tid = test2.id LEFT JOIN test4.qid = test1.id WHERE test1.id = {0} But it doesn't work for MS Access. I've tried to add parentheses around the LEFT JOIN , but it gives me syntax error in FROM clause. So how should this query look in order to work in MS Access? The Access DELETE requires a star (*): DELETE * FROM ... In addition, the joins must be nested by using parentheses: DELETE test1.*, test2.*, test3.*, test4.* FROM (

How to delete all records created today?

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I am dealing with a very big database ~ 6 Million records. I've added ~30,000 bad records today. How can I delete all of the records created today in MySQL? It seems created_at is a datetime. Try: delete from table where date(created_at) = curdate() Of course, run a select * prior to run this query and make sure the data you're going to delete is the one you really want to delete. DELETE FROM Table WHERE ( (DAY(CallDate) = DAY(GETDATE()) AND (MONTH(CallDate) = MONTH(GETDATE()) AND (YEAR(CallDate) = YEAR(GETDATE()) ) Try below: delete from table where left(created_at,10) =curdate() The

How to turn this MySQL SELECT query into a DELETE query?

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I want to delete certain items from the database. I have the following query: SELECT * FROM sheets, entries WHERE entries.sheetID = sheets.id AND sheets.clientID = 13 This works, and returns 2 results. Now I want to turn this SELECT query into a DELETE query. However, the following doesn't work: DELETE FROM sheets, entries WHERE entries.sheetID = sheets.id AND sheets.clientID = 13 MySQL throws the following error: 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE entries.sheetID = sheets.id AND

what does “delete from table where NULL = NULL” means?

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问题 What does delete from table where NULL = NULL mean? 回答1: That will delete nothing from the table. NULL does not equal NULL. Now delete from table where NULL is NULL would delete all rows from the table. 回答2: It means don't delete anything, because NULL is never equal to anything. Or maybe it means "don't delete anything unless the user's DBMS really sucks, in which case delete it all out of spite". Seriously though, that kind of construct usually comes about when a WHERE clause is

SQLite, check if TEXT field has any alphabetical chars in it

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Okay, so I have a huge list of entries, and in one of the columns (for simplicity let's call it num there's a number, something like 123456780000 (they are all the same length and format), but sometimes there are fields that look something like this 12345678E000 or 12345678H000 Now, I need to delete all the rows in which the num column is not entirely numeric. The type of num is TEXT , not INTEGER . So the above examples should be deleted, while 123456780000 should not. I have tried two solutions, of which one works but is inelegant and messy, and the other one doesn't work at all. The first

Delete multiple rows in MYSQL with info from python list

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问题 If list LL: LL = ['foo', bar', 'noo', 'boo',] is in a MySQL table, test in column ID with other ID's. I could use the following to delete all rows with ID's in LL: csr.execute("""DELETE FROM test.test WHERE ID = "Foo"; """) csr.execute("""DELETE FROM test.test WHERE ID = "bar"; """) csr.execute("""DELETE FROM test.test WHERE ID = "noo"; """) csr.execute("""DELETE FROM test.test WHERE ID = "boo"; """) How could I do it programatically? 回答1: You can do it with a single query: id_list = ['abc',

Deleting orphans from a table

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问题 I am trying to clean up a table where there are quite a few orphaned items. I am approaching this by checking to see if there is a relationship to another table by looking for null values. DELETE FROM table1 LEFT JOIN table2 ON table1.ID = table2.ID WHERE table2.ID IS NULL I get an error that the left outer join is not valid. I am looking for suggestions on other ways that I can delete these orphans from this broken relationship 回答1: try this: DELETE FROM table1 WHERE NOT EXISTS (SELECT NULL

MySQL insert on duplicate key; delete?

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Is there a way of removing record on duplicate key in MySQL? Say we have a record in the database with the specific primary key and we try to add another one with the same key - ON DUPLICATE KEY UPDATE would simply update the record, but is there an option to remove record if already exists? It is for simple in/out functionality on click of a button. It's a work-around, but it works: Create a new column and call it do_delete , or whatever, making it a tiny-int. Then do On Duplicate Key Update do_delete = 1; Depending on your MySQL version/connection, you can execute multiple queries in the

php sql delete button

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I'm trying to make a delete button from sql. I started a function called $del but I don't know how to complete it, in the form of a delete button echoe'd out beside the current echo statements. $con = mysql_connect("localhost", "user", "pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $sql = "INSERT INTO camps (city, map, park, day, details) VALUES ('$_POST[city]','$_POST[map]','$_POST[park]','$_POST[day]','$_POST[details]')"; if (!mysql_query($sql, $con)) { die('Error: ' . mysql_error()); } $number = 0; $del = mysql_query("DELETE FROM camps

How to delete records in DB with mySQL using group by [duplicate]

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Possible Duplicate: SQL Delete: can't specify target table for update in FROM clause I have one table only (call this table TAB), representing University exams. I have the following attributes: CourseName, CourseCode and year. I want to delete all courses that have a cardinality less than 100. If I type select CourseName from TAB group by CourseName having count(CourseName) < 100; I have an exact result. But if I want to delete this entries I try with delete from TAB where CourseName not in (select CourseName from TAB group by CourseName having count(CourseName) > 100); but the system returns