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  • SQL Join query help

    - by lostInTransit
    Hi I have 2 tables A and B with the following columns Table A - id,bId,aName,aVal Table B - id,bName where A.bId is the same as B.id. I want a result set from a query to get A.id, A.aName, B.bName where A.bId=B.id OR A.id, A.aName, "" when A.bId=0. In both cases, only those records should be considered where A.aVal LIKE "aVal" Can someone please help me with the query? I can use left join but how do I get the blank string if bId=0 and B.bName otherwise? Thanks

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  • Join using combined conditions on one join table

    - by Nathan Wienert
    I have join a table joining songs to genres. The table has a 'source' column that's used to identify where the genre was found. Genres are found from blogs, artists, tags, and posts. So, songs | song_genre | genres id | song_id, source, genre_id | id What I want to build is a song SELECT query that works something like this, given I already have a genre_id: IF exists song_genre with source='artist' AND a song_genre with source='blog' OR exists song_genre with source='artist' AND a song_genre with source='post' OR exists song_genre with source='tag' I'm was going to do it by doing a bunch of joins, but am sure I'm not doing it very well. Using Postgres 9.1.

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  • Objective-C : Fowler–Noll–Vo (FNV) Hash implementation

    - by Dough
    Hi ! I have a HTTP connector in my iPhone project and queries must have a parameter set from the username using the Fowler–Noll–Vo (FNV) Hash. I have a Java implementation working at this time, this is the code : long fnv_prime = 0x811C9DC5; long hash = 0; for(int i = 0; i < str.length(); i++) { hash *= fnv_prime; hash ^= str.charAt(i); } Now on the iPhone side, I did this : int64_t fnv_prime = 0x811C9DC5; int64_T hash = 0; for (int i=0; i < [myString length]; i++) { hash *= fnv_prime; hash ^= [myString characterAtIndex:i]; } This script doesn't give me the same result has the Java one. In first loop, I get this : hash = 0 hash = 100 (first letter is "d") hash = 1865261300 (for hash = 100 and fnv_prime = -2128831035 like in Java) Do someone see something I'm missing ? Thanks in advance for the help !

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  • MS SQL 2008, join or no join?

    - by Patrick
    Just a small question regarding joins. I have a table with around 30 fields and i was thinking about making a second table to store 10 of those fields. Then i would just join them in with the main data. The 10 fields that i was planning to store in a second table does not get queried directly, it's just some settings for the data in the first table. Something like: Table 1 Id Data1 Data2 Data3 etc ... Table 2 Id (same id as table one) Settings1 Settings2 Settings3 Is this a bad solution? Should i just use 1 table? How much performance inpact does it have? All entries in table 1 would also then have an entry in table 2. Small update is in order. Most of the Data fields are of the type varchar and 2 of them are of the type text. How is indexing treated? My plan is to index 2 data fields, email (varchar 50) and author (varchar 20). And yes, all records in Table 1 will have a record in Table 2. Most of the settings fields are of the bit type, around 80%. The rest is a mix between int and varchar. The varchars can be null.

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  • SQL Server 2008, join or no join?

    - by Patrick
    Just a small question regarding joins. I have a table with around 30 fields and i was thinking about making a second table to store 10 of those fields. Then i would just join them in with the main data. The 10 fields that i was planning to store in a second table does not get queried directly, it's just some settings for the data in the first table. Something like: Table 1 Id Data1 Data2 Data3 etc ... Table 2 Id (same id as table one) Settings1 Settings2 Settings3 Is this a bad solution? Should i just use 1 table? How much performance inpact does it have? All entries in table 1 would also then have an entry in table 2. Small update is in order. Most of the Data fields are of the type varchar and 2 of them are of the type text. How is indexing treated? My plan is to index 2 data fields, email (varchar 50) and author (varchar 20). And yes, all records in Table 1 will have a record in Table 2. Most of the settings fields are of the bit type, around 80%. The rest is a mix between int and varchar. The varchars can be null.

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  • Why do we need Hash by key? [migrated]

    - by Royi Namir
    (i'm just trying to find what am I missing...) Assuming John have a clear text message , he can create a regular hash ( like md5 , or sha256) and then encrypt the message. John can now send Paul the message + its (clear text)hash and Paul can know if the message was altered. ( decrypt and then compare hashes). Even if an attacker can change the encrpyted data ( without decrypt) - - when paul will open the message - and recalc the hash - it wont generate the same hash as the one john sent him. so why do we need hash by key ?

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  • Basics of Join Factorization

    - by Hong Su
    We continue our series on optimizer transformations with a post that describes the Join Factorization transformation. The Join Factorization transformation was introduced in Oracle 11g Release 2 and applies to UNION ALL queries. Union all queries are commonly used in database applications, especially in data integration applications. In many scenarios the branches in a UNION All query share a common processing, i.e, refer to the same tables. In the current Oracle execution strategy, each branch of a UNION ALL query is evaluated independently, which leads to repetitive processing, including data access and join. The join factorization transformation offers an opportunity to share the common computations across the UNION ALL branches. Currently, join factorization only factorizes common references to base tables only, i.e, not views. Consider a simple example of query Q1. Q1:    select t1.c1, t2.c2    from t1, t2, t3    where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c2 = 2 and t2.c2 = t3.c2   union all    select t1.c1, t2.c2    from t1, t2, t4    where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c3 = t4.c3; Table t1 appears in both the branches. As does the filter predicates on t1 (t1.c1 > 1) and the join predicates involving t1 (t1.c1 = t2.c1). Nevertheless, without any transformation, the scan (and the filtering) on t1 has to be done twice, once per branch. Such a query may benefit from join factorization which can transform Q1 into Q2 as follows: Q2:    select t1.c1, VW_JF_1.item_2    from t1, (select t2.c1 item_1, t2.c2 item_2                   from t2, t3                    where t2.c2 = t3.c2 and t2.c2 = 2                                  union all                   select t2.c1 item_1, t2.c2 item_2                   from t2, t4                    where t2.c3 = t4.c3) VW_JF_1    where t1.c1 = VW_JF_1.item_1 and t1.c1 > 1; In Q2, t1 is "factorized" and thus the table scan and the filtering on t1 is done only once (it's shared). If t1 is large, then avoiding one extra scan of t1 can lead to a huge performance improvement. Another benefit of join factorization is that it can open up more join orders. Let's look at query Q3. Q3:    select *    from t5, (select t1.c1, t2.c2                  from t1, t2, t3                  where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c2 = 2 and t2.c2 = t3.c2                 union all                  select t1.c1, t2.c2                  from t1, t2, t4                  where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c3 = t4.c3) V;   where t5.c1 = V.c1 In Q3, view V is same as Q1. Before join factorization, t1, t2 and t3 must be joined first before they can be joined with t5. But if join factorization factorizes t1 from view V, t1 can then be joined with t5. This opens up new join orders. That being said, join factorization imposes certain join orders. For example, in Q2, t2 and t3 appear in the first branch of the UNION ALL query in view VW_JF_1. T2 must be joined with t3 before it can be joined with t1 which is outside of the VW_JF_1 view. The imposed join order may not necessarily be the best join order. For this reason, join factorization is performed under cost-based transformation framework; this means that we cost the plans with and without join factorization and choose the cheapest plan. Note that if the branches in UNION ALL have DISTINCT clauses, join factorization is not valid. For example, Q4 is NOT semantically equivalent to Q5.   Q4:     select distinct t1.*      from t1, t2      where t1.c1 = t2.c1  union all      select distinct t1.*      from t1, t2      where t1.c1 = t2.c1 Q5:    select distinct t1.*     from t1, (select t2.c1 item_1                   from t2                union all                   select t2.c1 item_1                  from t2) VW_JF_1     where t1.c1 = VW_JF_1.item_1 Q4 might return more rows than Q5. Q5's results are guaranteed to be duplicate free because of the DISTINCT key word at the top level while Q4's results might contain duplicates.   The examples given so far involve inner joins only. Join factorization is also supported in outer join, anti join and semi join. But only the right tables of outer join, anti join and semi joins can be factorized. It is not semantically correct to factorize the left table of outer join, anti join or semi join. For example, Q6 is NOT semantically equivalent to Q7. Q6:     select t1.c1, t2.c2    from t1, t2    where t1.c1 = t2.c1(+) and t2.c2 (+) = 2  union all    select t1.c1, t2.c2    from t1, t2      where t1.c1 = t2.c1(+) and t2.c2 (+) = 3 Q7:     select t1.c1, VW_JF_1.item_2    from t1, (select t2.c1 item_1, t2.c2 item_2                  from t2                  where t2.c2 = 2                union all                  select t2.c1 item_1, t2.c2 item_2                  from t2                                                                                                    where t2.c2 = 3) VW_JF_1       where t1.c1 = VW_JF_1.item_1(+)                                                                  However, the right side of an outer join can be factorized. For example, join factorization can transform Q8 to Q9 by factorizing t2, which is the right table of an outer join. Q8:    select t1.c2, t2.c2    from t1, t2      where t1.c1 = t2.c1 (+) and t1.c1 = 1 union all    select t1.c2, t2.c2    from t1, t2    where t1.c1 = t2.c1(+) and t1.c1 = 2 Q9:   select VW_JF_1.item_2, t2.c2   from t2,             (select t1.c1 item_1, t1.c2 item_2            from t1            where t1.c1 = 1           union all            select t1.c1 item_1, t1.c2 item_2            from t1            where t1.c1 = 2) VW_JF_1   where VW_JF_1.item_1 = t2.c1(+) All of the examples in this blog show factorizing a single table from two branches. This is just for ease of illustration. Join factorization can factorize multiple tables and from more than two UNION ALL branches.  SummaryJoin factorization is a cost-based transformation. It can factorize common computations from branches in a UNION ALL query which can lead to huge performance improvement. 

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  • Time complexity to fill hash table (homework)?

    - by Heathcliff
    This is a homework question, but I think there's something missing from it. It asks: Provide a sequence of m keys to fill a hash table implemented with linear probing, such that the time to fill it is minimum. And then Provide another sequence of m keys, but such that the time fill it is maximum. Repeat these two questions if the hash table implements quadratic probing I can only assume that the hash table has size m, both because it's the only number given and because we have been using that letter to address a hash table size before when describing the load factor. But I can't think of any sequence to do the first without knowing the hash function that hashes the sequence into the table. If it is a bad hash function, such that, for instance, it hashes every entry to the same index, then both the minimum and maximum time to fill it will take O(n) time, regardless of what the sequence looks like. And in the average case, where I assume the hash function is OK, how am I suppossed to know how long it will take for that hash function to fill the table? Aren't these questions linked to the hash function stronger than they are to the sequence that is hashed? As for the second question, I can assume that, regardless of the hash function, a sequence of size m with the same key repeated m-times will provide the maximum time, because it will cause linear probing from the second entry on. I think that will take O(n) time. Is that correct? Thanks

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  • Performing Inner Join for Multiple Columns in the Same Table

    - by frankiefrank
    I have a scenario which I'm a bit stuck on. Let's say I have a survey about colors, and I have one table for the color data, and another for people's answers. tbColors color_code , color_name 1 , 'blue' 2 , 'green' 3 , 'yellow' 4 , 'red' tbAnswers answer_id , favorite_color , least_favorite_color , color_im_allergic_to 1 , 1 , 2 3 2 , 3 , 1 4 3 , 1 , 1 2 4 , 2 , 3 4 For display I want to write a SELECT that presents the answers table but using the color_name column from tbColors. I understand the "most stupid" way to do it naming tbColors three times in the FROM section, using a different alias for each column to replace. How would a non-stupid way look?

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  • Hibernate: Perform criteria query with Sub-Select AND Left-Outer join?

    - by Markos Fragkakis
    Can I perform a Criteria query with Sub-Select AND Left-Outer join? For example, I have A 1-many B 1-many C. With Criteria.createAlias ("b", "b", Criteria.LEFT_JOIN) I can perform Left Outer join. With Criteria.setFetchMode ("b", org.hibernate.FetchMode.DEFAULT) I can perform Join with the default fetching strategy. I assume that having set @org.hibernate.annotations.FetchMode.SUBSELECT in both A.B and B.C is enough (is it?). Question 1: Why does org.hibernate.FetchMode not have SUBSELECT option, whereas the org.hibernate.annotations.FetchMode does? Question 2: Can I perform a Criteria query with Sub-Select AND Left-Outer join?

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  • Can MySQL / SQL's short hand of "Using" be used without saying "Inner Join" ?

    - by Jian Lin
    The following 2 statements are to join using gifts.giftID = sentgifts.giftID: mysql> select * from gifts, sentgifts using (giftID); ERROR 1064 (42000): 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 'using (giftID)' at line 1 and the second one: mysql> select * from gifts INNER JOIN sentgifts using (giftID); +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ | giftID | name | filename | effectiveTime | sentID | whenSent | fromID | toID | trytryWhen | +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ | 2 | teddy bear | bear.jpg | 2010-04-24 04:36:03 | 4 | 2010-04-24 | NULL | 111 | 2010-04-24 03:10:42 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 6 | 2010-04-24 | 11 | 222 | 2010-04-24 03:54:49 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | 7 | 2010-04-24 | 1 | 2 | 2010-04-24 03:58:45 | +--------+------------+----------------+---------------------+--------+------------+--------+------+---------------------+ 4 rows in set (0.00 sec) Can the first statement also use the "using" shorthand? It seems that when it is used then the word "Inner Join" must be specified... but the first statement is actually an inner join?

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  • Perl Hash Slice, Replication x Operator, and sub params

    - by user210757
    Ok, I understand perl hash slices, and the "x" operator in Perl, but can someone explain the following code example from here (slightly simplified)? sub test{ my %hash; @hash{@_} = (undef) x @_; } Example Call to sub: test('one', 'two', 'three'); This line is what throws me: @hash{@_} = (undef) x @_; It is creating a hash where the keys are the parameters to the sub and initializing to undef, so: %hash: 'one' = undef, 'two' = undef, 'three' = undef The rvalue of the x operator should be a number; how is it that @_ is interpreted as the length of the sub's parameter array? I would expect you'd at least have to do this: @hash{@_} = (undef) x length(@_);

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  • 3 hash functions to best hash sliding window strings for a bloom filter with minimum collisions

    - by Duaa
    Hi all: I need 3 hash functions to hash strings of a sliding window moving over a text, to be used later to search within a bloom vector. I'm using C# in my programming I read something about rolling hash functions and cyclic polynomials, they are used for sliding window applications. But really, I did not find any codes, they are just descriptions So please, if anyone have any idea about 3 best C# hash functions to use with sliding window strings of fixed size (5-char), that consume less time and have minimum number of collisions, either they are rolling hash functions or others, please help me with some C# codes or links to hash functions names Duaa

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  • INNER JOIN vs LEFT JOIN performance in SQL Server

    - by Ekkapop
    I've created SQL command that use INNER JOIN for 9 tables, anyway this command take a very long time (more than five minutes). So my folk suggest me to change INNER JOIN to LEFT JOIN because the performance of LEFT JOIN is better, at first time its despite what I know. After I changed, the speed of query is significantly improve. I want to know why LEFT JOIN is faster than INNER JOIN? My SQL command look like below: SELECT * FROM A INNER JOIN B ON ... INNER JOIN C ON ... INNER JOIN D and so no

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  • How sql server evaluates the multiple different joins?

    - by ziang
    Hi, i have a general question about how sql server evaluates the joins.The query is SELECT * FROM TableA INNER JOIN TableB ON TableB.id = TableA.id LEFT JOIN TABLEC ON TABLEC.id = TABLEB.id Q1: What tables is the left join based on? I know it will based on the TABLEC but what is the other one? Is it the result of the first inner join or the TABLEB specified in the left join condition? Q2: Is "LEFT JOIN TABLEC ON TABLEC.id = TABLEB.id" equivalent to "LEFT JOIN TABLEC ON TABLEB.id = TABLEC.id" Q3: Is the query equivalent to the following one? (with TABLEB.id replaced by TABLEA.id?) SELECT * FROM TableA INNER JOIN TableB ON TableB.id = TableA.id LEFT JOIN TABLEC ON TABLEC.id = TABLEA.id Thank you!

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  • 3 SQL Join Concepts to Help You Choose the Right Join

    What do SQL joins and the "teach a man to fish" Chinese proverb have in common? SQL joins, like regular expressions, are one of those commonplace programming tasks in which true success is entirely dependent upon your ability to conceptualize the outcome. Fail to do so and you'll likely wind up spending a few hours in a frustrating round of trial and error. Like regular expressions, the proliferation of online examples has actually contributed to the frustration, providing the equivalent of a day's worth of fish rather than the proverbial fishing pool. The Future of SQL Server MonitoringMonitor wherever, whenever with Red Gate's SQL Monitor. See it live in action now.

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  • Duplicate Items Using Join in NHibernate Map

    - by Colin Bowern
    I am trying to retrieve the individual detail rows without having to create an object for the parent. I have a map which joins a parent table with the detail to achieve this: Table("UdfTemplate"); Id(x => x.Id, "Template_Id"); Map(x => x.FieldCode, "Field_Code"); Map(x => x.ClientId, "Client_Id"); Join("UdfFields", join => { join.KeyColumn("Template_Id"); join.Map(x => x.Name, "COLUMN_NAME"); join.Map(x => x.Label, "DISPLAY_NAME"); join.Map(x => x.IsRequired, "MANDATORY_FLAG") .CustomType<YesNoType>(); join.Map(x => x.MaxLength, "DATA_LENGTH"); join.Map(x => x.Scale, "DATA_SCALE"); join.Map(x => x.Precision, "DATA_PRECISION"); join.Map(x => x.MinValue, "MIN_VALUE"); join.Map(x => x.MaxValue, "MAX_VALUE"); }); When I run the query in NH using: Session.CreateCriteria(typeof(UserDefinedField)) .Add(Restrictions.Eq("FieldCode", code)).List<UserDefinedField>(); I get back the first row three times as opposed to the three individual rows it should return. Looking at the SQL trace in NH Profiler the query appears to be correct. The problem feels like it is in the mapping but I am unsure how to troubleshoot that process. I am about to turn on logging to see what I can find but I thought I would post here in case someone with experience mapping joins knows where I am going wrong.

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  • Standard way to hash an RSA key?

    - by Adam J.R. Erickson
    What's the algorithm for creating hash (sha-1 or MD5) of an RSA public key? Is there a standard way to do this? Hash just the modulus, string addition of both and then take a hash? Is SHA-1 or MD5 usually used? I want to use it to ensure that I got the right key (have the sender send a hash, and I calculate it myself), and log said hash so I always know which exact key I used when I encrypt the payload.

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  • Would shell command join cause out of memory?

    - by Hancy
    I have two file to join. FILE 1: a A1 a A2 a A3 ... c C1 c C2 ... FILE 2: a feature1_of_a a feature2_of_a ... a featureN_of_a ... ... c feature1_of_c c feature2_of_c ... after join, i could get File like this: A1 feature1_of_a A2 feature1_of_a A3 feature1_of_a A1 feature2_of_a A2 feature2_of_a A3 feature2_of_a ... A1 featureN_of_a A2 featureN_of_a A3 featureN_of_a ... In order to do that: i wrote shell command join -11 -21 -o1.2,2.2 file1 file2. But the problem is: number N might be huge. So if join read all feautre of a into memory at once, memory might not be enough. I don't know how join is implemented. WQould the momery become a problem? If so, is there any way to get what I want?

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  • SQL SERVER – Challenge – Puzzle – Why does RIGHT JOIN Exists

    - by pinaldave
    I had interesting conversation with the attendees of the my SQL Server Performance Tuning course. I was asked if LEFT JOIN can do the same task as RIGHT JOIN by reserving the order of the tables in join, why does RIGHT JOIN exists? The definitions are as following: Left Join – select all the records from the LEFT table and then pick up any matching records from the RIGHT table   Right Join – select all the records from the RIGHT table and then pick up any matching records from the LEFT table Most of us read from LEFT to RIGHT so we are using LEFT join. Do you have any explaination why RIGHT JOIN exists or can you come up with example, where RIGHT JOIN is absolutely required and the task can not be achieved with LEFT JOIN. Other Puzzles: SQL SERVER – Puzzle – Challenge – Error While Converting Money to Decimal SQL SERVER – Challenge – Puzzle – Usage of FAST Hint Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Pinal Dave, SQL, SQL Authority, SQL Puzzle, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • reference to specific hash key

    - by dave
    How do I create a reference to the value in a specific hash key. I tried the following but $$foo is empty. Any help is much appreciated. $hash->{1} = "one"; $hash->{2} = "two"; $hash->{3} = "three"; $foo = \${$hash->{1}}; $hash->{1} = "ONE"; #I want "MONEY: ONE"; print "MONEY: $$foo\n";

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  • Oracle Hash Cluster Overflow Blocks

    - by Andrew
    When inserting a large number of rows into a single table hash cluster in Oracle, it will fill up the block with any values that hash to that hash-value and then start using overflow blocks. These overflow blocks are listed as chained off the main block, but I can not find detailed information on the way in which they are allocated or chained. When an overflow block is allocated for a hash value, is that block exclusively allocated to that hash value, or are the overflow blocks used as a pool and different hash values can then start using the same overflow block. How is the free space of the chain monitored - in that, as data is continued to be inserted, does it have to traverse the entire chain to find out if it has some free space in the current overflow chain, and then if it finds none, it then chooses to allocate a new block?

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  • From string to hex MD5 hash and back

    - by Pablo Fernandez
    I have this pseudo-code in java: bytes[] hash = MD5.hash("example"); String hexString = toHexString(hash); //This returns something like a0394dbe93f bytes[] hexBytes = hexString.getBytes("UTF-8"); Now, hexBytes[] and hash[] are different. I know I'm doing something wrong since hash.length() is 16 and hexBytes.length() is 32. Maybe it has something to do with java using Unicode for chars (just a wild guess here). Anyways, the question would be: how to get the original hash[] array from the hexString. The whole code is here if you want to look at it (it's ~ 40 LOC) http://gist.github.com/434466 The output of that code is: 16 [-24, 32, -69, 74, -70, 90, -41, 76, 90, 111, -15, -84, -95, 102, 65, -10] 32 [101, 56, 50, 48, 98, 98, 52, 97, 98, 97, 53, 97, 100, 55, 52, 99, 53, 97, 54, 102, 102, 49, 97, 99, 97, 49, 54, 54, 52, 49, 102, 54] Thanks a lot!

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  • mysql join with conditional

    - by Conor H
    Hi There, I am currently working on a MySQL query that contains a table: TBL:lesson_fee -fee_type_id (PRI) -lesson_type_id (PRI) -lesson_fee_amount this table contains the fees for a particular 'lesson type' and there are different 'fee names' (fee_type). Which means that there can be many entries in this table for one 'lesson type' In my query I am joining this table onto the rest of the query via the 'lesson_type' table using: lesson_fee INNER JOIN (other joins here) ON lesson_fee.lesson_type_id = lesson_type.lesson_type_id The problem with this is that it is currently returning duplicate data in the result. 1 row for every duplicate entry in the 'lesson fee' table. I am also joining the 'fee type' table using this 'fee_type_id' Is there a way of telling MySQL to say "Join the lesson_fee table rows that have lesson_fee.lesson_type_id and fee_type_id = client.fee_type_id". UPDATE: Query: SELECT lesson_booking.lesson_booking_id,lesson_fee.lesson_fee_amount FROM fee_type INNER JOIN (lesson_fee INNER JOIN (color_code INNER JOIN (employee INNER JOIN (horse_owned INNER JOIN (lesson_type INNER JOIN (timetable INNER JOIN (lesson_booking INNER JOIN CLIENT ON client.client_id = lesson_booking.client_id) ON lesson_booking.timetable_id = timetable.timetable_id) ON lesson_type.lesson_type_id = timetable.lesson_type_id) ON horse_owned.horse_owned_id = lesson_booking.horse_owned_id) ON employee.employee_id = timetable.employee_id) ON employee.color_code_id = color_code.color_code_id) ON lesson_fee.lesson_type_id = lesson_type.lesson_type_id) ON lesson_fee.fee_type_id = client.fee_type_id WHERE booking_date = '2010-04-06' ORDER BY lesson_booking_id ASC How do I keep the format(indentation) of my query?

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