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LPAD('o',ROUND(AVG(ReturnedDate-CheckOutDate)/2,0),'o') AS Graph from BOOKSHELF_CHECKOUT group by Name order by AVG(ReturnedDate-CheckOutDate); Day 1 2 3 4 5 6 7 NAME DAYSOUT 0 0 0 0 0 0 0 DORAH TALBOT 13 ooooooo EMILY TALBOT 14 ooooooo JED HOPKINS 18 ooooooooo GERHARDT KENTGEN 19 ooooooooo PAT LAVAY 21 oooooooooo FRED FULLER 24 oooooooooooo ROLAND BRANDT 52 oooooooooooooooooooooooooo 7 rows selected. This use of LPAD is similar to what was done in Chapter 13, where the cows and bulls were shown in their family tree. Basically, a lowercase o is the column here, and it is padded on the left with a number of additional lowercase o ’s, to the maximum width determined by ROUND(AVG(ReturnedDate-CheckOutDate)/2,0). Notice that the scale of the column heading is in increments of two. A simple change to the SQL will produce a classic graph rather than a bar chart. The literal column is changed from an o to a lowercase x , and the padding on the left is done by spaces. select Name, AVG(ReturnedDate-CheckOutDate) DaysOut, LPAD('x',ROUND(AVG(ReturnedDate-CheckOutDate)/2,0),' ') AS Graph from BOOKSHELF_CHECKOUT group by Name order by AVG(ReturnedDate-CheckOutDate); Day 1 2 3 4 5 6 7 NAME DAYSOUT 0 0 0 0 0 0 0 DORAH TALBOT 13 x EMILY TALBOT 14 x JED HOPKINS 18 x GERHARDT KENTGEN 19 x PAT LAVAY 21 x FRED FULLER 24 x ROLAND BRANDT 52 x 7 rows selected. Chapter 16: Advanced Use of Functions and Variables 299 ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:299 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:21 PM Color profile: Generic CMYK printer profile Composite Default screen 300 Part II: SQL and SQL*Plus ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:300 Another way to graph the data is by its distribution rather than by person. First, a view is created that puts each number of days out into its decile. Thus 13, 14, 18, and 19 become 10; 20 through 29 become 20; 30 through 39 become 30; and so on: create or replace view DaysOutRange as select ROUND((ReturnedDate-CheckOutDate),-1) Decile, COUNT(*) Counter from BOOKSHELF_CHECKOUT group by ROUND((ReturnedDate-CheckOutDate),-1); Next, a column heading is set up, similar to the previous heading although shorter and in increments of one: column Graph Heading 'Count| 1 1| 5 0 5'- justify c column Graph format a15 The next SQL determines the count of values that are represented in each decile. select Decile, Counter, LPAD('o',Counter,'o') AS Graph from DAYSOUTRANGE; Count 1 1 DECILE COUNTER 5 0 5 10 8 oooooooo 20 5 ooooo 30 3 ooo 50 2 oo 60 1 o If one of the dates were NULL, the group by output would include a NULL group, with a count and display similar to those shown in this listing. If you want to customize how Oracle handles books with NULL ReturnedDate values, you can use the NVL function to replace the NULL value with one of your choosing (such as SysDate). For more complex data replacement logic, you can use DECODE and CASE, as introduced later in this chapter and more fully described in Chapter 17. Using TRANSLATE TRANSLATE converts characters in a string into different characters, based on a substitution plan you give it, from if to then : TRANSLATE( string , if , then ) P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:22 PM Color profile: Generic CMYK printer profile Composite Default screen Chapter 16: Advanced Use of Functions and Variables 301 ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:301 In the following SQL, the letters in a sentence are replaced. Any time an uppercase T is detected, it is replaced by an uppercase T . In effect, nothing changes. Any time an uppercase vowel is detected, however, it is replaced by a lowercase version of the same vowel. Any letter not in the TAEIOU string is left alone. When a letter is found in TAEIOU, its position is checked in the TAEIOU string, and the letter there is substituted. Thus, the letter E , at position 3 in TAEIOU, is replaced by e , in position 3 of Taeiou: select TRANSLATE('NOW VOWELS ARE UNDER ATTACK','TAEIOU','Taeiou') from DUAL; TRANSLATE('NOWVOWELSAREUNDE NoW VoWeLS aRe uNDeR aTTaCK Eliminating Characters Extending this logic, what happens if the if string is TAEIOU and the then string is only T ? Checking for the letter E (as in the word VOWELS) finds it in position 3 of TAEIOU. There is no position 3 in the then string (which is just the letter T ), so the value in position 3 is nothing. So E is replaced by nothing. This same process is applied to all of the vowels. They appear in the if string, but not in the then string. As a result, they disappear, as shown here: select TRANSLATE('NOW VOWELS ARE UNDER ATTACK','TAEIOU','T') from DUAL; TRANSLATE('NOWVOWE NW VWLS R NDR TTCK This feature of TRANSLATE, the ability to eliminate characters from a string, can prove very useful in cleaning up data. Recall the magazine titles in Chapter 7: select Title from MAGAZINE; TITLE THE BARBERS WHO SHAVE THEMSELVES. "HUNTING THOREAU IN NEW HAMPSHIRE" THE ETHNIC NEIGHBORHOOD RELATIONAL DESIGN AND ENTHALPY "INTERCONTINENTAL RELATIONS." The method used in Chapter 7 to clean out the periods and double quotes was a nested combination of LTRIM and RTRIM: select LTRIM( RTRIM(Title,'."') ,'"') from MAGAZINE; P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:22 PM Color profile: Generic CMYK printer profile Composite Default screen The same goal can be accomplished with a single use of TRANSLATE: select TRANSLATE(Title,'T".','T') AS TITLE from MAGAZINE; TITLE THE BARBERS WHO SHAVE THEMSELVES HUNTING THOREAU IN NEW HAMPSHIRE THE ETHNIC NEIGHBORHOOD RELATIONAL DESIGN AND ENTHALPY INTERCONTINENTAL RELATIONS In the listing, you have to include one character in the then string—in this case, the letter T translates to the letter T . All other characters in the if set are eliminated. Complex Cut and Paste The NAME table contains a list of names as you might receive them from a mailing list company or another application. First name, last name, and initials are all in one column: column Name format a25 select Name from NAME; NAME HORATIO NELSON VALDO MARIE DE MEDICIS FLAVIUS JOSEPHUS EDYTHE P. M. GAMMIERE Suppose you want to cut and paste these names and put them into a table with FirstName and LastName columns. How would you go about it? The technique you learned in Chapter 7 involved using INSTR to locate a space, and using the number it returns in a SUBSTR to clip out the portion up to that space. Here’s an attempt to do just that for the first name: select SUBSTR(Name,1,INSTR(Name,' ')) from NAME; SUBSTR(NAME,1,INSTR(NAME, HORATIO MARIE FLAVIUS EDYTHE 302 Part II: SQL and SQL*Plus ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:302 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:22 PM Color profile: Generic CMYK printer profile Composite Default screen Chapter 16: Advanced Use of Functions and Variables 303 ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:303 VALDO has vanished! The problem is that these names are not as consistent as the authors’ names were in Chapter 7. One of these names (probably a magician) is only one name long, so there is no space for the INSTR to find. When INSTR has an unsuccessful search, it returns a 0. The SUBSTR of the name VALDO, starting at position 1 and going for 0 positions, is nothing, so he disappears. This is solved with DECODE. In place of this: INSTR(Name,' ') you put the entire expression, like this: DECODE(INSTR(Name,' '),0,99,INSTR(Name,' ')) DECODE’s format is this: DECODE( value , if1 , then1 [, if2 , then2 , if3 , then3 ] , else ) In the preceding example, DECODE tests the value of INSTR(Name, ‘ ‘). If it is equal to 0, DECODE substitutes 99; otherwise, it returns the default value, which is also INSTR(Name, ‘ ‘). The choice of 99 as a substitute is arbitrary. It will create an effective SUBSTR function for VALDO that looks like this: SUBSTR('VALDO',1,99) This works because SUBSTR will clip out text from the starting number, 1, to the ending number or the end of the string, whichever comes first. With the DECODE function in place, the first names are retrieved correctly: select SUBSTR(Name,1, DECODE(INSTR(Name,' '),0,99,INSTR(Name,' '))) from NAME; SUBSTR(NAME,1,DECODE(INST HORATIO VALDO MARIE FLAVIUS EDYTHE How about the last names? You could use INSTR again to search for a space, and use the location of the space in the string, plus one (+1), as the starting point for SUBSTR. No ending point is required for SUBSTR, because you want it to go to the end of the name. This is what happens: select SUBSTR(Name,INSTR(Name,' ')+1) from NAME; P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:23 PM Color profile: Generic CMYK printer profile Composite Default screen 304 Part II: SQL and SQL*Plus ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:304 SUBSTR(NAME,INSTR(NAME,'' NELSON VALDO DE MEDICIS JOSEPHUS P. M. GAMMIERE This didn’t quite work. One solution is to use three INSTR functions, looking successively for the first, second, or third occurrence of a space in the name. Each of these INSTRs will return either the location where it found a space or a 0 if it didn’t find any. In a name with only one space, the second and third INSTRs will both return 0. The GREATEST function, therefore, will pick the number returned by the INSTR that found the space furthest into the name: select SUBSTR(Name, GREATEST(INSTR(Name,' '),INSTR(Name,' ',1,2), INSTR(Name,' ',1,3)) +1) from NAME; SUBSTR(NAME,GREATEST(INST NELSON VALDO MEDICIS JOSEPHUS GAMMIERE Except for the fact that you also got VALDO again, this worked. (GREATEST also could have been used similarly in place of DECODE in the previous example.) There is a second and simpler method: select SUBSTR(Name,INSTR(Name,' ',-1)+1) from NAME; SUBSTR(NAME,INSTR(NAME,'' NELSON VALDO MEDICIS JOSEPHUS GAMMIERE The -1 in the INSTR tells it to start its search in the final position and go backward, or right to left, in the Name column. When it finds the space, INSTR returns its position, counting from the left as usual. The -1 simply makes INSTR start searching from the end rather than from the beginning. A -2 would make it start searching from the second position from the end, and so on. The +1 in the SUBSTR command has the same purpose as in the previous example: Once the space is found, SUBSTR has to move one position to the right to begin clipping out the Name. If no space is found, the INSTR returns 0, and SUBSTR therefore starts with position 1. That’s why VALDO made the list. P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:23 PM Color profile: Generic CMYK printer profile Composite Default screen How do you get rid of VALDO? Add an ending position to the SUBSTR instead of its default (which goes automatically all the way to the end). The ending position is found by using this: DECODE(INSTR(Name,' '),0,0,LENGTH(Name)) which says, “Find the position of a space in the Name. If the position is 0, return 0; otherwise, return the LENGTH of the Name.” For VALDO, the DECODE produces 0 as the ending position for SUBSTR, so nothing is displayed. For any other name, because there is a space somewhere, the LENGTH of the Name becomes the ending position for the SUBSTR, so the whole last name is displayed. This is similar to the DECODE used to extract the first name, except that the value 99 used there has been replaced by LENGTH(Name), which will always work, whereas 99 would fail for a name longer than 99 characters. This won’t matter here, but in other uses of DECODE and SUBSTR, it could be important: select SUBSTR(Name, INSTR(Name,' ',-1)+1, DECODE(INSTR(Name,' '), 0, 0, LENGTH(Name))) from NAME; SUBSTR(NAME,INSTR(NAME,'' NELSON MEDICIS JOSEPHUS GAMMIERE This DECODE also could have been replaced by a GREATEST: select SUBSTR(Name, INSTR(Name,' ',-1)+1, GREATEST(INSTR(Name,' '),0)) from NAME; A third method to accomplish the same end uses RTRIM. Remember that RTRIM eliminates everything specified in its set from the right side of a string until it encounters any character not in its set . The RTRIM here effectively erases all the letters on the right until it hits the first space (just before the last name) or reaches the beginning of the string: select SUBSTR(Name,LENGTH(RTRIM(NAME,'ABCDEFGHIJKLMNOPQRSTUVWXYZ')) +1) from NAME; SUBSTR(NAME,LENGTH(RTRIM( NELSON Chapter 16: Advanced Use of Functions and Variables 305 ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:305 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:24 PM Color profile: Generic CMYK printer profile Composite Default screen MEDICIS JOSEPHUS GAMMIERE LENGTH then measures the resulting string (with the last name erased). This tells you the position of the space before the last name. Add 1 to this number, do a SUBSTR starting there, and you’ll get just the last name. Let’s create a table with FirstName and LastName columns (you’ll see complete details on creating tables in Chapter 18): create table TWONAME( FirstName VARCHAR2(25), LastName VARCHAR2(25) ); Table created. Now, use an insert with a subquery to load the table data with the first and last names from the NAME table: insert into TWONAME (FirstName, LastName) select SUBSTR(Name,1,DECODE(INSTR(Name,' '),0,99,INSTR(Name,' '))), SUBSTR(Name,LENGTH(RTRIM(NAME,'ABCDEFGHIJKLMNOPQRSTUVWXYZ'))+1) from NAME; Check the contents of the TWONAME table: select * from TWONAME; FIRSTNAME LASTNAME HORATIO NELSON VALDO MARIE MEDICIS FLAVIUS JOSEPHUS EDYTHE GAMMIERE You can use similar techniques to extract the middle initial or initials, and apply these methods elsewhere as well, such as to addresses, product descriptions, company names, and so on. When moving data from an old system to a new one, reformatting is frequently necessary and often difficult. The facilities exist in SQL, but they require some knowledge of how the functions work and some thoughtfulness to avoid the kinds of difficulties shown in the examples so far in this chapter. 306 Part II: SQL and SQL*Plus ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:306 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:24 PM Color profile: Generic CMYK printer profile Composite Default screen Chapter 16: Advanced Use of Functions and Variables 307 ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:307 Counting String Occurrences Within Larger Strings You can use a combination of the LENGTH and REPLACE functions to determine how many times a string (such as ABC) occurs within a larger string (such as ABCDEFABC). The REPLACE function replaces a character or characters in a string with zero or more characters. Thus, REPLACE('ADAH', 'A', 'BLAH') will evaluate the string ADAH. Everywhere an A is found, it will be replaced with the string BLAH. Thus, the function shown in this example will return the string BLAHDBLAHH. You can use the REPLACE function to eliminate characters from strings. For example, you can replace the character string you’re searching for with NULL values. Thus, REPLACE('GEORGE', 'GE', NULL) returns a value of OR. The two separate occurrences of GE in GEORGE were each set to NULL. You can use the REPLACE function to determine how many times a string (like GE) is found in a larger string (like GEORGE). First, replace the string with NULL values: select REPLACE('GEORGE', 'GE', NULL) from DUAL; The result of that query will be OR: RE OR More importantly, the LENGTH of the result of that query will provide information about how many times the string was found. The following query will tell you how long the resulting string is: select LENGTH(REPLACE('GEORGE', 'GE', NULL)) from DUAL; LENGTH(REPLACE('GEORGE','GE',NULL)) 2 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:24 PM Color profile: Generic CMYK printer profile Composite Default screen Now you can tell how many times the string was found. If you subtract the length of the shortened string from the original string and divide that difference by the length of the search string, the result will be the number of times the search string was found: select (LENGTH('GEORGE') - LENGTH(REPLACE('GEORGE', 'GE', NULL)) ) / LENGTH('GE') AS Counter from DUAL; COUNTER 2 This example uses strings instead of character columns in order to be simpler to understand; in real applications, you would replace the original string (GEORGE) with your column name. The length of the string GEORGE is six characters. The length of GEORGE after GE is replaced with NULL is two characters. Therefore, the difference in the lengths of the original and shortened strings is four characters. Dividing four characters by the length of the search string (two characters) tells you that the string was found twice. The only problem with this method occurs when the search string is zero characters in length (since you cannot divide by zero). Searching for a zero-length string may indicate a problem with the application logic that initiated the search. Additional Facts About Variables The command accept forces SQLPLUS to define the variable as equal to the entered value, and it can do this with a text message, with control over the datatype entered, and even with the response blanked out from viewing (such as for passwords; see the entry for accept in the Alphabetical Reference). You can pass arguments to a start file when it is started by embedding numbered variables in the select statements (rather than variables with names). To select all of the checkout records between one date and another, the select statement might look like this: select * from BOOKSHELF_CHECKOUT where CheckOutDate BETWEEN '&1' AND '&2'; That query can be saved as a file (such as checkout.sql) and started from within SQLPLUS: start checkout.sql 01-JAN-02 31-JAN-02 The start file checkout.sql would begin, with 01-JAN-02 substituted for &1, and 31-JAN-02 substituted for &2. As with other variables, character and DATE datatypes must be enclosed in single quotation marks in the SQL statement. One limitation of this is that each argument following the word start must be a single word without spaces. Variable substitutions are not restricted to the SQL statement. The start file also may use variables for such things as SQLPLUS commands. 308 Part II: SQL and SQL*Plus ORACLE Series TIGHT / Oracle9 i : The Complete Reference / Loney, Koch / 222521-1 / Chapter 16 Blind Folio 16:308 P:\010Comp\Oracle8\521-1\CD\Ventura\book.vp Friday, July 19, 2002 4:12:25 PM Color profile: Generic CMYK printer profile Composite Default screen [...]... Chapter 8 for examples of the ROUND function Value in insert Statement Actual Value in Table For precision of NUMBER (4, 1) 123 .4 123 .4 123 .44 123 .4 123 .45 123.5 123 .44 5 123 .4 12 34. 5 insert fails For precision of NUMBER (4) 123 .4 123 123 .44 123 123 .45 123 123 .44 5 123 12 34. 5 1235 12 345 insert fails For precision of NUMBER (4, -1) 123 .4 120 123 .44 120 123 .45 120 123 .44 5 120 125 130 12 34. 5 1230 TABLE 18-1 Results... Value in insert Statement Actual Value in Table 12 345 insert fails For precision of NUMBER 123 .4 123 .4 123 .44 123 .44 123 .45 123 .45 123 .44 5 123 .44 5 125 125 12 34. 5 12 34. 5 12 345 .6789012 345 678 12 345 .6789012 345 678 TABLE 18-1 Results of inserts into NUMBER Datatype Columns (continued) Constraints in create table The create table statement lets you enforce several different kinds of constraints on a table:... 21-MAR-01 40 -1.3 3.6 22-JUN-01 101 .4 86.2 1.63 23-SEP-01 92.9 79.6 1.00003 22-DEC-01 -17 .4 -10 .4 2 .4 Look at the first insert statement The value for Noon is 39.99 In the query, it is rounded to 40 Midnight in the insert is -1.31 In the query it is -1.3 Oracle rounds the number based on the digit just to the right of the allowed precision Table 18-1 shows the effects of precision in... Composite Default screen Series TIGHT / Oracle9 i: The Complete Reference / Loney, Koch / 222521-1 / Chapter 18 Blind Folio 18:329 Chapter 18: Creating, Dropping, and Altering Tables and Views Note that the error here is caused by the 101, not the 44 , because NUMBER(3,1) leaves only two positions available to the left of the decimal point The 44 will not cause the “value larger than specified precision”... (BC.ReturnedDate-BC.CheckoutDate - 14) *0.20 ) LateFee Reading through the preceding listing, the first DECODE function tells Oracle that if the first five letters of the CategoryName are ‘ADULT’, then perform the second DECODE function The second DECODE tells Oracle that if letters 6, 7, and 8 of the CategoryName are ‘FIC’, then the late fee is $0.60 per day after the twenty-first day; otherwise, it is $0.30 per day after the twenty-first... table The point of these constraints is to get Oracle to do most of the work in maintaining the integrity of your database The more constraints you add to a table definition, the less work you have to do in applications to maintain the data On the other hand, the more constraints there are in a table, the longer it takes to update the data There are two ways to specify constraints: as part of the column... and so on, for virtually as many if/then pairs as you can construct If value equals none of the ifs, then the result of the DECODE is else Each of the ifs, thens, and the else also can be a column or the result of a function or computation You can have up to 255 elements within the parentheses Let’s consider the checkout history for the bookshelf, as recorded in the BOOKSHELF_ CHECKOUT table: column... 101 .44 , 86.2, 1.63); insert into TROUBLE values ('PLEASANT LAKE','23-SEP-01', 92.85, 79.6, 1.00003); insert into TROUBLE values ('PLEASANT LAKE','22-DEC-01', -17 .44 5, -10 .4, 2 .4) ; select * from TROUBLE; P:\010Comp \Oracle8 \521-1\CD\Ventura\book.vp Friday, July 19, 2002 4: 12:33 PM 329 ORACLE Color profile: Generic CMYK printer profile Composite Default screen 330 Series TIGHT / Oracle9 i: The Complete Reference. .. -THE SHIPPING NEWS 59 THE DISCOVERERS 48 WEST WITH THE NIGHT 48 WONDERFUL LIFE 31 THE MISMEASURE OF MAN 31 JOHN ADAMS 28 INNUMERACY 21 THE MISMEASURE OF MAN 20 TRUMAN 19 ANNE OF GREEN GABLES 18 MY LEDGER 16 MIDNIGHT MAGIC 14 POLAR EXPRESS 14 GOOD DOG, CARL 14 TO KILL A MOCKINGBIRD 14 HARRY POTTER AND THE 11 GOBLET OF FIRE PAT LAVAY THE SHIPPING NEWS 10 DORAH TALBOT EITHER/OR 8 GERHARDT... CHILDRENFIC 5 5 CHILDRENPIC 14 14 P:\010Comp \Oracle8 \521-1\CD\Ventura\book.vp Friday, July 19, 2002 4: 12:27 PM ORACLE Color profile: Generic CMYK printer profile Composite Default screen Series TIGHT / Oracle9 i: The Complete Reference / Loney, Koch / 222521-1 / Chapter 17 Blind Folio 17:315 Chapter 17: DECODE and CASE: if, then, and else in SQL The output now shows the borrowers across the top—Fred Fuller’s . number times another, and so on. value is tested for each row. If value equals if1 , then the result of the DECODE is then1 ; if value equals if2 , then the result of the DECODE is then2 ; and so. as many if/then pairs as you can construct. If value equals none of the if s, then the result of the DECODE is else . Each of the if s, then s, and the else also can be a column or the result. SHIPPING NEWS 12-MAR-02 59.00 45 .00 9 THE DISCOVERERS 01-MAR-02 48 .00 34. 00 6.8 WEST WITH THE NIGHT 01-MAR-02 48 .00 34. 00 6.8 For books in the Adult categories, increase the allowed days late to 21