Try to use regular expressions to convert names in title case

I'm changing name to their appropriate case of the most common names in North America (especially the United States).
Some examples are in the comments of the code provided below.
My problem is that * retName = retName.replaceAll ("([^]) ([^] +)", "$1".toUpperCase () + "$2" ""); * does not work as I expect. It seems that calling toUpperCase method actually doesn't do anything for the identified group.
Everything else works as I hope.

I hope I did not browse through each character of the string, mounted the characters that follow the spaces.

All your RegEx expert help will be appreciated.

{code}
/**
* Converts names in some random case in name business. This method does not have the
* additional processing that would be required for the conversion of addresses.
* This method does not add or remove punctuation.
*
* Examples:
* DAN MARINO-> Dan Marino
' * old macdonald--> Old Macdonald & lt;-can capitalize it had "because of Ernst Mach
* ROY BLOUNT, JR.-> Roy Blount, Jr.
* CAROL mosely-BrAuN-> Carol Mosely-Braun
* Tom Jones-> Tom Jones
* ST. LOUIS-St. Louis >
* ST. LOUIS, MO-> St. Louis, Mo & lt;-avoid the names of cities and State Codes
*
* This is a work in progress that will have to be updated of new exceptions being proven.
*/
public static String toNameCase (String name) {}
/*
* Basic plan:
* 1. Strategically creating spaces double face of characters to capitalize
* 2. Take advantage of the characters with whom the above spaces
* 3. Remove double spaces.
*/
The string all lowercase
String retName = name.trim () .toLowerCase ();

Chains of collapse of spaces for simple spaces
retName = retName.replaceAll ("[] +","");

names of 'mc '.
retName = retName.replaceAll ("(mc) ', '$1'");

Ensure that there is a space after periods and commas
(retName_=_retName.replaceAll("(\\.|,) (^)", "$1 $2");

Add 2 spaces after points, commas, hyphens and apostrophes
(retName_=_retName.replaceAll("(\\.|,|-|') ', '$1');

Add a double space in front of the chain
retName = "" + retName;

MECHANISM of each character is preceded by a space
For some reason that it does not work
retName = retName.replaceAll ("([^]) ([^] +)", "$1".toUpperCase () + "$2" "");

Remove double spaces
retName = retName.replaceAll ("", "");

Return retName;
}

Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:56 AM

Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:57 AM

It looks like you want the equivalent to Java

echo -e "DAN MARINO\nold macdonald\nROY BLOUNT, JR.\nCAROL mosely-BrAuN\nTom Jones\nST.LOUIS\nST.LOUIS, MO"|perl -pe's#\b(\w+)\b#\u\L$1#g'

Unfortunately, the implementation of Java does not support

The pre-treatment operations \l \u, \L and \U.

http://download.Oracle.com/javase/6/docs/API/Java/util/regex/pattern.html#JCC

Tags: Java

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Hello

user12240205 wrote:

We get a response XML from a WEB SERVICE.

I convert it to VARCHAR2.

Why? XML has its own native ways of analysis; Why not use them?

If you use regular expressions, then REGEXP_REPLACE, as shown above, is a good option in Oracle 10, but starting in Oracle 11.1, you can use REGEXP_SUBSTR like this:

REGEXP_SUBSTR (v_1

, '(.+?)'

1

1

NULL

1

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The 6th argument is like a backreference; "He tells REGEXP_SUBSTR did not return to the entire organization, but only the part inside the 1st left '(' et correspondant à sa droite).

The '?' to make it non-greedy is necessary only if v_1 can contain more than one response.

Now, I want to get the real answer that is inside the tag of the response.

I tried this:

DECLARE

V_1 VARCHAR2 (30000): = "" |

'Good';

v_response VARCHAR2 (100);

BEGIN

DBMS_OUTPUT. PUT_LINE (V_1);

v_response: = REGEXP_SUBSTR (v_1, ' /(. +?)) <\ esponse="">/');

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That's because it's looking for a slash ("/") before the '' tag and another after the ' tag.

The backslash ("\") is not necessary here, but it is not nothing wrong.

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Hello

Game 6' - 8 "wrote:
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Published by: release 6' - 8 "August 11, 2009 20:04

Serving what you ordered!

Originally, you said that you are looking for something that produces the same result as

where   c1 in (8, 1)
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in other words, the c1s could be coupled with any of the c2s.
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where     (     c1 = 8
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in other words, c1 = 8 and c2 = 1250 is not good; is not c1 = 1 and c2 = 1706.

In this case, try

WHERE     REGEXP_LIKE ( s
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The input string: 01| ^ | ABCD| ^ | 111| * | 02| ^ | efgh| ^ | 222

Separators: | * | is divided into lines, | ^ | is divided into columns

Expected result:

col1 col2 col3

Row1 - > 01 abcd 111

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Simply put, take a next

The input string: 01| ^ | ABCD |^| 111 |*| 02 |^| efgh |^| 222

Separator: | * |

Result:

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02. ^ | efgh | ^ | 222

How can I achieve this using regular expressions?
Kind regards
Euntaek

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/

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-------------------------------------
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STR                                  
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