Check that a string is a string

This may seem like a trivial question. I want to check that a string is a string. I thought to use a string method on a variable in a function and catch the exception. I did a little research around Java Platform SE 7. This seems to be a strange thing to do and I thought everything stable and if they fail, it must be a string, but I'd like something specific.

832844 wrote:
It's the thought. I wrote a recursive descent parser and verify a constant-> int | text | float

Ah. So what you want is to find out if your string contains an integer, float or whatever.

Assuming I'm right, and that the 'text' is equivalent to 'none of the above', can help the Java wrappers:

public enum Type {INTEGER, FLOAT, TEXT);

public boolean isType(String text, Type type) {
   try {
      switch (type) {
         case INTEGER:
            long i = Long.parseLong(text);
            break;
         case FLOAT:
            double d = Double.parseDouble(text);
            break;
         case TEXT:
            return true;
         default:
            throw new IllegalArgumentException();
      }
   } catch (Exeption) {
      return false;
   }
   return true;
}

Then just run above against all types except TEXT (or TEXT).

It is likely to be slow, but its extensible for any type of "analysable" you want to check for, and you can do as specific as you want (the above example is rather crude).
And I think that it is fairly easy to follow; at least I hope so.

Winston

Tags: Java

Similar Questions

Check that a string contains data only numbers

Hello

I'm trying to find an elegant way to test an input string "serial number" format A00000000 (where A can be any letter and 0 can be a number any). My first thought was to check the following features:

(1) total string length = 9.

(2) using the "Decimal String to the Number.vi", result = 0 (which indicates that the string contains no numbers).

(3) using the 'Decimal string to Number.vi' with an offset of 1, the result is a number.

Unfortunately, that does not differentiate between the following strings:

A00005123 valid

Invalid A5123

Invalid A5123B123

Invalid A005123BC

All the world is this recently? Is there a way to analyze non-specific chips (for example separating the groups of letters, numbers or other?) Is there a complete way to do this without breaking the chain in 9 individual items and do a comparison of 26 item for the letter and 10 element for each number comparisons?

Thank you

Sean

You can use a regular expression to match the pattern.

How do I check the particluar string in the xml data
Dear oracle Experts.
I'm using the following oracle database.

Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
PL/SQL Release 10.2.0.4.0 - Production
CORE 10.2.0.4.0 Production
AMT for Linux: release 10.2.0.4.0 - Production
NLSRTL Version 10.2.0.4.0 - Production

I have a following xml data...
p_msg_in CLOB: =.
< DATA >
< FLD FNR = DTE = '26 SEP 12' "ZZ0584" SRC = "DXB" OF the "DAC" = / >
< AAA LBP = ETK '22334455' = ' 1234567/4' ACT = "123" / >
< AAA LBP = ETK "223344" = "2345678/1" / >
< AAA LBP = ETK "223344" = ' 1/123456' ACT = "345" / >
< / DATA >

then im go get the header like this information...

v_msg_xml: = xmltype (p_msg_in);

I'm IN multicur (v_msg_xml, ' / DATA/FLD ') LOOP
V_1: = i.xml.extract('//@FNR').getstringval ();
v_4: = to_date(i.xml.extract('//@DTE').getstringval(),'DDMONYY');
v_5: = i.xml.extract('//@SRC').getstringval ();
v_6: = i.xml.extract('//@DES').getstringval ();
END LOOP;

After that, I need all the real records one by one, use it for a loop in a loop. Here for each iteration, I should check that a string of characters (for example law) is there or not. in the example above, 1.3 records have value of law in it. Well, I need to check something like that. If instr ("< AAA LBP = ETK"223344 "=" 123456/1 "ACT =" 345 "/ > ',' law") > 0 before performing the step below.
How to get there.
I appreciate your help.
Thank you.
C IN multicur (v_msg_xml, ' / DATA/AAA ') LOOP
v_7: = c.xml.extract('//@LBP').getstringval ();
v_8: = c.xml.extract('//@ETK').getstringval ();
If I understood corrrectly you must give the LBP and values ETK tags AAA who attribute the Act. If so, all you need is:
```
declare
    p_msg_in CLOB :=
'




';
    v_msg_xml xmltype;
    cursor v_cur
      is
        select  lbp,
                etk
          from  xmltable(
                         '/DATA/AAA[@ACT]'
                         passing v_msg_xml
                         columns
                           lbp varchar2(10) path '@LBP',
                           etk varchar2(10) path '@ETK'
                        );
begin
    v_msg_xml := xmltype(p_msg_in);
    for v_rec in v_cur loop
      dbms_output.put_line('LBP = "' || v_rec.lbp || '" ETK = "' || v_rec.etk || '"');
    end loop;
end;
/
LBP = "22334455" ETK = "1234567/4"
LBP = "223344" ETK = "123456/1"

PL/SQL procedure successfully completed.

SQL>
```
SY.

could not find a server reports and analysis hyperion running on localhost port 6800. Please check your connection string server and verify that the server is

Hi all
We have properly installed and configured Hyperion Reporting and analysis 11.1.2.4 in windows 2012 server however when I try to logint to the workspace, I get an error like "start-up of the specified document does not exist in the repository." "Select a new start under the General Preferences tab document" monitoring of "could not find a server reports and analysis hyperion running on localhost at port 6800. Please check your connection string server and verify that the server is up '.

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com.brio.one.services.globalservicemanager.GSMException: GsmNotFound
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at weblogic.t3.srvr.SubsystemRequest.run(SubsystemRequest.java:64)
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In the workspace, it shows that the Service is pending.
EMP Diagnostic Report:
Can help kindly out how should I solve this problem.
I restarted the services and rebooted the Machine but still getting the same error.
Kind regards
VJ

Hello

I've reconfigured the Reporting, analysis, and everything by, I chose "Drop table" and the issue was resolved.

Kind regards

VJ

Check if the string is a number
I need to check if a string is entered in a textinput control is not a number and do something. I tried

If ((object.text is Number) == false) {}
error = true;
}

But this does not work, I tried google, but didn't find anything useful.
I think I'll go with the regex solution, even if it still does not work: s

It always returns true even if it has an 'a' or so in it. It seems to only do that when the 'a' is NOT in front of the string.

EDIT: found on google: var expression: RegExp = / ^ [0-9] + $/;

Thank you for you answer and help me :)

Check if a string is a number
Is there a way to check if a string value can be converted to a number?

I try to add a filter to a report that will check when the data in a column is a number. This column can contain either numbers or strings. For example: I don't want to see '423' and '6562' but I wouldn't see 'ABC' or '23TSDF '.

I'm not aware of any type of isInt() of the functions in the report designer - or maybe try a CAST statement and being able to catch the thrown error?

Thank you
Hello

You can try to use this expression to check whether it is digital or Non-digital

BOX WHEN EVALUATE ('LENGTH (TRIM (TRANSLATE (%1, %2, %3)))' DOUBLE PRECISION)
'' Activity Metrics. '' "" activities ","+-. 0123456789', ' ') IS NULL THEN 'Numeric' OTHER
"Non-Numeric" END

Replace "activity Metrics." "" activities "with your own column that you want to check

It may be useful

-John CRMIT

How to check if a string exists in varray or not

Hi all

How to check if a string exists in varray or not.

Version Details 


BANNER
----------------------------------------------------------------
Oracle Database 10g Express Edition Release 10.2.0.1.0 - Product
PL/SQL Release 10.2.0.1.0 - Production
CORE    10.2.0.1.0      Production
TNS for 32-bit Windows: Version 10.2.0.1.0 - Production
NLSRTL Version 10.2.0.1.0 - Production

SQL> get test
  1  DECLARE
  2     TYPE dnames_var IS VARRAY(7) OF VARCHAR2(30);
  3     dept_names dnames_var := dnames_var('Shipping','Sales','Finance','Payroll');
  4  BEGIN
  5    if dept_names.exists('Shipping')
  6    then
  7       dbms_output.put_line('Exists ................');
  8    end if ;
  9  /*   DBMS_OUTPUT.PUT_LINE('dept_names has ' || dept_names.COUNT
 10                          || ' elements now');
 11     DBMS_OUTPUT.PUT_LINE('dept_names''s type can hold a maximum of '
 12                           || dept_names.LIMIT || ' elements');
 13     DBMS_OUTPUT.PUT_LINE('The maximum number you can use with '
 14         || 'dept_names.EXTEND() is ' || (dept_names.LIMIT - dept_names.COUNT));
 15  */
 16* END;
 17  
 18  /
DECLARE
*
ERROR at line 1:
ORA-06502: PL/SQL: numeric or value error: character to number conversion error
ORA-06512: at line 5


SQL> 

Any help in this regard is appreciated ...

Thank you
Prakash P

Published by: prakash on April 29, 2012 05:42

EXISTS checks for the existence of an element, not a value. Since you're using VARRAY (btw, it is not recommended), your only choice is a loop through:

SQL> DECLARE
  2       TYPE dnames_var IS VARRAY(7) OF VARCHAR2(30);
  3       dept_names dnames_var := dnames_var('Shipping','Sales','Finance','Payroll');
  4  BEGIN
  5      for i in 1..dept_names.count loop
  6        if dept_names(i) = 'Shipping'
  7          then
  8            dbms_output.put_line('Exists ................');
  9            exit;
 10        end if;
 11      end loop;
 12  END;
 13  /
Exists ................

PL/SQL procedure successfully completed.

SQL>

If you use the nested table, you can use MEMBER, SUMBULTISET, MULTISET EXCEPT:

SQL> DECLARE
  2       TYPE dnames_var IS TABLE OF VARCHAR2(30);
  3       dept_names dnames_var := dnames_var('Shipping','Sales','Finance','Payroll');
  4  BEGIN
  5      if 'Shipping' member of dept_names
  6      then
  7         dbms_output.put_line('Exists ................');
  8      end if ;
  9  END;
 10  /
Exists ................

PL/SQL procedure successfully completed.

SQL> DECLARE
  2       TYPE dnames_var IS TABLE OF VARCHAR2(30);
  3       dept_names dnames_var := dnames_var('Shipping','Sales','Finance','Payroll');
  4  BEGIN
  5      if dnames_var('Shipping') submultiset dept_names
  6      then
  7         dbms_output.put_line('Exists ................');
  8      end if ;
  9  END;
 10  /
Exists ................

PL/SQL procedure successfully completed.

SQL> DECLARE
  2       TYPE dnames_var IS TABLE OF VARCHAR2(30);
  3       dept_names dnames_var := dnames_var('Shipping','Sales','Finance','Payroll');
  4  BEGIN
  5      if dnames_var('Shipping') multiset except dept_names = dnames_var()
  6      then
  7         dbms_output.put_line('Exists ................');
  8      end if ;
  9  END;
 10  /
Exists ................

PL/SQL procedure successfully completed.

SQL>

SY.

Check that this class has been marked with the annotation @Entity
Hello

We have a level of the shared lib App that makes all the APP associated with request for an application. And each application can use the lib goes to get the related JPA query to do.
To define the scope of the Joint Parliamentary Assembly between App. We have created the EMF for each application.

When an application to run and any associated APP action that it works fine, but when trying to run a second application we are seeing the following error message when you try to do a specific action JPA.

the @Entity annotation.
[2012 05-23 T 04: 08:16.839 - 07:00] [WC_Spaces] [ERROR] [] [oracle.webcenter.spaces] [tid: [ASSETS].] [ExecuteThread: '3' for the queue: "(self-adjusting) weblogic.kernel.Default"] [ecid: 5825b814-2931-4ad5-8dc3-3e18f66992b7-00000004,0] [APP: webcenterCustom] []
java.lang.IllegalArgumentException: unknown entity bean class: oracle.webcenter.spaces.internal.repository.WcSpaceHeader class, please check that this class has been marked with the @Entity annotation.
at org.eclipse.persistence.internal.jpa.EntityManagerImpl.find(EntityManagerImpl.java:648)
at org.eclipse.persistence.internal.jpa.EntityManagerImpl.find(EntityManagerImpl.java:532)
at oracle.webcenter.spaces.internal.repository.SpaceRepositoryUtils.refreshSpaceRows(SpaceRepositoryUtils.java:1791)

Where WcSpaceHeader is an entity.

To create emf by App, this is the code

private static EntityManagerFactory getEntityManagerFactory()
{
String appName = Utility.getApplicationName ();
EntityManagerFactory emf = sEntityMgrFactory.get (appName);

if(EMF==null)
{
EMF = Persistence.createEntityManagerFactory ("SpacesReposPUnit");
sEntityMgrFactory.put (appName, emf);
SpacesConstants.LOGGER.info ("Caching" +)
("EMF for" + appName);
}

return of the emf;
}

My persistence.xml

<? XML version = "1.0" encoding = "US-ASCII"? >
< persistence xmlns: xsi = "http://www.w3.org/2001/XMLSchema-instance".
xsi: schemaLocation = "http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd".
version = "1.0" xmlns = "http://java.sun.com/xml/ns/persistence" >
< name of persistence - unit = "SpacesReposPUnit" transaction-type = "RESOURCE_LOCAL" >
<>provider
org.eclipse.persistence.jpa.PersistenceProvider
< / provider >
oracle.webcenter.spaces.internal.repository.WcSpaceHeader < class > < / class >
oracle.webcenter.spaces.internal.repository.WcSpaceUsrDetail < class > < / class >
oracle.webcenter.framework.service.jpa.WcCommonXlationEntity < class > < / class >
oracle.webcenter.spaces.internal.repository.WCNavigationActivity < class > < / class >
Properties of <>
< name = "eclipselink.session.customizer property"
value="Oracle.WebCenter.spaces.internal.repository.SpacesEclipselinkSessionCustomizer"/ >
< / properties >
< / persistence - unit >
< / persistence >

Another problem I see in this is that. When I create an EMF EM, (which is different for different app) and do a query, I get a class cast Exception.

[(self-adjusting)'] [ecid: 5825b814-2931-4ad5-8dc3-3e18f66992b7-00000004,0] [APP: webcenterCustom] []
java.lang.IllegalArgumentException: unknown entity bean class: oracle.webcenter.spaces.internal.repository.WcSpaceHeader class, please check that this class has been marked with the @Entity annotation.
at org.eclipse.persistence.internal.jpa.EntityManagerImpl.find(EntityManagerImpl.java:648)
at org.eclipse.persistence.internal.jpa.EntityManagerImpl.find(EntityManagerImpl.java:532)
It is a problem of class loader, somehow you have deployed the same classes in two places, have two versions of the classes.

How, exactly made things, where are your domain classes?

The questions is more likely that,
Persistence.createEntityManagerFactory ("SpacesReposPUnit");
always returns the same plant once it has been deployed.

If you pass a map properties for createEntityManagerFactory and set the property "name eclipselink.session" a single value, then you should get a new plant.

There seems to be a display problem. Please check that your monitor is HDCP and is not mirrored by using Airplay.

There seems to be a display problem. Please check that your monitor is HDCP and is not mirrored by using Airplay.

Apparently, you're trying to use AirPlay to display content that is protected in a way that not only allows the copy.

How can I check that my Iphone 5 is unlocked

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Two ways-

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ieSpell is for Internet Explorer and various shells for IE.

I guess you're a fan of the AMC series.

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Anyone know of one?

https://addons.Mozilla.org/en-us/Firefox/addon/CheckPlaces/

My Mac has said that I don't have the necessary permissions, (although I repaired permissions and checked that I am an administrator of this computer) for the upgrade - what can you tell me what the problem is? Mac 10.5.8

My Mac has said that I don't have the necessary permissions, (although I repaired permissions and checked that I am an administrator of this computer) for the upgrade - what can you tell me what the problem is? Mac 10.5.8

This has happened

Each time Firefox opened

I tried to upgrade to version 2.0

User Agent

Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_5_8; en - us) AppleWebKit/533.16 (KHTML, like Gecko) Safari/533.16 Version/5.0

See http://kb.mozillazine.org/Installing_Firefox#Mac_OS_X
.....
If you have Mac OS X 10.5 or newer then follow these steps:

Download a new copy of the Firefox program: http://www.mozilla.com/firefox/all.html
Trash the current demand for Firefox to do a clean reinstall.
Install the new version you downloaded.
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Qosmio X 770-107 - check that the 4 GB of NAND flash is optimized OK for Win7?

Hello

I put the Bios on the laptop Qosmio of X 770-107 to use Fast Boot, but I think that it is other parameters for the use of the Flash 4 GB on HDD1.

How to check that this setting is configured OK?

Thank you
Kirk

What are the settings do you mean exactly?
I think you can change only those settings that are available in the BIOS directly.

I can't install it last update for itunes on my pc. When I try and download it and it's over, he said, "Apple mobile device service couldn't start. Check that you have sufficient privileges to start system services. »

I already tried uninstalling and reinstalling several times and that does not work, I simply get the same result.

I start to download, through all the steps and when it is ready to be finished install it says in a pop-up, "Apple mobile device service couldn't start. Check that you have sufficient privileges to start system services. »

Right-click on the downloaded installer and use run as administrator.

For general advice, see troubleshooting problems with iTunes for Windows updates.

The steps described in the second case are a guide to remove everything related to iTunes and then rebuild what is often a good starting point, unless the symptoms indicate a more specific approach.

Review the other boxes and other support documents list to the bottom of the page, in case one of them applies.

More information area has direct links with the current and recent buildings if you have problems to download, must revert to an older version or want to try the version of iTunes for Windows (64-bit-for old video cards) as a workaround for problems with installation or operation, or compatibility with QuickTime software or a third party.

Backups of your library and device should be affected by these measures but there are links to backup and recovery advice there.

TT2

Check that a string is a string

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