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  • NHibernate / ORM - Child Update over Web Service

    - by tyndall
    What is the correct way to UPDATE a child object with NHibernate but not have to "awake" the parent object. Lets say you would like to try to avoid this because the parent object is large or expensive to initiate. Lets assume classes are called Author(parent) and Book(child). (still, trying to avoid instantiating Author) Book comes back over a web service as XML. It gets deserialized back into a CLR object. Book has an AuthorId property which allows this to happen. But it also has a Author property. Problem, comes when you try to SaveOrUpdate() Book and the author_id in the database gets wiped out because the Author was null when the object gets deserialized. This seems like this would be a common problem. What is the workaround? Also, if you instantiate the Author and it has a Books property. The book you are trying to update is already one of these books (List<Book>). We have also run into the "a different object with the same identifier value was already associated with the session" problems. What is the standard process to update a child over a web service?

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  • Trying to import SQL file in a xampp server returns error

    - by Victor_J_Martin
    I have done a ER diagram in Mysql Workbench, and I am trying load in my server with phpMyAdmin, but it returns me the next error: Error SQL Query: -- ----------------------------------------------------- -- Table `BDA`.`UG` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`UG` ( `numero_ug` INT NOT NULL, `nombre` VARCHAR(45) NOT NULL, `segunda_firma_autorizada` VARCHAR(45) NOT NULL, `fecha_creacion` DATE NOT NULL, `nombre_depto` VARCHAR(140) NOT NULL, `dni` INT NOT NULL, `anho_contable` INT NOT NULL, PRIMARY KEY (`numero_ug`), INDEX `nombre_depto_idx` (`nombre_depto` ASC), INDEX `dni_idx` (`dni` ASC), INDEX `anho_contable_idx` (`anho_contable` ASC), CONSTRAINT `nombre_depto` FOREIGN KEY (`nombre_depto`) REFERENCES `BDA`.`Departamento` (`nombre_depto`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `dni` FOREIGN KEY (`dni`) REFERENCES `BDA`.`Trabajador` (`dni`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `anho_contable` FOREIGN KEY (`anho_contable`) REFERENCES `BDA`.`Capitulo_Contable` (`anho_contable`) [...] MySQL said: Documentation #1022 - Can't write; duplicate key in table 'ug' I export the result of the diagram from Mysql Workbench to a SQL file, and this file is what I'm trying to upload. This is the file. I can not find the duplicate key. SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0; SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0; SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL,ALLOW_INVALID_DATES'; CREATE SCHEMA IF NOT EXISTS `BDA` DEFAULT CHARACTER SET utf8 COLLATE utf8_general_ci ; USE `BDA` ; -- ----------------------------------------------------- -- Table `BDA`.`Departamento` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Departamento` ( `nombre_depto` VARCHAR(140) NOT NULL, `area_depto` VARCHAR(140) NOT NULL, PRIMARY KEY (`nombre_depto`)) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Trabajador` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Trabajador` ( `dni` INT NOT NULL, `direccion` VARCHAR(140) NOT NULL, `nombre` VARCHAR(45) NOT NULL, `apellidos` VARCHAR(140) NOT NULL, `fecha_nacimiento` DATE NOT NULL, `fecha_contrato` DATE NOT NULL, `titulacion` VARCHAR(140) NULL, `nombre_depto` VARCHAR(45) NOT NULL, PRIMARY KEY (`dni`), INDEX `nombre_depto_idx` (`nombre_depto` ASC), CONSTRAINT `nombre_depto` FOREIGN KEY (`nombre_depto`) REFERENCES `BDA`.`Departamento` (`nombre_depto`) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Capitulo_Contable` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Capitulo_Contable` ( `anho_contable` INT NOT NULL, `numero_ug` INT NOT NULL, `debe` DOUBLE NOT NULL, `haber` DOUBLE NOT NULL, PRIMARY KEY (`anho_contable`), INDEX `numero_ug_idx` (`numero_ug` ASC), CONSTRAINT `numero_ug` FOREIGN KEY (`numero_ug`) REFERENCES `BDA`.`UG` (`numero_ug`) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`UG` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`UG` ( `numero_ug` INT NOT NULL, `nombre` VARCHAR(45) NOT NULL, `segunda_firma_autorizada` VARCHAR(45) NOT NULL, `fecha_creacion` DATE NOT NULL, `nombre_depto` VARCHAR(140) NOT NULL, `dni` INT NOT NULL, `anho_contable` INT NOT NULL, PRIMARY KEY (`numero_ug`), INDEX `nombre_depto_idx` (`nombre_depto` ASC), INDEX `dni_idx` (`dni` ASC), INDEX `anho_contable_idx` (`anho_contable` ASC), CONSTRAINT `nombre_depto` FOREIGN KEY (`nombre_depto`) REFERENCES `BDA`.`Departamento` (`nombre_depto`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `dni` FOREIGN KEY (`dni`) REFERENCES `BDA`.`Trabajador` (`dni`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `anho_contable` FOREIGN KEY (`anho_contable`) REFERENCES `BDA`.`Capitulo_Contable` (`anho_contable`) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Cliente` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Cliente` ( `cif_cliente` INT NOT NULL, `nombre_cliente` VARCHAR(140) NOT NULL, PRIMARY KEY (`cif_cliente`)) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Ingreso` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Ingreso` ( `id` INT NOT NULL, `concepto` VARCHAR(45) NOT NULL, `importe` DOUBLE NOT NULL, `fecha` DATE NOT NULL, `cif_cliente` INT NOT NULL, `numero_ug` INT NOT NULL, PRIMARY KEY (`id`), INDEX `cif_cliente_idx` (`cif_cliente` ASC), INDEX `numero_ug_idx` (`numero_ug` ASC), CONSTRAINT `cif_cliente` FOREIGN KEY (`cif_cliente`) REFERENCES `BDA`.`Cliente` (`cif_cliente`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `numero_ug` FOREIGN KEY (`numero_ug`) REFERENCES `BDA`.`UG` (`numero_ug`) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Proveedor` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Proveedor` ( `cif_proveedor` INT NOT NULL, `nombre_proveedor` VARCHAR(140) NOT NULL, PRIMARY KEY (`cif_proveedor`)) ENGINE = InnoDB; -- ----------------------------------------------------- -- Table `BDA`.`Gasto` -- ----------------------------------------------------- CREATE TABLE IF NOT EXISTS `BDA`.`Gasto` ( `id` INT NOT NULL, `concepto` VARCHAR(45) NOT NULL, `importe` DOUBLE NOT NULL, `fecha` DATE NOT NULL, `factura` INT NOT NULL, `cif_proveedor` INT NOT NULL, `numero_ug` INT NOT NULL, PRIMARY KEY (`id`), INDEX `cif_proveedor_idx` (`cif_proveedor` ASC), INDEX `numero_ug_idx` (`numero_ug` ASC), CONSTRAINT `cif_proveedor` FOREIGN KEY (`cif_proveedor`) REFERENCES `BDA`.`Proveedor` (`cif_proveedor`) ON DELETE NO ACTION ON UPDATE NO ACTION, CONSTRAINT `numero_ug` FOREIGN KEY (`numero_ug`) REFERENCES `BDA`.`UG` (`numero_ug`) ON DELETE NO ACTION ON UPDATE NO ACTION) ENGINE = InnoDB; SET SQL_MODE=@OLD_SQL_MODE; SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS; SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS; Thanks for your advices.

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  • How can I decrypt encrypted files using a PEM private key?

    - by Phil Cole
    I have files which have either been encrypted with a public key and the Blowfish algorithm, or a public key and the AES-256 algorithm. I'm looking to put together a Perl script that would be able to use the private keys (which I do have) to decrypt the files. The public and private key files are all in PEM format, and while I can find ways of reading the PEM files, and ways of decrypting data with a key, I haven't yet found a way of going from PEM - key. Any suggestions?

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  • JPA @OneToMany and composite PK

    - by Fleuri F
    Good Morning, I am working on project using JPA. I need to use a @OneToMany mapping on a class that has three primary keys. You can find the errors and the classes after this. If anyone has an idea! Thanks in advance! FF javax.persistence.PersistenceException: No Persistence provider for EntityManager named JTA_pacePersistence: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory: javax.persistence.PersistenceException javax.persistence.PersistenceException: Exception [TOPLINK-28018] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.EntityManagerSetupException Exception Description: predeploy for PersistenceUnit [JTA_pacePersistence] failed. Internal Exception: Exception [TOPLINK-7220] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.ValidationException Exception Description: The @JoinColumns on the annotated element [private java.util.Set isd.pacepersistence.common.Action.permissions] from the entity class [class isd.pacepersistence.common.Action] is incomplete. When the source entity class uses a composite primary key, a @JoinColumn must be specified for each join column using the @JoinColumns. Both the name and the referenceColumnName elements must be specified in each such @JoinColumn. at oracle.toplink.essentials.internal.ejb.cmp3.EntityManagerSetupImpl.predeploy(EntityManagerSetupImpl.java:643) at oracle.toplink.essentials.ejb.cmp3.EntityManagerFactoryProvider.createEntityManagerFactory(EntityManagerFactoryProvider.java:196) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:110) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:83) at isd.pacepersistence.common.DataMapper.(Unknown Source) at isd.pacepersistence.server.MainServlet.getDebugCase(Unknown Source) at isd.pacepersistence.server.MainServlet.doGet(Unknown Source) at javax.servlet.http.HttpServlet.service(HttpServlet.java:718) at javax.servlet.http.HttpServlet.service(HttpServlet.java:831) at org.apache.catalina.core.ApplicationFilterChain.servletService(ApplicationFilterChain.java:411) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:290) at org.apache.catalina.core.StandardContextValve.invokeInternal(StandardContextValve.java:271) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:202) There is the source code of my classes : Action : @Entity @Table(name="action") public class Action { @Id @GeneratedValue(strategy=GenerationType.IDENTITY) private int num; @ManyToOne(cascade= { CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH }) @JoinColumn(name="domain_num") private Domain domain; private String name; private String description; @OneToMany @JoinTable(name="permission", joinColumns= { @JoinColumn(name="action_num", referencedColumnName="action_num", nullable=false, updatable=false) }, inverseJoinColumns= { @JoinColumn(name="num") }) private Set<Permission> permissions; public Action() { } Permission : @SuppressWarnings("serial") @Entity @Table(name="permission") public class Permission implements Serializable { @EmbeddedId private PermissionPK primaryKey; @ManyToOne @JoinColumn(name="action_num", insertable=false, updatable=false) private Action action; @ManyToOne @JoinColumn(name="entity_num", insertable=false, updatable=false) private isd.pacepersistence.common.Entity entity; @ManyToOne @JoinColumn(name="class_num", insertable=false, updatable=false) private Clazz clazz; private String kondition; public Permission() { } PermissionPK : @SuppressWarnings("serial") @Entity @Table(name="permission") public class Permission implements Serializable { @EmbeddedId private PermissionPK primaryKey; @ManyToOne @JoinColumn(name="action_num", insertable=false, updatable=false) private Action action; @ManyToOne @JoinColumn(name="entity_num", insertable=false, updatable=false) private isd.pacepersistence.common.Entity entity; @ManyToOne @JoinColumn(name="class_num", insertable=false, updatable=false) private Clazz clazz; private String kondition; public Permission() { }

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  • JPA @ManyToOne and composite PK

    - by Fleuri F
    Good Morning, I am working on project using JPA. I need to use a @ManyToOne mapping on a class that has three primary keys. You can find the errors and the classes after this. If anyone has an idea! Thanks in advance! FF javax.persistence.PersistenceException: No Persistence provider for EntityManager named JTA_pacePersistence: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory: javax.persistence.PersistenceException javax.persistence.PersistenceException: Exception [TOPLINK-28018] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.EntityManagerSetupException Exception Description: predeploy for PersistenceUnit [JTA_pacePersistence] failed. Internal Exception: Exception [TOPLINK-7220] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.ValidationException Exception Description: The @JoinColumns on the annotated element [private java.util.Set isd.pacepersistence.common.Action.permissions] from the entity class [class isd.pacepersistence.common.Action] is incomplete. When the source entity class uses a composite primary key, a @JoinColumn must be specified for each join column using the @JoinColumns. Both the name and the referenceColumnName elements must be specified in each such @JoinColumn. at oracle.toplink.essentials.internal.ejb.cmp3.EntityManagerSetupImpl.predeploy(EntityManagerSetupImpl.java:643) at oracle.toplink.essentials.ejb.cmp3.EntityManagerFactoryProvider.createEntityManagerFactory(EntityManagerFactoryProvider.java:196) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:110) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:83) at isd.pacepersistence.common.DataMapper.(Unknown Source) at isd.pacepersistence.server.MainServlet.getDebugCase(Unknown Source) at isd.pacepersistence.server.MainServlet.doGet(Unknown Source) at javax.servlet.http.HttpServlet.service(HttpServlet.java:718) at javax.servlet.http.HttpServlet.service(HttpServlet.java:831) at org.apache.catalina.core.ApplicationFilterChain.servletService(ApplicationFilterChain.java:411) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:290) at org.apache.catalina.core.StandardContextValve.invokeInternal(StandardContextValve.java:271) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:202) There is the source code of my classes : Action : @Entity @Table(name="action") public class Action { @Id @GeneratedValue(strategy=GenerationType.IDENTITY) private int num; @ManyToOne(cascade= { CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REFRESH }) @JoinColumn(name="domain_num") private Domain domain; private String name; private String description; @OneToMany @JoinTable(name="permission", joinColumns= { @JoinColumn(name="action_num", referencedColumnName="action_num", nullable=false, updatable=false) }, inverseJoinColumns= { @JoinColumn(name="num") }) private Set<Permission> permissions; public Action() { } Permission : @SuppressWarnings("serial") @Entity @Table(name="permission") public class Permission implements Serializable { @EmbeddedId private PermissionPK primaryKey; @ManyToOne @JoinColumn(name="action_num", insertable=false, updatable=false) private Action action; @ManyToOne @JoinColumn(name="entity_num", insertable=false, updatable=false) private isd.pacepersistence.common.Entity entity; @ManyToOne @JoinColumn(name="class_num", insertable=false, updatable=false) private Clazz clazz; private String kondition; public Permission() { } PermissionPK : @SuppressWarnings("serial") @Entity @Table(name="permission") public class Permission implements Serializable { @EmbeddedId private PermissionPK primaryKey; @ManyToOne @JoinColumn(name="action_num", insertable=false, updatable=false) private Action action; @ManyToOne @JoinColumn(name="entity_num", insertable=false, updatable=false) private isd.pacepersistence.common.Entity entity; @ManyToOne @JoinColumn(name="class_num", insertable=false, updatable=false) private Clazz clazz; private String kondition; public Permission() { }

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  • How to retrieve from two tables with same foreign key repeated more than once?

    - by Sarenya
    How to display the data of tables that are linked by a primary key and foreign key where the foreign key of the data repeats? For ex. I have two tables, ParentTable and Childtable. The primary key of ParentTable acts as the foreign key of ChildTable. There are more than one record with same ParentId in ChildTable. How to retrieve them and display in a single Grid or List or any type of view?

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  • Windows 7 Activation: Can I use OEM key for retail version? [closed]

    - by Joudicek Jouda
    Possible Duplicate: Windows 7 and Vista Activation FAQ: How do language, version, 64-bit or 32-bit, and source affect ability to install and transfer Windows licenses? I have a Lenovo notebook with pre-installed Windows 7 Professional x64. I just did a clean reinstall with retail installation Win7 Professional x64 disk I got from MSDNAA programme I'm member. I wanted to install the OS as Lenovo OEM with the same key I had before, but when I tried to activate the newly installed Windows, it gave me 0xc004e003 error. So the question is, can I use my OEM key for Win7 installed as retail? If not, can I somehow change the installed OS to OEM? The other think is, the OEM key I have tried is just some key I found in my notes, because I don't have any Windows sticker on beneath my laptop as it used to be on all my previous ones. Can I somehow find out "what is the origin" of the supposed key I'm trying to use?

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  • Cannot open simple script application on mac

    - by streetpc
    Mac OS X 10.6 I created a very simple app, which is only a wrapper of a shell script (so that I can select this script in application selectors, like startup apps). I try to launch it and yesterday it worked, but today I changed the executable script's content and name (with something that perfeclty works in a shell script launched in the Terminal) and it will only display a Finder-iconed dialog saying Cannot open the application because it is not supported on this kind of Mac. I restored the previous script (content/name) but I still get the error! Same when re-bundling the app from scratch, or completely changing the bundle identifier… If I try to open it in the Terminal using open My.app, I get The application cannot be opened because it has an incorrect executable format. But when I executes directly the Contents/MacOS/Script, it allways works (iwth both contents). Also, it is displayed with correct icon and meta-information in the Finder (so I guess the Info.plist is understood). The app's file tree is: Contents/ Info.plist MacOS/ Script (executable bit set, works when launched directly) PkgInfo Resources/ AppIcon.icns Here is the Info.plist content: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"> <plist version="1.0"> <dict> <key>CFBundleExecutable</key> <string>Script</string> <key>CFBundleIconFile</key> <string>AppIcon</string> <key>CFBundleIdentifier</key> <string>asdf.ScriptApp</string> <key>CFBundleInfoDictionaryVersion</key> <string>6.0</string> <key>CFBundleName</key> <string>My script</string> <key>CFBundlePackageType</key> <string>APPL</string> <key>CFBundleShortVersionString</key> <string>1.0</string> <key>CFBundleSignature</key> <string>????</string> <key>CFBundleVersion</key> <string>1</string> <key>LSMinimumSystemVersion</key> <string>10.4</string> </dict> </plist> And the PkgInfo file only contains APPL????. I tested the Script with a simple echo "ok" and echo "ok" >/tmp/test (plus #!/bin/sh header). So my questions are: Is there some kind of validity caching for applications ? based on what ? how do I flush it ? Where does this message come from ? I tried to google it but all I get is a page talking about 32/64 bits Java…

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  • Cannot open simplest mac application

    - by streetpc
    I created a very simple app, which is only a wrapper of a shell script (so that I can select this script in application selectors, like startup apps). I try to launch it and yesterday it worked, but today I changed the executable script's content and name (with something that perfeclty works in a shell script launched in the Terminal) and it will only display a Finder-iconed dialog saying Cannot open the application because it is not supported on this kind of Mac. I restored the previous script (content/name) but I still get the error! Same when re-bundling the app from scratch, or completely changing the bundle identifier… If I try to open it in the Terminal using open My.app, I get The application cannot be opened because it has an incorrect executable format. But when I executes directly the Contents/MacOS/Script, it allways works (iwth both contents). Also, it is displayed with correct icon and meta-information in the Finder (so I guess the Info.plist is understood). The app's file tree is: Contents/ Info.plist MacOS/ Script (executable bit set, works when launched directly) PkgInfo Resources/ AppIcon.icns Here is the Info.plist content: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"> <plist version="1.0"> <dict> <key>CFBundleExecutable</key> <string>Script</string> <key>CFBundleIconFile</key> <string>AppIcon</string> <key>CFBundleIdentifier</key> <string>asdf.ScriptApp</string> <key>CFBundleInfoDictionaryVersion</key> <string>6.0</string> <key>CFBundleName</key> <string>My script</string> <key>CFBundlePackageType</key> <string>APPL</string> <key>CFBundleShortVersionString</key> <string>1.0</string> <key>CFBundleSignature</key> <string>????</string> <key>CFBundleVersion</key> <string>1</string> <key>LSMinimumSystemVersion</key> <string>10.4</string> </dict> </plist> And the PkgInfo file only contains APPL????. I tested the Script with a simple echo "ok" and echo "ok" >/tmp/test (plus #!/bin/sh header). So my questions are: * Is there some kind of validity caching for applications ? based on what ? how do I flush it ? * Where does this message come from ? I tried to google it but all I get is a page talking about 32/64 bits Java…

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  • Cannot open simplest script mac application

    - by streetpc
    I created a very simple app, which is only a wrapper of a shell script (so that I can select this script in application selectors, like startup apps). I try to launch it and yesterday it worked, but today I changed the executable script's content and name (with something that perfeclty works in a shell script launched in the Terminal) and it will only display a Finder-iconed dialog saying Cannot open the application because it is not supported on this kind of Mac. I restored the previous script (content/name) but I still get the error! Same when re-bundling the app from scratch, or completely changing the bundle identifier… If I try to open it in the Terminal using open My.app, I get The application cannot be opened because it has an incorrect executable format. But when I executes directly the Contents/MacOS/Script, it allways works (iwth both contents). Also, it is displayed with correct icon and meta-information in the Finder (so I guess the Info.plist is understood). The app's file tree is: Contents/ Info.plist MacOS/ Script (executable bit set, works when launched directly) PkgInfo Resources/ AppIcon.icns Here is the Info.plist content: <?xml version="1.0" encoding="UTF-8"?> <!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"> <plist version="1.0"> <dict> <key>CFBundleExecutable</key> <string>Script</string> <key>CFBundleIconFile</key> <string>AppIcon</string> <key>CFBundleIdentifier</key> <string>asdf.ScriptApp</string> <key>CFBundleInfoDictionaryVersion</key> <string>6.0</string> <key>CFBundleName</key> <string>My script</string> <key>CFBundlePackageType</key> <string>APPL</string> <key>CFBundleShortVersionString</key> <string>1.0</string> <key>CFBundleSignature</key> <string>????</string> <key>CFBundleVersion</key> <string>1</string> <key>LSMinimumSystemVersion</key> <string>10.4</string> </dict> </plist> And the PkgInfo file only contains APPL????. I tested the Script with a simple echo "ok" and echo "ok" >/tmp/test (plus #!/bin/sh header). So my questions are: Is there some kind of validity caching for applications ? based on what ? how do I flush it ? Where does this message come from ? I tried to google it but all I get is a page talking about 32/64 bits Java…

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  • Can I re-attach SSH key forwarding through a disconnected Screen session?

    - by David Mackintosh
    I have a laptop on which I have pageant (the PuTTy SSH key agent) running. If I ssh to a system and launch screen, the ssh key forwarding works properly. However, if I disconnect from that screen session, log off, then later reconnect -- the key forwarding doesn't work any more. I am presuming that this is because when I reconnect the key forwarding is set up on different ports for the new ssh session than was the old one. Is there a way to teach an individual screen window to reconnect to the agent forwarding so that I can use my key to forward again?

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  • How can one create a bootable linux usb key that works on Mac (Intel 64 bit CPU) hardware ?

    - by user3621
    Hi, I'm trying to create a bootable usb key with linux (debian) and that can be booted on Macintel hardware. I have read that MAC's EFI can only boot GPT GUID formatted disks. I'm desperately trying to find a good tutorial which explains how to create such a key. Here what I have done so far: create a GUID partition on te key using linux GNU parted create a HFS+ or ext3 partition on the key, with the boot flag on install a linux .iso with unetbootin While all steps were successfull and in some cases I could even boot on a PC, the step of booting on Macintel software failed (on a macbook). I need to precise that I holded the "alt" key while booting the mac and the only visible bootable disk was the hard disk. Thanks for any advice. PS: I have tried with rEFIt as well. In one case I had a "windows" icon but it then failed to boot with a message like "no system found"

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  • Why does select SCOPE_IDENTITY() return a decimal instead of an integer?

    - by Earlz
    So I have a table with an identity column as the primary key, so it is an integer. So, why does SCOPE_IDENTITY() always return a decimal value instead of an int to my C# application? This is really annoying since decimal values will not implicitly convert to integers in C#, which means I now have to rewrite a bunch of stuff and have a lot of helper methods because I use SQL Server and Postgres, which Postgres does return an integer for the equivalent function.. Why does SCOPE_IDENTITY() not just return a plain integer? Are there people out there that commonly use decimal/non-identity values for primary keys?

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  • Need strategy to phone home to a Pyton web app and check licensing information for a commercial Wind

    - by Cornish
    What's a good strategy for building licensing checking into a Windows desktop app using a Python web application? This is a very open ended question because I don't have the slightest clue how to start to build this feature. What I do have is a number of general concerns: I have developed a commercial Windows desktop application and I want to make money from it but I don't want to build the licensing into the app since it's inevitable that someone will create a keygen or a crack, circulate it online and then it's 'game over' for me. So my idea was to create a web application where people could purchase a license key that is generated by the web app and every time the desktop application is started up, it will 'phone home' to the web app to check whether the license is valid and whether it seems to be in use at multiple locations. I'm just not sure how to do this. Would appreciate any general technical strategies and/or pointers to libraries/modules that might be of use.

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  • Is there any legitimate use for bare strings in PHP?

    - by Robert
    This question got me thinking about bare strings. When PHP sees a string that's not enclosed in quotes, it first checks to see if it's a constant. If not, it just assumes it's a string and goes on anyway. So for example if I have echo $foo[bar]; If there's a constant called bar it uses that for the array key, but if not then it treats bar as a bare string, so it behaves just like echo $foo["bar"]; This can cause all kinds of problems if at some future date a constant is added with the same name. My question is, is there any situation in which it actually makes sense to use a bare string?

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  • What one-time-password devices are compatible with mod_authn_otp?

    - by netvope
    mod_authn_otp is an Apache web server module for two-factor authentication using one-time passwords (OTP) generated via the HOTP/OATH algorithm defined in RFC 4226. The developer's has listed only one compatible device (the Authenex's A-Key 3600) on their website. If a device is fully compliant with the standard, and it allows you to recover the token ID, it should work. However, without testing, it's hard to tell whether a device is fully compliant. Have you ever tried other devices (software or hardware) with mod_authn_otp (or other open source server-side OTP program)? If yes, please share your experience :)

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  • Choice of primary index for mysql innoDB

    - by Saif Bechan
    I have an auction website where users can place a bid on a product. Now i have a primary index on the bid table for easy access of the last places bid on the product. This index is just a unique auto incrementing value. During the week this number becomes huge!! I was wondering if this is a good setup for the primary key in an innoDB table. The bids table exist of the following important fields: table: bids fields: user_id,product_id,bid So what i want to do is make the primary of these 3 fields combined. Is this a good idea or is this just too much for innoDB keys.

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  • MySQL ALTER TABLE on very large table - is it safe to run it?

    - by Timothy Mifsud
    I have a MySQL database with one particular MyISAM table of above 4 million rows. I update this table about once a week with about 2000 new rows. After updating, I then perform the following statement: ALTER TABLE x ORDER BY PK DESC i.e. I order the table in question by the primary key field in descending order. This has not given me any problems on my development machine (Windows with 3GB memory), but, even though 3 times I have tried it successfully on the production Linux server (with 512MB RAM - and achieving the resulted sorted table in about 6 minutes each time), the last time I tried it I had to stop the query after about 30 minutes and rebuild the database from a backup. I have started to wonder whether a 512MB server can cope with that statement (on such a large table) as I have read that a temporary table is created to perform the ALTER TABLE command?! And, if it can be safely run, what should be the expected time for the alteration of the table? Thanks in advance, Tim

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  • Filtering foreign keys with AJAX in Django admin

    - by cnobile
    I have most of this figured out already. I have AJAX returning the region/state/province when a country is selected. The correct foreign key is saved to the database, however, when the record is viewed afterwards the selected state is not shown in the select nor are any states for the selected country. I understand why this is happening as, the admin view is not aware of the relation between the state and the country. So here is the question. Is there a hook in the admin view that will allow me to load the correct states for the country and set the selected attribute on the option in the select tag? Or how can I override the admin view for any forms that require the country and region/state/province set? I am using jQuery and Djando-1.1. Thanks

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  • Loading Fact Table + Lookup / UnionAll for SK lookups.

    - by Nev_Rahd
    I got to populate FactTable with 12 lookups to dimension table to get SK's, of which 6 are to different Dim Tables and rest 6 are lookup to same DimTable (type II) doing lookup to same natural key. Ex: PrimeObjectID = lookup to DimObject.ObjectID = get ObjectSK and got other columns which does same OtherObjectID1 = lookup to DimObject.ObjectID = get ObjectSK OtherObjectID2 = lookup to DimObject.ObjectID = get ObjectSK OtherObjectID3 = lookup to DimObject.ObjectID = get ObjectSK OtherObjectID4 = lookup to DimObject.ObjectID = get ObjectSK OtherObjectID5 = lookup to DimObject.ObjectID = get ObjectSK for such multiple lookup how should go in my SSIS package. for now am using lookup / unionall foreach lookup. Is there a better way to this.

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  • ^+Left Arrow and ^+Right Arrow suddenly stopped working on OS X

    - by user293261
    Hello. I'm not really sure what to make of this. The key combination of ^← and ^→ have stopped working for one of the two users on my OS X installation. I use these keys all the time (switching tabs in terminal, IntelliJ primarily), and it's driving me crazy. On one user account, it works fine. On the other, it doesn't. This happened today and nothing significant comes to mind that would have caused some weird keybinding issue. If anyone has heard of or experienced anything like this, I would very much appreciate your advice! Thanks.

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  • Custom keys for Google App Engine models (Python)

    - by Cameron
    First off, I'm relatively new to Google App Engine, so I'm probably doing something silly. Say I've got a model Foo: class Foo(db.Model): name = db.StringProperty() I want to use name as a unique key for every Foo object. How is this done? When I want to get a specific Foo object, I currently query the datastore for all Foo objects with the target unique name, but queries are slow (plus it's a pain to ensure that name is unique when each new Foo is created). There's got to be a better way to do this! Thanks.

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  • how so select similarities in MySQL?

    - by mysqllearner
    Currently, I am doing a search function. Lets say in my database, I have this data: Keyword1 Keyword2 Keyword3 Keysomething Key and the user entered: "Key" as the keyword to search. This is my current query: Q1: SELECT * FROM data WHERE (data_string LIKE '$key%' OR data_string LIKE '%$key%' OR data_string LIKE '%$key') Basically, I have 2 questions: How do I sort by (order by) similarity. From above example, I should have "Key" as my first result. My current result is: Keyword1, Keyword2, Keyword3, Keysomething and Key My SQL query only search by the "data_string" column, what if I want to seach others column? Do I need to do something like this: Q2: SELECT * FROM data WHERE (data_string LIKE '$key%' OR data_string LIKE '%$key%' OR data_string LIKE '%$key') OR (data_other LIKE '$key%' OR data_other LIKE '%$key%' OR data_other LIKE '%$key') ... Is there any better/faster query than Q2?

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  • Should I use "id" or "unique username"?

    - by roa3
    I am using PHP, AS3 and mysql. I have a website. A flash(as3) website. The flash website store the members' information in mysql database through php. In "members" table, i have "id" as the primary key and "username" as a unique field. Now my situation is: When flash want to display a member's profile. My questions: Should Flash pass the member "ID" or "username" to php to process the mysql query? Is there any different passing the "id" or "username"? Which one is more secure? Which one you recommend? I would like to optimize my website in terms of security and performance.

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