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  • Data Guard - Snapshot Standby Database??

    - by Jian Zhang-Oracle
    ?? -------- ?????,??standby?????mount??????????REDO??,??standby????????????????????,???????read-only???open????,????ACTIVE DATA GUARD,????standby?????????(read-only)??(????????),????standby???????????(read-write)? ?????,?????????????Real Application Testing(RAT)??????????,?????????standby??????snapshot standby?????????,??snapshot standby??????????,???????????(read-write)??????snapshot standby??????????????,?????????,??????????,????????,?????????snapshot standby?????standby???,????????? ?? ---------  1.??standby?????? SQL> Alter system set db_recovery_file_dest_size=500M; System altered. SQL> Alter system set db_recovery_file_dest='/u01/app/oracle/snapshot_standby'; System altered. 2.??standby?????? SQL> alter database recover managed standby database cancel; Database altered. 3.??standby???snapshot standby,??open snapshot standby SQL> alter database convert to snapshot standby; Database altered. SQL> alter database open;    Database altered. ??snapshot standby??????SNAPSHOT STANDBY,open???READ WRITE: SQL> select DATABASE_ROLE,name,OPEN_MODE from v$database; DATABASE_ROLE    NAME      OPEN_MODE ---------------- --------- -------------------- SNAPSHOT STANDBY FSDB      READ WRITE 4.?snapshot standby???????????Real Application Testing(RAT)????????? 5.?????,??snapshot standby???physical standby,?????????? SQL> shutdown immediate; Database closed. Database dismounted. ORACLE instance shut down. SQL> startup mount; ORACLE instance started. Database mounted. SQL> ALTER DATABASE CONVERT TO PHYSICAL STANDBY; Database altered. SQL> shutdown immediate; ORA-01507: database not mounted ORACLE instance shut down. SQL> startup mount; ORACLE instance started. Database mounted. SQL>ALTER DATABASE RECOVER MANAGED STANDBY DATABASE DISCONNECT FROM SESSION; Database altered. 5.?????standby?,???????PHYSICAL STANDBY,open???MOUNTED SQL> select DATABASE_ROLE,name,OPEN_MODE from v$database; DATABASE_ROLE    NAME      OPEN_MODE ---------------- --------- -------------------- PHYSICAL STANDBY FSDB      MOUNTED 6.??????????????? ????: SQL> select ads.dest_id,max(sequence#) "Current Sequence",            max(log_sequence) "Last Archived"        from v$archived_log al, v$archive_dest ad, v$archive_dest_status ads        where ad.dest_id=al.dest_id        and al.dest_id=ads.dest_id        and al.resetlogs_change#=(select max(resetlogs_change#) from v$archived_log )        group by ads.dest_id;    DEST_ID Current Sequence Last Archived ---------- ---------------- -------------      1              361           361      2              361           362 --???? SQL>    select al.thrd "Thread", almax "Last Seq Received", lhmax "Last Seq Applied"       from (select thread# thrd, max(sequence#) almax           from v$archived_log           where resetlogs_change#=(select resetlogs_change# from v$database)           group by thread#) al,          (select thread# thrd, max(sequence#) lhmax           from v$log_history           where resetlogs_change#=(select resetlogs_change# from v$database)           group by thread#) lh      where al.thrd = lh.thrd;     Thread Last Seq Received Last Seq Applied ---------- ----------------- ----------------          1               361              361 ??????????,???blog,???????????,??"??:Data Guard - Snapshot Standby Database??" 

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  • Oracle Applications Day 2012. Experience the Global Innovation of Management Applications

    - by antonella.buonagurio
    Iscriviti subito all’Oracle Applications Day 2012 e partecipa al concorso fotografico Oracle I.M.A.G.E. Pochi i giorni rimasti per partecipare al CONCORSO, molte le possibilità di vincere il tuo iPad (*)! Hai tempo fino al 5 OTTOBRE per inviare le tue fotografie Oracle I.M.A.G.E. e vincere uno dei 5 iPad(*) in palio per ciascuna delle due città! Non perdere quest’occasione, scatta le immagini che per te descrivono i cinque concept dell’evento e inviale per e-mail a [email protected] indicando: •  nell’oggetto della mail, il tema della fotografia: Innovation, Management, Applications, Global, Experience; •  nel corpo della mail, il tuo nome e cognome e città nella quale parteciperai all’Applications Day 2012 Milano o Roma. 10 ottobre 2012 – Milano, East End Studios | 17 ottobre 2012 - Roma, Officine Farneto L’evento per condividere con Clienti e Partner Oracle le soluzioni più innovative e le esperienze più significative sulle scelte strategiche per affrontare le sfide attuali e future. Iscriviti all’evento sul sito

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  • Hazlii cu politisti

    - by interesante
    Mor 6 politisti si se ancheteaza cazul:-Pai...Trei dintre ei erau cu barca pe lac si doi si-au aprins cate-o tigara.Unul si-a adus aminte ca a uitat sa stinga chibritul si a sarit in apa sa-l stinga si sa-necat.Al doilea uitase se-si stinga chistocul si a sarit si el si sa-necat.-Si al treilea?-Nu pornea barca si s-a dat jos s-o-mpinga si sa-necat.-Bine, dar ceilalti trei ?-Ei au murit la reconstituire.....Distreaza-te copios si cu jocuri flash de pe un site cu jocuri online.Doua sotii de politisti stau de vorba. Una zice:- Draga, sotul meu are post langa o florarie. Niciodata nu mi-a adus vreo floare...- Si ce? Al meu are post langa conservator. O conserva n-am vazut pana acum...

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  • Simple PHP upload form not working

    - by Lea
    Hi all, I seem to run into so many issues when dealing with writing to directories. If someone could please look over this script, and tell me why it's not working, I'd be so appreciative. Upon uploading the file from the form, I don't get anything.. It doesnt output any errors, it simply just refreshes. Thanks, Lea <?php include ('config.php'); if( isset($_POST['submit']) ) { $target = ''.$_SERVER['DOCUMENT_ROOT'].'/images/'; $target = $target . basename( $_FILES['photo']['name']) ; $url = basename( $_FILES['photo']['name']) ; if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { $moved = "File has been moved to location $target"; $name = mysql_real_escape_string($_POST['photoname']); mysql_query("INSERT INTO photos (photo_name, photo_image) VALUES ('$name', '$url' )") or die(mysql_error()); $success = "Photo has been added!"; } else { $moved = "File has not been moved to $target"; $success = "Failed to upload:("; } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Photo Upload</title> <meta name="robots" content="index, follow" /> <meta name="keywords" content="" /> <meta name="description" content="" /> <link href="<?php echo $globalurl; ?>styles.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="holder"> <br /><br /> <b>Add a new photo</b> <hr /> <br /> <b><?php echo "$success<br />"; ?> <?php echo "$moved<br />"; ?></b> <form enctype="multipart/form-data" method="post" action="<?php echo $PHP_SELF; ?>"> <table cellspacing="0" cellpadding="0" width="500px"> <tbody> <tr> <td valign="top">Photo Name:</td> <td valign="top"> <input type="text" name="photoname" /><br /> <br /> </td> </tr> <tr> <td valign="top">Photo:</td> <td valign="top"> <input type="file" name="photo"><br /> <br /> </td> </tr> </tbody> </table> <input type="submit" value="submit" /> </form> </div> </body> </html>

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  • Laptop forgets second monitor when it goes to sleep

    - by al c
    I recently updated my nvidia driver... now I have to re-establish the connection to my second monitor everytime the laptop goes to sleep. When I open the NVIDIA control panel, the second monitor is listed there but the checkbox beside it has been cleared. Oddly my laptop screen is identified as display #3 and the external screen as display #4. Is there a registry key that I can clear to get them set back to #1 & #2? Will that fix the problem? Or is there a simpler solution. (It's a Dell Studio XPS laptop with GeForce 9500M running Windows Vista 64-bit) TIA Al

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  • reading from File in assembly

    - by Natasha
    i am trying to read a username and a password from a file in x86 assembly for the perpose of authentication obviously the file consists of two lines , the user name and the password how can i read the two lines seperately and compare them? My attempt: proc read_file mov ah,3dh lea dx,file_name int 21h mov bx, ax xor si,si repeat: mov ah, 3fh lea dx, buffer mov cx, 100 int 21h mov si, ax mov buffer[si], '$' mov ah, 09h int 21h ;print on screen cmp si, 100 je repeat jmp stop;jump to end stop: RET read_file ENDP

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  • MBR Booting from DOS

    - by eflukx
    For a project I would like to invoke the MBR on the first harddisk directly from DOS. I've written a small assembler program that loads the MBR in memory at 0:7c00h an does a far jump to it. I've put my util on a bootable floppy. The disk (HD0, 0x80) i'm trying to boot has a TrueCrypt boot loader on it. It shows up the TrueCrypt screen, but after typing in the password it crashes the system. When I run my little utlility (w00t.com) on a normal WinXP machine it seams to crash immedealty. Apparently I'm forgetting some crucial stuff the BIOS normally does, my guess is it's something trivial. Can someone with better bare-metal DOS and BIOS experience help me out? Heres my code: .MODEL tiny .386 _TEXT SEGMENT USE16 INCLUDE BootDefs.i ORG 100h start: ; http://vxheavens.com/lib/vbw05.html ; Before DOS has booted the BIOS stores the amount of usable lower memory ; in a word located at 0:413h in memory. We going to erase this value because ; we have booted dos before loading the bootsector, and dos is fat (and ugly). ; fake free memory ;push ds ;push 0 ;pop ds ;mov ax, TC_BOOT_LOADER_SEGMENT / 1024 * 16 + TC_BOOT_MEMORY_REQUIRED ;mov word ptr ds:[413h], ax ;ax = memory in K ;pop ds ;lea si, memory_patched_msg ;call print ;mov ax, cs mov ax, 0 mov es, ax ; read first sector to es:7c00h (== cs:7c00) mov dl, 80h mov cl, 1 mov al, 1 mov bx, 7c00h ;load sector to es:bx call read_sectors lea si, mbr_loaded_msg call print lea si, jmp_to_mbr_msg call print ;Set BIOS default values in environment cli mov dl, 80h ;(drive C) xor ax, ax mov ds, ax mov es, ax mov ss, ax mov sp, 0ffffh sti push es push 7c00h retf ;Jump to MBR code at 0:7c00h ; Print string print: xor bx, bx mov ah, 0eh cld @@: lodsb test al, al jz print_end int 10h jmp @B print_end: ret ; Read sectors of the first cylinder read_sectors: mov ch, 0 ; Cylinder mov dh, 0 ; Head ; DL = drive number passed from BIOS mov ah, 2 int 13h jnc read_ok lea si, disk_error_msg call print read_ok: ret memory_patched_msg db 'Memory patched', 13, 10, 7, 0 mbr_loaded_msg db 'MBR loaded', 13, 10, 7, 0 jmp_to_mbr_msg db 'Jumping to MBR code', 13, 10, 7, 0 disk_error_msg db 'Disk error', 13, 10, 7, 0 _TEXT ENDS END start

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  • How to Lookup For something main VBA

    - by gorPweN
    Hi, I wanna know how to Lookup For many number in a column A corresponding to a name in Column B By coding in VBA..And Write it in column C??? Thanks in advance..Im new in this... Exemple A B C 200-333 Jack 200-345 Lea 200-346 Fresh 200-347 Tide Jack 200-323 Tide Lea Fresh 200-344 Tide

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  • Permissions restoring from Time Machine - Finder copy vs "cp" copy

    - by Ben Challenor
    Note: this question was starting to sprawl so I rewrote it. I have a folder that I'm trying to restore from a Time Machine backup. Using cp -R works fine, but certain folders cannot be restored with either the Time Machine UI or Finder. Other users have reported similar errors and the cp -R workaround was suggested (e.g. Restoring from Time Machine - Permissions Error). But I wanted to understand: Why cp -R works when the Finder and the Time Machine UI do not. Whether I could prevent the errors by changing file permissions before the backup. There do indeed seem to be some permissions that Finder works with and some that it does not. I've narrowed the errors down to folders with the user ben (that's me) and the group wheel. Here's a simplified reproduction. I have four folders with the owner/group combinations I've seen so far: ben ~/Desktop/test $ ls -lea total 16 drwxr-xr-x 7 ben staff 238 27 Nov 14:31 . drwx------+ 17 ben staff 578 27 Nov 14:29 .. 0: group:everyone deny delete -rw-r--r--@ 1 ben staff 6148 27 Nov 14:31 .DS_Store drwxr-xr-x 3 ben staff 102 27 Nov 14:30 ben-staff drwxr-xr-x 3 ben wheel 102 27 Nov 14:30 ben-wheel drwxr-xr-x 3 root admin 102 27 Nov 14:31 root-admin drwxr-xr-x 3 root wheel 102 27 Nov 14:31 root-wheel Each contains a single file called file with the same owner/group: ben ~/Desktop/test $ cd ben-staff ben ~/Desktop/test/ben-staff $ ls -lea total 0 drwxr-xr-x 3 ben staff 102 27 Nov 14:30 . drwxr-xr-x 7 ben staff 238 27 Nov 14:31 .. -rw-r--r-- 1 ben staff 0 27 Nov 14:30 file In the backup, they look like this: ben /Volumes/Deimos/Backups.backupdb/Ben’s MacBook Air/Latest/Macintosh HD/Users/ben/Desktop/test $ ls -leA total 16 -rw-r--r--@ 1 ben staff 6148 27 Nov 14:34 .DS_Store 0: group:everyone deny write,delete,append,writeattr,writeextattr,chown drwxr-xr-x@ 3 ben staff 102 27 Nov 14:51 ben-staff 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 ben wheel 102 27 Nov 14:51 ben-wheel 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 root admin 102 27 Nov 14:52 root-admin 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown drwxr-xr-x@ 3 root wheel 102 27 Nov 14:52 root-wheel 0: group:everyone deny add_file,delete,add_subdirectory,delete_child,writeattr,writeextattr,chown Of these, ben-staff can be restored with Finder without errors. root-wheel and root-admin ask for my password and then restore without errors. But ben-wheel does not prompt for my password and gives the error: The operation can’t be completed because you don’t have permission to access “file”. Interestingly, I can restore the file from this folder by dragging it directly to my local drive (instead of dragging its parent folder), but when I do so its permissions are changed to ben/staff. Here are the permissions after the restore for the three folders that worked correctly, and the file from ben-wheel that was changed to ben/staff. ben ~/Desktop/test-restore $ ls -leA total 16 -rw-r--r--@ 1 ben staff 6148 27 Nov 14:46 .DS_Store drwxr-xr-x 3 ben staff 102 27 Nov 14:30 ben-staff -rw-r--r-- 1 ben staff 0 27 Nov 14:30 file drwxr-xr-x 3 root admin 102 27 Nov 14:31 root-admin drwxr-xr-x 3 root wheel 102 27 Nov 14:31 root-wheel Can anyone explain this behaviour? Why do Finder and the Time Machine UI break with the ben / wheel permissions? And why does cp -R work (even without sudo)?

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  • Oracle anuncia la disponibilidad de Oracle Knowledge 8.5

    - by Noelia Gomez
    Normal 0 21 false false false ES X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-family:"Calibri","sans-serif";} El lanzamiento más completo de la gestión del conocimiento de Oracle ayuda a las organizaciones a ofrecer las respuestas correctas en el momento adecuado a los Agentes y Clientes Normal 0 21 false false false ES X-NONE X-NONE MicrosoftInternetExplorer4 -"/ /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-family:"Calibri","sans-serif";} Continuando con su compromiso de ayudar a las organizaciones a ofrecer la mejor experiencia del cliente con la exploración de los datos empresariales, Oracle anunció Oracle Knowledge 8.5, el software líder en la industria del conocimiento que soporta la gestión web de autoservicio, servicio asistido por agente y comunidades de clientes. Oracle Knowledge 8.5 es la versión más completa desde la adquisición de InQuira en octubre de 2011. Se introduce importantes mejoras de productos, análisis de mejora y los avances en el rendimiento y la escalabilidad. Actualmente, las organizaciones entienden la necesidad imperiosa de ofrecer un servicio al cliente consistente y de alta calidad, a través de diferentes canales. Además, las organizaciones reconocen la necesidad de información que sirva para este proceso. Oracle Knowledge 8,5 proactivamente brinda conocimientos relevantes y contextuales en el punto de interacción con los agentes, con los trabajadores del conocimiento y con los clientes - ayudando a aumentar la lealtad del cliente y reducir costes. Al permitir búsquedas a través de una amplia variedad de fuentes, Oracle Knowledge 8,5 amplifica el acceso al conocimiento, una vez oculto en los sistemas múltiples, aplicaciones y bases de datos utilizadas para almacenar contenido empresarial. Nuevas capacidades: AnswerFlow : Oracle Knowledge 8.5 introduce AnswerFlow, una nueva aplicación para la resolución de problemas guiada, diseñado para mejorar la eficiencia, reducir los costes de servicio y proporcionar experiencias de servicio al cliente personalizado. Mejora la analítica del conocimiento: estandarizado en Oracle Business Intelligence Enterprise Edition, el líder en la industria de las soluciones de Business Intelligence, Oracle Knowledge 8.5 proporciona una funcionalidad analítica robusta que se puede adaptar para satisfacer las necesidades únicas de cada negocio. Soporte de idiomas mejorada: Con las capacidades mejoradas en varios idiomas disponibles en Oracle Knowledge 8.5, incluyendo soporte Natural Language Search con 16 Idiomas y búsqueda por palabra clave mejorado para la mayoría de las otras lenguas , las empresas pueden llegar rápidamente a nuevos clientes y, al mismo tiempo, reducir los costes de hacerlo. Mejora la funcionalidad iConnect:Oracle Knowledge 8.5 ofrece iConnect mejorada para una mayor facilidad de uso y rendimiento. Esta aplicación especializada en conocimiento ofrece de forma proactiva un conocimiento contextualizado directamente en las aplicaciones de CRM, permitiendo a los empleados reducir el esfuerzo para servir a los clientes. Estandarización de Plataforma y Tecnología: Oracle Knowledge 8.5 ha sido certificado en tecnologías Oracle, incluyendo Oracle WebLogic Server, Oracle Business Intelligence, Oracle Exadata Database Machine y Oracle Exalogic Elastic Cloud, reduciendo el coste y la complejidad para los clientes a administrar los activos de todas las plataformas."Hemos realizado importantes inversiones de desarrollo desde nuestra adquisición de InQuira, y ahora estamos muy orgullosos de anunciar la disponibilidad de estos esfuerzos con el lanzamiento de Oracle Knowledge 8.5, por lo que es más fácil para las organizaciones obtener una ventaja diferencial a un coste total de propiedad significativamente menor ". dijo David Vap,Vicepresident productos de Oracle. Puedes encontrar más información aquí: Oracle Knowledge 8.5 Oracle Customer Experience Oracle WebLogic Server Oracle Business Intelligence Enterprise Edition

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  • La búsqueda de la eficiencia como Santo Grial de las TIC sanitarias

    - by Eloy M. Rodríguez
    Las XVIII Jornadas de Informática Sanitaria en Andalucía se han cerrado el pasado viernes con 11.500 horas de inteligencia colectiva. Aunque el cálculo supongo que resulta de multiplicar las horas de sesiones y talleres por el número de inscritos, lo que no sería del todo real ya que la asistencia media calculo que andaría por las noventa personas, supongo que refleja el global si incluimos el montante de interacciones informales que el formato y lugar de celebración favorecen. Mi resumen subjetivo es que todos somos conscientes de que debemos conseguir más eficiencia en y gracias a las TIC y que para ello hemos señalado algunas pautas, que los asistentes, en sus diferentes roles debiéramos aplicar y ayudar a difundir. En esa línea creo que destaca la necesidad de tener muy claro de dónde se parte y qué se quiere conseguir, para lo que es imprescindible medir y que las medidas ayuden a retroalimentar al sistema en orden de conseguir sus objetivos. Y en este sentido, a nivel anecdótico, quisiera dejar una paradoja que se presentó sobre la eficiencia: partiendo de que el coste/día de hospitalización es mayor al principio que los últimos días de la estancia, si se consigue ser más eficiente y reducir la estancia media, se liberarán últimos días de estancia que se utilizarán para nuevos ingresos, lo que hará que el número de primeros días de estancia aumente el coste económico total. En este caso mejoraríamos el servicio a los ciudadanos pero aumentaríamos el coste, salvo que se tomasen acciones para redimensionar la oferta hospitalaria bajando el coste y sin mejorer la calidad. También fue tema destacado la posibilidad/necesidad de aprovechar las capacidades de las TIC para realizar cambios estructurales y hacer que la medicina pase de ser reactiva a proactiva mediante alarmas que facilitasen que se actuase antes de ocurra el problema grave. Otro tema que se trató fue la necesidad real de corresponsabilizar de verdad al ciudadano, gracias a las enormes posibilidades a bajo coste que ofrecen las TIC, asumiendo un proceso hacia la salud colaborativa que tiene muchos retos por delante pero también muchas más oportunidades. Y la carpeta del ciudadano, emergente en varios proyectos e ideas, es un paso en ese aspecto. Un tema que levantó pasiones fue cuando la Directora Gerente del Sergas se quejó de que los proyectos TIC eran lentísimos. Desgraciadamente su agenda no le permitió quedarse al debate que fue bastante intenso en el que salieron temas como el larguísimo proceso administrativo, las especificaciones cambiantes, los diseños a medida, etc como factores más allá de la eficiencia especifica de los profesionales TIC involucrados en los proyectos. Y por último quiero citar un tema muy interesante en línea con lo hablado en las jornadas sobre la necesidad de medir: el Índice SEIS. La idea es definir una serie de criterios agrupados en grandes líneas y con un desglose fino que monitorice la aportación de las TIC en la mejora de la salud y la sanidad. Nos presentaron unas versiones previas con debate aún abierto entre dos grandes enfoques, partiendo desde los grandes objetivos hasta los procesos o partiendo desde los procesos hasta los objetivos. La discusión no es sólo académica, ya que influye en los parámetros a establecer. La buena noticia es que está bastante avanzado el trabajo y que pronto los servicios de salud podrán tener una herramienta de comparación basada en la realidad nacional. Para los interesados, varios asistentes hemos ido tuiteando las jornadas, por lo que el que quiera conocer un poco más detalles puede ir a Twitter y buscar la etiqueta #jisa18 y empezando del más antiguo al más moderno se puede hacer un seguimiento con puntos de vista subjetivos sobre lo allí ocurrido. No puedo dejar de hacer un par de autocríticas, ya que soy miembro de la SEIS. La primera es sobre el portal de la SEIS que no ha tenido la interactividad que unas jornadas como estas necesitaban. Pronto empezará a tener documentos y análisis de lo allí ocurrido y luego vendrán las crónicas y análisis más cocinados en la revista I+S. Pero en la segunda década del siglo XXI se necesita bastante más. La otra es sobre la no deseada poca presencia de usuarios de las TIC sanitarias en los roles de profesionales sanitarios y ciudadanos usuarios de los sistemas de información sanitarios. Tenemos que ser proactivos para que acudan en número significativo, ya que si no estamos en riesgo de ser unos TIC-sanitarios absolutistas: todo para los usuarios pero sin los usuarios. Tweet

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  • How to write a buffer-overflow exploit in windows XP,x86?

    - by Mask
    void function(int a, int b, int c) { char buffer1[5]; char buffer2[10]; int *ret; ret = buffer1 + 12; (*ret) += 8;//why is it 8?? } void main() { int x; x = 0; function(1,2,3); x = 1; printf("%d\n",x); } The above demo is from here: http://insecure.org/stf/smashstack.html But it's not working here: D:\test>gcc -Wall -Wextra hw.cpp && a.exe hw.cpp: In function `void function(int, int, int)': hw.cpp:6: warning: unused variable 'buffer2' hw.cpp: At global scope: hw.cpp:4: warning: unused parameter 'a' hw.cpp:4: warning: unused parameter 'b' hw.cpp:4: warning: unused parameter 'c' 1 And I don't understand why it's 8 though the author thinks: A little math tells us the distance is 8 bytes. My gdb dump as called: Dump of assembler code for function main: 0x004012ee <main+0>: push %ebp 0x004012ef <main+1>: mov %esp,%ebp 0x004012f1 <main+3>: sub $0x18,%esp 0x004012f4 <main+6>: and $0xfffffff0,%esp 0x004012f7 <main+9>: mov $0x0,%eax 0x004012fc <main+14>: add $0xf,%eax 0x004012ff <main+17>: add $0xf,%eax 0x00401302 <main+20>: shr $0x4,%eax 0x00401305 <main+23>: shl $0x4,%eax 0x00401308 <main+26>: mov %eax,0xfffffff8(%ebp) 0x0040130b <main+29>: mov 0xfffffff8(%ebp),%eax 0x0040130e <main+32>: call 0x401b00 <_alloca> 0x00401313 <main+37>: call 0x4017b0 <__main> 0x00401318 <main+42>: movl $0x0,0xfffffffc(%ebp) 0x0040131f <main+49>: movl $0x3,0x8(%esp) 0x00401327 <main+57>: movl $0x2,0x4(%esp) 0x0040132f <main+65>: movl $0x1,(%esp) 0x00401336 <main+72>: call 0x4012d0 <function> 0x0040133b <main+77>: movl $0x1,0xfffffffc(%ebp) 0x00401342 <main+84>: mov 0xfffffffc(%ebp),%eax 0x00401345 <main+87>: mov %eax,0x4(%esp) 0x00401349 <main+91>: movl $0x403000,(%esp) 0x00401350 <main+98>: call 0x401b60 <printf> 0x00401355 <main+103>: leave 0x00401356 <main+104>: ret 0x00401357 <main+105>: nop 0x00401358 <main+106>: add %al,(%eax) 0x0040135a <main+108>: add %al,(%eax) 0x0040135c <main+110>: add %al,(%eax) 0x0040135e <main+112>: add %al,(%eax) End of assembler dump. Dump of assembler code for function function: 0x004012d0 <function+0>: push %ebp 0x004012d1 <function+1>: mov %esp,%ebp 0x004012d3 <function+3>: sub $0x38,%esp 0x004012d6 <function+6>: lea 0xffffffe8(%ebp),%eax 0x004012d9 <function+9>: add $0xc,%eax 0x004012dc <function+12>: mov %eax,0xffffffd4(%ebp) 0x004012df <function+15>: mov 0xffffffd4(%ebp),%edx 0x004012e2 <function+18>: mov 0xffffffd4(%ebp),%eax 0x004012e5 <function+21>: movzbl (%eax),%eax 0x004012e8 <function+24>: add $0x5,%al 0x004012ea <function+26>: mov %al,(%edx) 0x004012ec <function+28>: leave 0x004012ed <function+29>: ret In my case the distance should be - = 5,right?But it seems not working..

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  • How to write a buffer-overflow exploit in GCC,windows XP,x86?

    - by Mask
    void function(int a, int b, int c) { char buffer1[5]; char buffer2[10]; int *ret; ret = buffer1 + 12; (*ret) += 8;//why is it 8?? } void main() { int x; x = 0; function(1,2,3); x = 1; printf("%d\n",x); } The above demo is from here: http://insecure.org/stf/smashstack.html But it's not working here: D:\test>gcc -Wall -Wextra hw.cpp && a.exe hw.cpp: In function `void function(int, int, int)': hw.cpp:6: warning: unused variable 'buffer2' hw.cpp: At global scope: hw.cpp:4: warning: unused parameter 'a' hw.cpp:4: warning: unused parameter 'b' hw.cpp:4: warning: unused parameter 'c' 1 And I don't understand why it's 8 though the author thinks: A little math tells us the distance is 8 bytes. My gdb dump as called: Dump of assembler code for function main: 0x004012ee <main+0>: push %ebp 0x004012ef <main+1>: mov %esp,%ebp 0x004012f1 <main+3>: sub $0x18,%esp 0x004012f4 <main+6>: and $0xfffffff0,%esp 0x004012f7 <main+9>: mov $0x0,%eax 0x004012fc <main+14>: add $0xf,%eax 0x004012ff <main+17>: add $0xf,%eax 0x00401302 <main+20>: shr $0x4,%eax 0x00401305 <main+23>: shl $0x4,%eax 0x00401308 <main+26>: mov %eax,0xfffffff8(%ebp) 0x0040130b <main+29>: mov 0xfffffff8(%ebp),%eax 0x0040130e <main+32>: call 0x401b00 <_alloca> 0x00401313 <main+37>: call 0x4017b0 <__main> 0x00401318 <main+42>: movl $0x0,0xfffffffc(%ebp) 0x0040131f <main+49>: movl $0x3,0x8(%esp) 0x00401327 <main+57>: movl $0x2,0x4(%esp) 0x0040132f <main+65>: movl $0x1,(%esp) 0x00401336 <main+72>: call 0x4012d0 <function> 0x0040133b <main+77>: movl $0x1,0xfffffffc(%ebp) 0x00401342 <main+84>: mov 0xfffffffc(%ebp),%eax 0x00401345 <main+87>: mov %eax,0x4(%esp) 0x00401349 <main+91>: movl $0x403000,(%esp) 0x00401350 <main+98>: call 0x401b60 <printf> 0x00401355 <main+103>: leave 0x00401356 <main+104>: ret 0x00401357 <main+105>: nop 0x00401358 <main+106>: add %al,(%eax) 0x0040135a <main+108>: add %al,(%eax) 0x0040135c <main+110>: add %al,(%eax) 0x0040135e <main+112>: add %al,(%eax) End of assembler dump. Dump of assembler code for function function: 0x004012d0 <function+0>: push %ebp 0x004012d1 <function+1>: mov %esp,%ebp 0x004012d3 <function+3>: sub $0x38,%esp 0x004012d6 <function+6>: lea 0xffffffe8(%ebp),%eax 0x004012d9 <function+9>: add $0xc,%eax 0x004012dc <function+12>: mov %eax,0xffffffd4(%ebp) 0x004012df <function+15>: mov 0xffffffd4(%ebp),%edx 0x004012e2 <function+18>: mov 0xffffffd4(%ebp),%eax 0x004012e5 <function+21>: movzbl (%eax),%eax 0x004012e8 <function+24>: add $0x5,%al 0x004012ea <function+26>: mov %al,(%edx) 0x004012ec <function+28>: leave 0x004012ed <function+29>: ret In my case the distance should be - = 5,right?But it seems not working.. Why function needs 56 bytes for local variables?( sub $0x38,%esp )

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  • Multiple synonym dictionary matches in PostgreSQL full text searching

    - by Ryan VanMiddlesworth
    I am trying to do full text searching in PostgreSQL 8.3. It worked splendidly, so I added in synonym matching (e.g. 'bob' == 'robert') using a synonym dictionary. That works great too. But I've noticed that it apparently only allows a word to have one synonym. That is, 'al' cannot be 'albert' and 'allen'. Is this correct? Is there any way to have multiple dictionary matches in a PostgreSQL synonym dictionary? For reference, here is my sample dictionary file: bob robert bobby robert al alan al albert al allen And the SQL that creates the full text search config: CREATE TEXT SEARCH DICTIONARY nickname (TEMPLATE = synonym, SYNONYMS = nickname); CREATE TEXT SEARCH CONFIGURATION dxp_name (COPY = simple); ALTER TEXT SEARCH CONFIGURATION dxp_name ALTER MAPPING FOR asciiword WITH nickname, simple; What am I doing wrong? Thanks!

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  • Problem with stack based implementation of function 0x42 of int 0x13

    - by IceCoder
    I'm trying a new approach to int 0x13 (just to learn more about the way the system works): using stack to create a DAP.. Assuming that DL contains the disk number, AX contains the address of the bootable entry in PT, DS is updated to the right segment and the stack is correctly set, this is the code: push DWORD 0x00000000 add ax, 0x0008 mov si, ax push DWORD [ds:(si)] push DWORD 0x00007c00 push WORD 0x0001 push WORD 0x0010 push ss pop ds mov si, sp mov sp, bp mov ah, 0x42 int 0x13 As you can see: I push the dap structure onto the stack, update DS:SI in order to point to it, DL is already set, then set AX to 0x42 and call int 0x13 the result is error 0x01 in AH and obviously CF set. No sectors are transferred. I checked the stack trace endlessly and it is ok, the partition table is ok too.. I cannot figure out what I'm missing... This is the stack trace portion of the disk address packet: 0x000079ea: 10 00 adc %al,(%bx,%si) 0x000079ec: 01 00 add %ax,(%bx,%si) 0x000079ee: 00 7c 00 add %bh,0x0(%si) 0x000079f1: 00 00 add %al,(%bx,%si) 0x000079f3: 08 00 or %al,(%bx,%si) 0x000079f5: 00 00 add %al,(%bx,%si) 0x000079f7: 00 00 add %al,(%bx,%si) 0x000079f9: 00 a0 07 be add %ah,-0x41f9(%bx,%si) I'm using qemu latest version and trying to read from hard drive (0x80), have also tried with a 4bytes alignment for the structure with the same result (CF 1 AH 0x01), the extensions are present.

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  • NASM and a question about ADC - ASM 8086

    - by Tal
    Hello,I study assembly on High-school and I would like to try to make assembly programs at home. I downloaded NASM but I don't understand how to run the .s files with it - if you can write a simple way here to run it I'd glad :-) and in addition I have a question: when I use ADC for exmaple: AL = 01 and BL = 02, and CF = 1, when I make this: ADC AL,BL Will AL be 3 or 4? (with the CF addition or without?) Thank you!!

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  • 8086 programming using TASM: pc to pc communication

    - by Komal
    .model small .stack 100 .data .code mov ah,00h mov al,0e3h mov dx,00h int 14h back: nop l1: mov ah,03h mov dx,00h int 14h and ah,01h cmp ah,01h jne l1 mov ah,02h mov dx,00h int 21h mov dl,al mov ah,02h int 21h jmb back mov ah,4ch int 21h end this a pc to pc commnication receiver program.i would like to know why have we used the mov dx,00h function and what is the meaning of mov al,0e3h this ?

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  • Question about ADD on ASM 8086

    - by Tal
    Hello, I'm studying ASM 8086 theoretically on highschool. (that means that I study ASM 8086 on a notebook, and never got to run it over a computer). And I don't understand - what will happen if I do this: MOV AL, F2h ADD AL, 20h What will the computer do? (what will be the value of AL,AX, CF,ZF?) and what will happen if I do this: MOV AH,F2h ADD AH,20h Thank you !!

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  • vb.net -- Using arraylist as key in dictionary

    - by Zahid
    Dim dct As New Dictionary(Of ArrayList, ArrayList) ' Populate Dictionary dct.Add(New ArrayList({"Dot", "0"}), New ArrayList({20, 30, 40, 50})) dct.Add(New ArrayList({"Dot", "1"}), New ArrayList({120, 130, 140, 150})) ' Search in dictionary Dim al As New ArrayList({"Dot", "2"}) If dct.ContainsKey(al) Then *' does not work* MessageBox.Show("Found: " & al(0).ToString) End If

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  • Problem with asm program (nasm)

    - by GLeBaTi
    org 0x100 SEGMENT .CODE mov ah,0x9 mov dx, Msg1 int 0x21 ;string input mov ah,0xA mov dx,buff int 0x21 mov ax,0 mov al,[buff+1]; length ;string UPPERCASE mov cl, al mov si, buff cld loop1: lodsb; cmp al, 'a' jnb upper loop loop1 ;output mov ah,0x9 mov dx, buff int 0x21 exit: mov ah, 0x8 int 0x21 int 0x20 upper: sub al,32 jmp loop1 SEGMENT .DATA Msg1 db 'Press string: $' buff db 254,0 this code perform poorly. I think that problem in "jnb upper". This program make small symbols into big symbols.

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  • easy asm program(nasm)

    - by GLeBaTi
    org 0x100 SEGMENT .CODE mov ah,0x9 mov dx, Msg1 int 0x21 ;string input mov ah,0xA mov dx,buff int 0x21 mov ax,0 mov al,[buff+1]; length ;string UPPERCASE mov cl, al mov si, buff cld loop1: lodsb; cmp al, 'a' jnb upper loop loop1 ;output mov ah,0x9 mov dx, buff int 0x21 exit: mov ah, 0x8 int 0x21 int 0x20 upper: sub al,32 jmp loop1 SEGMENT .DATA Msg1 db 'Press string: $' buff db 254,0 this code perform poorly. I think that problem in "jnb upper". This program make small symbols into big symbols.

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  • Scan for first zero bit (Assembly)?

    - by cthulhu
    I have some numbers in AH, AL, BL, BH registers. I need to check whether there is 0 bit in each of the registers in left half of the number. If yes, then put into check variable value 10 else -10. How can I do this? I tried something like that: org 100h check dw 0 mov ah, 11111111b mov al, 11111111b mov bl, 11111111b mov bh, 11111111b mov check, -10 shr ah, 4 shr al, 4 shr bl, 4 shr bh, 4 cmp ah, 0Fh jz first first: cmp al, 0Fh jz second second: cmp bl, 0Fh jz third third: cmp bh, 0Fh jz final final: mov check, 10 ret

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  • Autocomplete for generic types in Eclipse

    - by AvrDragon
    "Refer to objects by their interfaces" is a good practise, as mentioned in Effective Java. So for example i prefer List<String> al = new ArrayList<String>(); over ArrayList<String> al = new ArrayList<String>(); in my code. One annoying thing is that if i type ArrayList<String> al = new and then hit Ctrl+Space in Eclipse i get ArrayList<String>() as propostal. But if i type List al = new and then hit Ctrl+Space i will get only propostal to define anonymous inner class, but not propostals such as new ArrayList<String>(), what is 99% the case, or for example new Vector<String>(). Is there any way to get the subclasses as propostals for generic types?

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