Factory HP restoration asks for Disc 3 when there are only two disks

ORIGINAL TITLE: Fail factory restore

While trying to do a disk offering factory restore to my computer, I was prompted to insert disk 3. Problem, I have only 2 disks, so restore failed and on reboot I get BSOD. Any ideas? The computer is a hp pavilion dv6000

You will need to contact HP and get the good system restore disks from them. Only they can help with this problem unless you supported your system yourself.

I hope this helps. Good luck.

Tags: Windows

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Hi Experts.,.

Is it possible to hide a column in the report when there are only zeros.

Thanks in advance
Vincent
Vincent Krishna wrote:
Hi Experts.,.

Is it possible to hide a column in the report when there are only zeros.

Thanks in advance
Vincent

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Hi all

I have a small problem:

I want to find the right cause of partitions of the partition key. I have a partition and a Subpartition (sample - table in annex).

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When I try to find the right partiton name, I would use generally not:

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Okay, I get 2 rows, I'm expecting one, but cause the result are the subpartions, I get this:

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But I don't want the subparts, I want the PARTITON!

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    "UPDATES" NUMBER(8,0) DEFAULT 0, 
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Any ideas?

Thanks for Berlin

First of all - the DOF you posted is INVALID - you do not specify a delayed segment creation for partitions. Specify you for the table.

Also, you must remove the DEFAULT value - 1 for the partitioning column. The value '-1' cannot be converted to a valid date for the virtual partitioning column value.» The default value is useless and will give confusion around the year exception err when the exception should actually provide a NULL value for a column not null.

The documentation explains how to use the EXCHANGE PARTITION and has the code example

Maintenance of Partitions

Exchange a Partition of a partitioned Table interval
You can exchange of range partitions in a partitioned table interval. However, you must make sure that the partition of the interval was created before that you can swap partition. You can leave the database create the partition in the partition of the interval of locking.

The following example shows an Exchange partition for the interval_sales table, range-partitioned using the monthly from partitions of 1 January 2004. This example shows how to add data to the table using charge Exchange partition for June 2007. Assume there is only a local index on the interval_sales table and equivalent index have been created on the interval_sales_june_2007 table.
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Note the use of the FOR syntax to identify a partition that has been generated by the system. The name of the partition can be used by querying the *_TAB_PARTITIONS view of data dictionary to find the partition system-generated name.
Given that the last paragraph? That's what showed Solomon.

NOTE:-special attention to the FIRST paragraph. If you use segment carried OVER creating the partition you want to Exchange could NOT have been created yet.

Make SURE that you lock the partition as the example shows, or you will get an exception if it has not been created.

Also - we know NOT what PROBLEM you're trying to solve.

If you exchange a PARTITION of a table subpartitioned your working table should actually be partitioned using the SAME parttioning as the subparts in the source table:

See my response in this thread:

https://forums.Oracle.com/forums/thread.jspa?threadID=2482090&TSTART=30

In this thread I have provide the code example illustrating how the table should be created and made Exchange.

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Could not put up the question correctly to the last channel, my problem is with the join when there are several tables are involved, this is just one example of the task that I have to carry.

Tab1 aura model id retailer_id information for all the weeks (from the first Monday) of the month of JUNE with cost and Helen

Tab1

model_id	retailer_id	sell_date	cost	Helene
1	12	June 3, 13	100	40
1

June 10, 13

200

17 June 13

300

24 June 13

400

June 3, 13

300

June 10, 13

200

17 June 13

300

24 June 13

400

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each retailer belongs to a dealer, under the table has the same information

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12	100

100

101

Tab 3

There is a third layer where each dealership is having a garage band

Dealer_id	Dealer_group
100	1001

101

1001

102

2001

103

2001

104

3001

105

3001

Tab4:

Of this table for each model and dealer discount information for the month of June (every week)

model_id

Dealer_group

discount_date

discount

1001

June 3, 13

1001

June 10, 13

1001

17 June 13

1001

24 June 13

1001

June 3, 13

1001

June 10, 13

1001

17 June 13

1001

24 June 13

2001

June 3, 13

2001

June 10, 13

2001

17 June 13

2001

24 June 13

Master_info:

It's the main table which is the master table for model /retailer information

Model_id	retailer_id
1

Output

model_id

retailer_id

sell_date

cost

Helene

Final (cost-helene-discount)

June 3, 13

100

June 10, 13

200

160

17 June 13

300

270

24 June 13

400

350

June 3, 13

300

280

June 10, 13

200

160

17 June 13

300

270

24 June 13

400

350

June 3, 13

June 10, 13

17 June 13

24 June 13

June 3, 13

June 10, 13

17 June 13

24 June 13

June 3, 13

June 10, 13

17 June 13

24 June 13

June 3, 13

June 10, 13

17 June 13

24 June 13

For highted above records (model_id / retailer_id combination) there is no record in tab1 but they have entered in master_info then the recordings should come for all model_id/retailer_id with all the 0 values

Hello

Thanks for posting the sample data.

It is unclear what dates you want to include in the output. The following query shows how you can generate every Monday in a given range. If you only want to include the dates that are actually present in tabl1 and/or tab4, you can simplify this a bit.

WITH date_range AS

(

SELECT TRUNC (TO_DATE (' 3 June 2013', 'DD-Mon-YYYY'))

, 'IW '.

) AS first_monday

, TRUNC (TO_DATE (24 June 2013 ', 'DD-Mon-YYYY') + 6)

, 'IW '.

) AS last_monday

OF the double

)

all_mondays AS

(

First_monday SELECT + (7 * (LEVEL - 1)) AS sell_date

OF date_range

CONNECT BY LEVEL<= 1="" +="" (="" (last_monday="" -="">

/ 7

)

)

SELECT mi.model_id

mi.retailer_id

am.sell_date

, Cost of NVL (t1.cost, 0) AS

, NVL (t1.rebat, 0) IN the refund

, NVL (t1.cost, 0)

-(NVL (t1.rebat, 0))

+ NVL (t4.discount, 0)

) AS final

E master_info

CROSS JOIN all_mondays am

LEFT OUTER JOIN tab1 t1 ON t1.model_id = mi.model_id

AND t1.retailer_id = mi.retailer_id

AND t1.sell_date = am.sell_date

LEFT OUTER JOIN tab2 t2 ON t2.retailer_id = mi.retailer_id

LEFT OUTER JOIN tab 3 t3 ON t3.dealer_id = t2.dealer_id

LEFT OUTER JOIN tab4 t4 ON t4.model_id = t1.model_id

AND t4.dealer_group = t3.dealer_group

AND t4.discount_date = t1.sell_date

ORDER BY mi.retailer_id

mi.model_id

am.sell_date

;

The results are not exactly what said you you wanted. I suspect it's because of typos in that you posted.

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Hello
How to create a business model when there is only a single source table in the physical layer

Concerning
Swathi
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