Row can not be located for updating

Of course make a backup of the database first before deleting anything.In order to remove the attributes, the corresponding METAOBJATTRRELATIONS row must be removed first, then the METAATTRIBUTES row can be deleted.First make sure the is properly registered by doing the following steps on the LANDesk Core Server: 1. In the command prompt browse to \Program Files\Landesk\Management Suite3. Ensure it states the dll register succeeded Next run dbrepair to clean up unmodeled data and coredbutil to check the database structure. Additional Troubleshooting: If the error description contains a table other than Unmodeled data then it is possible there are metaattributes pointing to nonexistant columns.

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The BDE considers any SQL join to be read-only because inserting, updating, and deleting rows in a join is ambiguous.Or alternatively, as the message says, add an "is not empty" filter on the field that has the lat lon. What I am doing is actually using the Feeds module to import some business listings. I don't know what is wrong, but I keep getting this error message after importing the listing.Out of a total of 53 result rows in view Map List Business (Block Maps ), 38 rows did not have their field_coordinates set and therefore could not be shown on the map. Have you inspected (i.e Viewed or Edited) some of those failed nodes? Then you will probably see that some of the latitude & longitude fields are blank or have wrong data in them. Thanks Joe I must disagree as this is not a bug in My ODBC! scid=kb; EN-GB;q190727&GSSNB=1 you will find all the ways ADO uses to update dataset: ad Criteria Key = 0 Uses only the primary key ad Criteria All Cols = 1 Uses all columns in the recordset ad Criteria Upd Cols = 2 (Default) Uses only the columns in the recordset that have been modified ad Criteria Time Stamp= 3 Uses the timestamp column (if available) in the recordset From what I see in logs, and by common sense, I can say neither 1 nor 2 (Default) will work as you can't match FLOATs with "=": UPDATE `test`.`testbug11970` SET `f`=1.23123001098632810e 002 WHERE `id`=1 AND `f`=1.44977005004882810e 002 Changing update criteria to 0 (PK only) corrects the situation: UPDATE `test`.`testbug11970` SET `f`=1.23123001098632810e 002 WHERE `id`=1 Consider following, and working, example: Private Sub bn Bug11970_Click(By Val sender As System.

Execute("insert into testbug11970(f) values(144.977477477477)") Dim p_rs As New ADODB. To determine if that is true run the following Microsoft SQL script to compare the metadata with system columns.

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