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Getting ERROR 000840 from arcpy.SelectLayerByAttribute_management?



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Listing Elements or Datasets of Layer gives ERROR 000840 value is not Feature Layer?ERROR 000840: The value is not a Raster|Raster Catalog|Mosaic Layer from arcpy.AddJoin_management()?Select By Atribute using python script on .Shp file getting error 000840 the value is not a raster layerExecuteError: Failed to execute. Parameters are not valid. ERROR 000732: Layer Name or Table View: Dataset L does not existUsing arcpy.MakeFeatureLayer_management on Shapefile and Feature ClassArcpy Network analyst - Perform OD Matrix Cost in loopArcpy.AddJoin errorSelect By Attributes Based on Unique Value in Shapefile gives ERROR 000840?ExecuteError: Failed to execute. Parameters are not valid error when SelectLayerByAttribute run in Python 2.7Cannot find results of MakeFeatureLayer, cannot use SelectFeatureByAttribute



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0















I am trying to select layer by attribute and for that, I am using arcpy.SelectLayerByAttribute_management function . My command is as follows:



arcpy.SelectLayerByAttribute_management(union, 'NEW_SELECTION', '"gridcode" =1 AND "area" >=25000000*0.4') 


Here, union is the shapefile, from which I need to select layers by attribute. My SQL query is "gridcode" =1 AND "area" >=25000000*0.4



I am getting the following error:




ExecuteError: Failed to execute. Parameters are not valid. The value
cannot be a feature class ERROR 000840: The value is not a Raster
Layer. ERROR 000840: The value is not a Mosaic Layer. Failed to
execute (SelectLayerByAttribute).`











share|improve this question
























  • couldn't you test "area" >= 10000000?

    – Vince
    Mar 13 at 11:34











  • yes i just changed it but still it gives error

    – lsr729
    Mar 13 at 11:35






  • 1





    Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

    – BERA
    Mar 13 at 12:02












  • We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

    – KHibma
    Mar 13 at 12:22






  • 1





    Yes it worked after making a layer first. Thank

    – lsr729
    Mar 13 at 12:47

















0















I am trying to select layer by attribute and for that, I am using arcpy.SelectLayerByAttribute_management function . My command is as follows:



arcpy.SelectLayerByAttribute_management(union, 'NEW_SELECTION', '"gridcode" =1 AND "area" >=25000000*0.4') 


Here, union is the shapefile, from which I need to select layers by attribute. My SQL query is "gridcode" =1 AND "area" >=25000000*0.4



I am getting the following error:




ExecuteError: Failed to execute. Parameters are not valid. The value
cannot be a feature class ERROR 000840: The value is not a Raster
Layer. ERROR 000840: The value is not a Mosaic Layer. Failed to
execute (SelectLayerByAttribute).`











share|improve this question
























  • couldn't you test "area" >= 10000000?

    – Vince
    Mar 13 at 11:34











  • yes i just changed it but still it gives error

    – lsr729
    Mar 13 at 11:35






  • 1





    Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

    – BERA
    Mar 13 at 12:02












  • We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

    – KHibma
    Mar 13 at 12:22






  • 1





    Yes it worked after making a layer first. Thank

    – lsr729
    Mar 13 at 12:47













0












0








0








I am trying to select layer by attribute and for that, I am using arcpy.SelectLayerByAttribute_management function . My command is as follows:



arcpy.SelectLayerByAttribute_management(union, 'NEW_SELECTION', '"gridcode" =1 AND "area" >=25000000*0.4') 


Here, union is the shapefile, from which I need to select layers by attribute. My SQL query is "gridcode" =1 AND "area" >=25000000*0.4



I am getting the following error:




ExecuteError: Failed to execute. Parameters are not valid. The value
cannot be a feature class ERROR 000840: The value is not a Raster
Layer. ERROR 000840: The value is not a Mosaic Layer. Failed to
execute (SelectLayerByAttribute).`











share|improve this question
















I am trying to select layer by attribute and for that, I am using arcpy.SelectLayerByAttribute_management function . My command is as follows:



arcpy.SelectLayerByAttribute_management(union, 'NEW_SELECTION', '"gridcode" =1 AND "area" >=25000000*0.4') 


Here, union is the shapefile, from which I need to select layers by attribute. My SQL query is "gridcode" =1 AND "area" >=25000000*0.4



I am getting the following error:




ExecuteError: Failed to execute. Parameters are not valid. The value
cannot be a feature class ERROR 000840: The value is not a Raster
Layer. ERROR 000840: The value is not a Mosaic Layer. Failed to
execute (SelectLayerByAttribute).`








arcpy select-by-attribute error-000840






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 13 at 12:29









artwork21

31.1k555120




31.1k555120










asked Mar 13 at 11:26









lsr729lsr729

294




294












  • couldn't you test "area" >= 10000000?

    – Vince
    Mar 13 at 11:34











  • yes i just changed it but still it gives error

    – lsr729
    Mar 13 at 11:35






  • 1





    Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

    – BERA
    Mar 13 at 12:02












  • We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

    – KHibma
    Mar 13 at 12:22






  • 1





    Yes it worked after making a layer first. Thank

    – lsr729
    Mar 13 at 12:47

















  • couldn't you test "area" >= 10000000?

    – Vince
    Mar 13 at 11:34











  • yes i just changed it but still it gives error

    – lsr729
    Mar 13 at 11:35






  • 1





    Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

    – BERA
    Mar 13 at 12:02












  • We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

    – KHibma
    Mar 13 at 12:22






  • 1





    Yes it worked after making a layer first. Thank

    – lsr729
    Mar 13 at 12:47
















couldn't you test "area" >= 10000000?

– Vince
Mar 13 at 11:34





couldn't you test "area" >= 10000000?

– Vince
Mar 13 at 11:34













yes i just changed it but still it gives error

– lsr729
Mar 13 at 11:35





yes i just changed it but still it gives error

– lsr729
Mar 13 at 11:35




1




1





Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

– BERA
Mar 13 at 12:02






Try creating a feature layer using MakeFeatureLayer, and pass this to SelectLayerByAttributes.

– BERA
Mar 13 at 12:02














We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

– KHibma
Mar 13 at 12:22





We'd need to see more of your script to know. Whatever union is appears to be the problem. It's not a layer that the tool wants.

– KHibma
Mar 13 at 12:22




1




1





Yes it worked after making a layer first. Thank

– lsr729
Mar 13 at 12:47





Yes it worked after making a layer first. Thank

– lsr729
Mar 13 at 12:47










1 Answer
1






active

oldest

votes


















0














You are trying to select on a feature class when it should be a feature layer. Use MakeFeatureLayer to create a layer then pass this to SelectLayerByAttributes:




Creates a feature layer from an input feature class or layer file




(It is also possible to use a where_clause in MakeFeatureLayer and then there is no need for SelectLayerByAttributes.)






share|improve this answer

























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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0














    You are trying to select on a feature class when it should be a feature layer. Use MakeFeatureLayer to create a layer then pass this to SelectLayerByAttributes:




    Creates a feature layer from an input feature class or layer file




    (It is also possible to use a where_clause in MakeFeatureLayer and then there is no need for SelectLayerByAttributes.)






    share|improve this answer





























      0














      You are trying to select on a feature class when it should be a feature layer. Use MakeFeatureLayer to create a layer then pass this to SelectLayerByAttributes:




      Creates a feature layer from an input feature class or layer file




      (It is also possible to use a where_clause in MakeFeatureLayer and then there is no need for SelectLayerByAttributes.)






      share|improve this answer



























        0












        0








        0







        You are trying to select on a feature class when it should be a feature layer. Use MakeFeatureLayer to create a layer then pass this to SelectLayerByAttributes:




        Creates a feature layer from an input feature class or layer file




        (It is also possible to use a where_clause in MakeFeatureLayer and then there is no need for SelectLayerByAttributes.)






        share|improve this answer















        You are trying to select on a feature class when it should be a feature layer. Use MakeFeatureLayer to create a layer then pass this to SelectLayerByAttributes:




        Creates a feature layer from an input feature class or layer file




        (It is also possible to use a where_clause in MakeFeatureLayer and then there is no need for SelectLayerByAttributes.)







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 13 at 13:59

























        answered Mar 13 at 13:53









        BERABERA

        17.4k62044




        17.4k62044



























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