I'm a newbie to rails.I have a model called OfflineExport. In the table I have data like this
#<OfflineExport id: 2,
parameters:
{"project_id"=>"3",
"type"=>"submissions",
"filters"=>{"task_type"=>"",
"corrections"=>"", "grade"=>"",
"min_duration"=>"", "after"=>"",
"max_duration"=>"", "reviews"=>"",
"before"=>""},
"send_email"=>"true",
"options"=>{"offline_record_id"=>2}}>
Am trying to fetch parameters["project_id"] in where clause like
OfflineExport.where("parameters[project_id] = '3'")
But am getting error like:
ActiveRecord::StatementInvalid: PGError: ERROR: cannot subscript type text because it is not an array
can anyone help me in solving this?
It seems that you have a serialized column in your model. Something like this: serialize :parameters. This means the data is stored in the database not in an easily readable format, which leads that you can not query on it. Same as here.
Solution: extract the field you want to query on, and make a column for it.
Related
DATA: t_mara type STANDARD TABLE OF mara WITH EMPTY KEY.
DATA(t_data1) = VALUE ty_data( FOR s_mara IN t_mara ( s_mara–matnr ) ).
I am trying to implement a similar code using FOR statement but I am getting an error that the field is unknown in the work area even though it would be declared inline.
Can you please let me know what went wrong? This is my first time I am facing this error on FOR loop.
Not sure because you are not providing too much detail but try this:
DATA t_mara type STANDARD TABLE OF mara WITH EMPTY KEY.
DATA(t_data1) = VALUE ty_data( FOR s_mara IN t_mara ( matnr = s_mara-matnr ) ).
I am having trouble working with the JSONB structure in PostgreSQL. currently my data is saved as follows:
"{\"Hello\":\"World\",\"idx\":0}"
Which obviously is not correct 😀 so I am trying to "repair" this and get the actual JSON representation for querying with:
SELECT regexp_replace(trim('"' FROM json_data::text), '\\"', '"', 'g')::jsonb FROM My_table
However when trying this, I get the following error:
ERROR: invalid input syntax for type json
DETAIL: Token "Рыба" is invalid.
CONTEXT: JSON data, line 1: ...х, как : Люди X, Пароль \\"Рыба...
SQL state: 22P02
So I am thinking that this is due to the character encoding that is not being accepted by the JSONB standard.
My main question then is though, how can I repair this kind of table so that I am still able to query it? I tried utilizing conver_from and convert_to but am unable to figure out how to fix this error... did anyone encounter this already?
Update: Found it! (thanks to Convert JSON string to JSONB), utilizing
SELECT (json_data#>>'{}')::jsonb FROM my_table
fixed it
I am working on SQL query, I have pre-created SOQL query, I am looking for a way to convert it to SQL query. The query is:
SELECT CronJobDetail.Name, Id, CreatedDate, State
FROM CronTrigger
WHERE CronjobDetail.JobType = 'bla bla'
AND CronJobDetail.Name LIKE '%bla bla2%'
But It does not run on terminal when I try to create monitoring script in Ruby. The error that I get:
(Got exception: INVALID_FIELD: No such relation 'CronJobDetail' on
entity 'CronTrigger'. If you are attempting to use a custom field, be
sure to append the '__c' after the custom field name. Please reference
your WSDL or the describe call for the appropriate names. in
/Users/gakdugan/.rvm/gems/ruby-1.9.3-p547/gems/restforce-2.2.0/lib/restforce/middleware/raise_error.rb:18:in
`on_complete'
Do you have any idea how can I fix it and make it run on SQL?
You are trying to access a relation without adding it to your FROM clause. Alternatively, if that's a custom field name, then do what the error message suggests you to do (add __c after the custom field name).
You probably want to do something like this:
SELECT CronJobDetail.Name, CronTrigger.Id, CronTrigger.CreatedDate, CronTrigger.State
FROM CronTrigger
INNER JOIN CronJobDetail ON CronJobDetail.id = CronTrigger.foreign_id // this you have to do yourself
WHERE CronjobDetail.JobType = 'bla bla'
AND CronJobDetail.Name LIKE '%bla bla2%'
We are using Hibernate to connect to AS/400. We are having issues with a query on the AS/400
with the LIKE clause.
The following error is shown:
java.sql.SQLException: [SQL0131] Operands of LIKE not compatible or not valid
My query is its auto generated by Hibernate:
select tab_parame0_.C1IMCD as C1_560_, tab_parame0_.C1NINB as C2_560_,
tab_parame0_.C1JXCD as C3_560_, tab_parame0_.C1HLTX as C4_560_, tab_parame0_.C1HMTX as C5_560_,
tab_parame0_.C1HDST as C6_560_, tab_parame0_.C1NGNB as C7_560_, tab_parame0_.C1NJNB as C8_560_,
tab_parame0_.C1NFNB as C9_560_, tab_parame0_.C1NHNB as C10_560_, tab_parame0_.C1HCST as C11_560_
from RYC1REP tab_parame0_
where lower(tab_parame0_.C1HLTX) like lower(?)
order by tab_parame0_.C1IMCD asc
fetch first 10 rows only
SQL0131 indicates a type mismatch.
What datatype is tab_parame0_.C1HLTX? What datatype is your query parameter?
Please include your HQL/JPQL query source code for comparison.
You may have to set up an SQL trace to see exactly what the AS/400 is receiving.
See How do I obtain trace information from my Java program using the Toolbox?
I recommend you change LIKE LOWER(:parameter) to LIKE :parameter in your source query and use .toLowerCase() when you set the parameter and see how that works.
I have been struggling with this, i have two models and showing data in Cgridview with one model, this model contains some id's whose values are in different table
So, i have added
'value'=> 'TblAreaoflaw::model()->FindByPk($data->typeoflaw)->areaoflaw'
which is giving this error
"Trying to get property of non-object"
Might be due to this reason that the some records doesn't exist in the TblAreaoflaw. Can't we check in this line through isset?
When i put static value, it work well, like
'value'=> 'TblAreaoflaw::model()->FindByPk(5)->areaoflaw',
Could anyone please help
thanks a lot
The error you get is because this expression TblAreaoflaw::model()->FindByPk($data->typeoflaw) is returning null. This means that you are effectively trying to get null->areaoflaw which won't work (this is what the error message "Trying to get property of non-object" clarifies).
My best guess is that $data->typeoflaw returns a non-existing primary key for the TblAreaoflaw model.
Make sure :
TblAreaoflaw is actually a model, I doubt its Areaoflaw
You have database specified primary key which is the id (5) you are passing
Try:
'value'=> '(TblAreaoflaw::model()->FindByPk($data->typeoflaw)->areaoflaw) ?
: "default or null value"'
Obviously substitute the null string to whatever you want. You may need to adjust the condition to use !empty() or similar, but see how it goes. (And if you do that or aren't using PHP 5.3, use the full ternary expression.)