I'm trying to use Query Variables in Grafana, the panel query source is PostgreSQL for QuestDB.
I have added the variable without any issue, but I'm unable to use the variable in Panel query since the variable values contains the spaces (SENSOR01 ON_OFF), also I'm unable to figure-out how to add single quote escape.
Following are the scenarios I tried:
Scenario1: this indicates due to space in the Variable value, on_off considered as separate word
where sensor_name = $sensor
db query error: pq: unexpected token: on_off
.
.
Scenario2: tried to add single quotes explicitly for the variable value, but there is generic error from source DB (QuestDB)
where sensor_name = concat('''', $sensor, '''')
db query error: pq: dangling expression
When tried Scenario2 approach directly in query of Variable, getting the same error
..
Scenario3: Hard-coded the variable value with space and with single quotes, but this giving me error with first part of the variable, looks like the hard-coded single quotes not passed here!
Error (Scenario3):
Is there any way/workaround to tackle this issue?
Could you just add the quotes directly in the query?
where sensor_name = '$sensor'
I have a similar grafana panel querying a questDB database using a variable and it works for me. This is my query:
select device_type, avg(duration_ms) as avg_duration_ms, avg(speed) as avg_speed, avg(measure1) as avg_m1, avg(measure2) as avg_m2 from ilp_test
WHERE
$__timeFilter(timestamp) and device_type = '$deviceType'
A rather hacky workaround would be to do:
where sensor_name = concat(cast(cast('&' as int) + 1 as char), $sensor, cast(cast('&' as int) + 1 as char))
This should work, but I'm pretty sure there is a better solution. Let me find it and get back to you.
Update. We may support Postgres syntax (which is '' escaping for a single quote char) in one of upcoming versions. For now, you'd have to use the above workaround.
Related
I am trying to run this query:
select *
from my_table
where column_one=${myValue}
I get the following error in Datagrip:
[42883] ERROR: operator does not exist: character varying = bigint Hint: No operator matches the given name and argument types. You might need to add explicit type casts.
Now, I have found this question, and I can fix the error by putting a string like this:
select *
from my_table
where column_one='123'
What I need is a way to pass in the '123' as a parameter. I usually do this ${myValue} and it works, but I am not sure how to keep my variable there as an input so I can run dynamic queries in code and let Postgres understand I want to pass in a string and not a number.
Any suggestions?
Here's a screenshot of how I am putting the parameter value in DataGrip...:
Ok, so, I just tried to put quotes in the data grip parameters input field for myValue #thirumal's answer things work. I didn't know I have to quote the value for it to work.
This is what it looks like:
Type cast ${myValue} using SQL Standard,
cast(${myValue} AS varchar)
or using Postgres Syntax:
${myValue}::varchar
I need to be able to use variables in table names - I basically have the same set of tables used for different types of data, so I would like to just have one dashboard and swapping between all types instead of always having to set up multiple identical dashboards.
My query is something like:
select * from table_$variable_name;
Where my list of possible variable is something like cat, dog, bird
I can seem to make this work, if I only put the variable as shown above I get the following error
Error 1146: Table 'table_$variable_name' doesn't exist
If I enclose it in curly brackets, I get this error instead.
Error 1064: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '{bird}' at line 1
(i.e. with the selected variable actually being visible this time)
I'm not sure if the issue is having underscores in the table names, I tried putting underscores around my variables too to check and I had no luck with that.
Another thing I tried was gradually adding on to the table name, so e.g.
select * from table_$variable;
Still returns an error, but I can see the table name starting to form correctly
Error 1146: Table 'table_bird_' doesn't exist
However, as soon as I add another underscore, the variable is not picked up abymore
```Error 1146: Table 'table_$variable_' doesn't exist``
I'm sure it's something silly I am missing in the syntax of the query - anyone has any suggestions?
Using this https://grafana.com/docs/grafana/latest/variables/templates-and-variables/ for reference
As #arturomp suggests, use
${var:raw}
At least in my case, this was the solution that worked.
I found double square brackets work. e.g.
Rather than
select * from table_$variable_name;
use
select * from table_[[variable_name]];
I'm trying to use update_all to update any records that is missing a key in a JSON stored in a table cell. ids is the ids of those records and I've tried the below...
User.where(id: ids).
update_all(
"preferences = jsonb_set(preferences, '{some_key}', 'true'"
)
Where the error returns is...
Caused by PG::SyntaxError: ERROR: syntax error at or near "WHERE"
LINE 1: ...onb_set(preferences, '{some_key}', 'true' WHERE "user...
The key takes a string value so not sure why the query is failing.
UPDATE:
Based on what was mentioned, I added the parentheses and also added / modified the last two arguments...
User.where(id: ids).
update_all(
"preferences = jsonb_set(preferences, '{some_key}', 'true'::jsonb, true)"
)
still running into issues and this time it seems related to the key I'm passing
I know this key doesn't currently exist for the set of ids
I added true for create_missing so that 1 isn't an issue
I get this error now...
Caused by PG::UndefinedFunction: ERROR: function jsonb_set(hstore, unknown, jsonb, boolean) does not exis
some_key should be a key in preferences
You're passing in raw SQL so you are 100% responsible for ensuring that is actually valid SQL. What you have there isn't. Check your parentheses:
User.where(id: ids).
update_all(
"preferences = jsonb_set(preferences, '{some_key}', 'true')"
)
If you look more closely at the error message it was telling you there was a problem precisely at the introduction of the WHERE clause, and right after ...true' so that was a good place to look for problems.
Syntax errors like this can be really annoying, but don't forget your database will usually do its best to pin down the place where the problem occurs.
I've been using MySQL for close to 10 years, but have recently jumped into a project that's exposed me to Postgres for the first time. Most of the differences aren't a bit deal, but I have been running into some small issues along the way that are throwing me off.
My latest one just has me stuck. I'm sure it's a simple work-around, but I haven't been able to find it. I need to update a set of records and the column name is, "view" - which is more than likely the cause of this issue, and my own dump fault. But I can't figure out how to get around it.
Here's my query:
UPDATE rv_template_fields SET view = "display_type_1"
WHERE rv_template_fields.view = "display_type_2"
It's causing the error:
ERROR: column "display_type_1" does not exist
It's clearly jumping past the column named "view". I'm not sure how to specify that as a string and not a reserved word.
For string literals, you should you single quote instead of double quote:
UPDATE rv_template_fields SET view = 'display_type_1' WHERE rv_template_fields.view = 'display_type_2'
Double quotes are for quoting identifiers of fields and relations, like, for instance view, so that you could write also:
UPDATE rv_template_fields SET "view" = 'display_type_1' WHERE "view" = 'display_type_2'
It has nothing to do with view: In postgres, double quotes (") are like backticks in mysql - if you code "display_type_1", you're telling postgres to use the identifier display_type_1.
Use single quotes for string literals:
UPDATE rv_template_fields SET view = 'display_type_1'
WHERE rv_template_fields.view = 'display_type_2'
Use double quotes when you have a (poorly named) identifier that's reserved word, like select * from "join" if your table name is literally join etc.
I'm trying to convert this NVARCHAR value into a periodeId.
The Raw data could be '12-02'.
My solution for this was first to try this
(1000+CONVERT(INT,LEFT(2,T1.PERIOD_NAME)))*100+CONVERT(Int,RIGHT(2,t1.PERIOD_NAME))
But i get the same error message here and could find any quick solution for it.
I also tried to just do a simple
LEFT(2,T1.PERIOD_NAME) to see if it was the formula itself that crashed it, but the same error came up.
If you want '12-02' to be 1202, then use replace() to remove the hyphen before conversion:
select cast(replace(period_name, '-', '') as int)
In SQL Server 2012+, you should use try_convert(), in case there are other unexpected values.
You can try:
SELECT (1000+CONVERT(INT,LEFT(t1.PERIOD_NAME,2)))*100+CONVERT(Int,RIGHT(t1.PERIOD_NAME,2))
The character_expression that LEFT operates on is at first place, whereas the integer expression that specifies how many characters of the character_expression will be returned, comes at second place.