What I am trying to do
https://i.stack.imgur.com/QzsCT.png
SELECT message, "Need to add user here"
FROM database WHERE source="Device Thresholds"
You could create another table with possible responses, based on your question. Joining your table with "message" in one table and joining it with another table with "possible responses" may work.
Something like this would be ok but cant get the syntax correct
SELECT message, SUBSTRING(message,7,6)
FROM database WHERE source="Device Thresholds"
Where the number 6 is for number of characters I would like to find part of the srting as number of charachters which in my case would be "' "
SELECT message, SUBSTRING(message,7,find("' ",message))
FROM database WHERE source="Device Thresholds"
But this brings back an error
Related
The question I have is similar to a few I've seen, but the solutions don't seem to work for me, so I'll add a new one:
I am trying to access a numeric value (the highest user ID) from an SQLite table into my QT Program, with the following query:
SELECT Name, MAX(ID) FROM User_lib;
This works from the command shell, with the expected result.
I then attempt to access the maximal ID value with:
QSqlQuery qry_ID;
qry_ID.prepare("SELECT Name, MAX(User_lib.ID) FROM User_lib");
qry_ID.exec()
qDebug() << qry_ID.record().value("ID").toInt();
The result is zero (and empty if I access "Name" with toString())
Also, as a test
qDebug() << "Number of Rows: " << qry_ID.size();
which I've seen in several other answers gives me -1.
On the other hand,
qDebug() << "Number of columns: " << qry_ID.record().count();
gives me "Number of columns 2" as expected. It appears that the query gives no results.
At first I thought that I had a problem with my query, but the same thing is happening when I attempt to count the rows in queries that clearly work correctly (table is displayed, I can add elements correctly, etc), so I think I must be doing something wrong in QT.
Let's see what you actually get in the command-line shell:
sqlite> .mode columns
sqlite> .header on
sqlite> SELECT Name, MAX(ID) FROM User_lib;
Name MAX(ID)
---------- ----------
user30248 8562493
So in the output that you get from the query, the second column is not named ID but MAX(ID).
The documentation actually says:
The name of a result column is the value of the "AS" clause for that column, if there is an AS clause. If there is no AS clause then the name of the column is unspecified and may change from one release of SQLite to the next.
So use AS, or access the column by its index.
SQLite computes output rows on demand, so it is not possible to give a count of rows before they have all been read. Therefore, the QuerySize feature is not supported by the SQLite driver.
I haven't checked CL's answer above, which I am sure will work as well, but I have managed to find an alternative solution that worked for me and may be interesting. Since QsqlQuery's internal pointer is set to one field ahead of the first record, I have to advance with next. This is the code that worked:
QSqlQuery qry_ID;
qry_ID.prepare("SELECT Name, MAX(ID) FROM User_lib");
qry_ID.exec();
qry_ID.next();
QString name = qry_ID.value(0).toString();
int IDmax = qry_ID.value(1).toInt();
If you need to access various rows, a while loop is required, but that is a different question.
before i use alias for table i get the error:
: Integrity constraint violation: 1052 Column 'id' in field list is ambiguous
Then i used aliases and i get this error:
unknown index a
I am trying to get a list of category name ( dependant to a translation) and the associated category id which is unique. Since i need to put them in a select, i see that i should use the lists.
$categorie= DB::table('cat as a')
->join('campo_cat as c','c.id_cat','=','a.id')
->join('campo as d','d.id','=','c.id_campo')
->join('cat_nome as nome','nome.id_cat','=','a.id')
->join('lingua','nome.id_lingua','=','lingua.id')
->where('lingua.lingua','=','it-IT')
->groupby('nome.nome')
->lists('nome.nome','a.id');
The best way to debug your query is to look at the raw query Laravel generates and trying to run this raw query in your favorite SQL tool (Navicat, MySQL cli tool...), so you can dump it to log using:
DB::listen(function($sql, $bindings, $time) {
Log::info($sql);
Log::info($bindings);
});
Doing that with yours I could see at least one problem:
->where('lingua.lingua','=','it-IT')
Must be changed to
->where('lingua.lingua','=',"'it-IT'")
As #jmail said, you didn't really describe the problem very well, just what you ended up doing to get around (part of) it. However, if I read your question right you're saying that originally you did it without all the aliases you got the 'ambiguous' error.
So let me explain that first: this would happen, because there are many parts of that query that use id rather than a qualified table`.`id.
if you think about it, without aliases you query looks a bit like this: SELECT * FROM `cat` JOIN `campo_cat` ON `id_cat` = `id` JOIN `campo` ON `id` = `id_campo`; and suddenly, MySQL doesn't know to which table all these id columns refer. So to get around that all you need to do is namespace your fields (i.e. use ... JOIN `campo` ON `campo`.`id` = `campo_cat`.`id_campo`...). In your case you've gone one step further and aliased your tables. This certianly makes the query a little simpler, though you don't need to actually do it.
So on to your next issue - this will be a Laravel error. And presumably happening because your key column from lists($valueColumn, $keyColumn) isn't found in the results. This is because you're referring to the cat.id column (okay in your aliased case a.id) in part of the code that's no longer in MySQL - the lists() method is actually run in PHP after Laravel gets the results from the database. As such, there's no such column called a.id. It's likely it'll be called id, but because you don't request it specifically, you may find that the ambiguous issue is back. My suggestion would be to select it specifically and alias the column. Try something like the below:
$categories = DB::table('cat as a')
->join('campo_cat as c','c.id_cat','=','a.id')
->join('campo as d','d.id','=','c.id_campo')
->join('cat_nome as nome','nome.id_cat','=','a.id')
->join('lingua','nome.id_lingua','=','lingua.id')
->where('lingua.lingua','=','it-IT')
->groupby('nome.nome')
->select('nome.nome as nome_nome','a.id as a_id') // here we alias `.id as a_id
->lists('nome_nome','a_id'); // here we refer to the actual columns
It may not work perfectly (I don't use ->select() so don't know whether you pass an array or multiple parameters, also you may need DB::raw() wrapping each one in order to do the aliasing) but hopefully you get my meaning and can get it working.
hi i am executing nested "select" query in mysql .
the query is
SELECT `btitle` FROM `backlog` WHERE `bid` in (SELECT `abacklog_id` FROM `asprint` WHERE `aid`=184 )
I am not getting expected answer by the above query. If I execute:
SELECT abacklog_id FROM asprint WHERE aid=184
separately
I will get abacklog_id as 42,43,44,45;
So if again I execute:
SELECT `btitle` FROM `backlog` WHERE `bid` in(42,43,44,45)
I will get btitle as scrum1 scrum2 scrum3 msoffice
But if I combine those queries I will get only scrum1 remaining 3 atitle will not get.
You Can Try As Like Following...
SELECT `age_backlog`.`ab_title` FROM `age_backlog` LEFT JOIN `age_sprint` ON `age_backlog`.`ab_id` = `age_sprint`.`as_backlog_id` WHERE `age_sprint`.`as_id` = 184
By using this query you will get result with loop . You will be able to get all result with same by place with comma separated by using IMPLODE function ..
May it will be helpful for you... If you get any error , Please inform me...
What you did is to store comma separated values in age_sprint.as_backlog_id, right?
Your query actually becomes
SELECT `ab_title` FROM `age_backlog` WHERE `ab_id` IN ('42,43,44,45')
Note the ' in the IN() function. You don't get separate numbers, you get one string.
Now, when you do
SELECT CAST('42,43,44,45' AS SIGNED)
which basically is the implicit cast MySQL does, the result is 42. That's why you just get scrum1 as result.
You can search for dozens of answers to this problem here on SO.
You should never ever store comma separated values in a database. It violates the first normal form. In most cases databases are in third normal form or BCNF or even higher. Lower normal forms are just used in some special cases to get the most performance, usually for reporting issues. Not for actually working with data. You want 1 row for every as_backlog_id.
Again, your primary goal should be to get a better database design, not to write some crazy functions to get each comma separated number out of the field.
I'm trying to insert a ton of rows into my MySQL database. I have a query like this, but with about 700 more repetitive entries in it but for some reason the query is only inserting the first row to the database. In this case it would be '374','4957','0'.
INSERT INTO table VALUES ('374','4957','0'),('374','3834','0'),('374','4958','0'),('374','5076','0'),('374','4921','0'),('374','3835','0'),('374','4922','0'),('374','3836','0'),('374','3837','0'),('374','4879','0'),('374','3838','0')
I can't figure out what I'm doing wrong.
Thank you in advance.
Don't mean to state the obvious, but if the first field '374' is your primary key field, than this is the issue.
Otherwise, are there any error messages received from the database? That is always a good place to look for bugs.
For better understanding why something is not working next time use code like this:
$sql = "INSERT INTO table VALUES ('374','4957','0'),('374','3834','0')";
if (!mysqli_query($link, $sql)) {
printf("Errormessage: %s\n", mysqli_error($link));
}
That should display error message returned from MySQL.
More information: PHP manual - mysqli_error
Try to write the column names before values.
For example:
INSERT INTO table (column1,column2,column3) VALUES ...
This may be a futile question, but I will ask anyway. I have now learned that it is bad practice to use a question mark at the end of a field name, as is the case with the Paid? field in the following statement:
$sql = "INSERT INTO `tblAppeals`
(
`#`,
`Year`,
`Property#`,
`Paid?`,
`Outcome`,
`ResolvedBy`,
`AppealCategory`
)
VALUES (?,?,?,?,?,?,?)";
When I try to run the statement, I get an error because the question mark is not handled correctly. I haven't been able to find any workarounds to avoid having to go back and change the field name.
Is there any way I can keep the field name the same, Paid?, and still use it in the INSERT statement? Thanks.
It looks like its an issue with your query layer and not MySQL itself. That is, whatever is doing the bind params handling is eagerly looking for all ? in the SQL and not just whats in the VALUES part of the clause.
What database drive / query framework are you using?