Query Error when trying to run query
Please bear with me as I am relatively new to SQL. I am using MySQL and trying to run a query against a DB with two tables ('Network', and 'Server'). When I trying a portion of the query it executes without fail, however, when I try and add to it, I receive nothing but errors ranging from 1046 to 1140 to 1241. I am sure it is a matter of syntax and formatting, but I cannot figure it out. Please help. Below is my query:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS V,
(SELECT CONCAT(ROUND(SUM(server.Tech=2)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=2)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS J
(SELECT CONCAT(ROUND(SUM(server.Tech=4)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=4)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS K
mysql-5.7
New contributor
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Please bear with me as I am relatively new to SQL. I am using MySQL and trying to run a query against a DB with two tables ('Network', and 'Server'). When I trying a portion of the query it executes without fail, however, when I try and add to it, I receive nothing but errors ranging from 1046 to 1140 to 1241. I am sure it is a matter of syntax and formatting, but I cannot figure it out. Please help. Below is my query:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS V,
(SELECT CONCAT(ROUND(SUM(server.Tech=2)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=2)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS J
(SELECT CONCAT(ROUND(SUM(server.Tech=4)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=4)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS K
mysql-5.7
New contributor
add a comment |
Please bear with me as I am relatively new to SQL. I am using MySQL and trying to run a query against a DB with two tables ('Network', and 'Server'). When I trying a portion of the query it executes without fail, however, when I try and add to it, I receive nothing but errors ranging from 1046 to 1140 to 1241. I am sure it is a matter of syntax and formatting, but I cannot figure it out. Please help. Below is my query:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS V,
(SELECT CONCAT(ROUND(SUM(server.Tech=2)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=2)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS J
(SELECT CONCAT(ROUND(SUM(server.Tech=4)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=4)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS K
mysql-5.7
New contributor
Please bear with me as I am relatively new to SQL. I am using MySQL and trying to run a query against a DB with two tables ('Network', and 'Server'). When I trying a portion of the query it executes without fail, however, when I try and add to it, I receive nothing but errors ranging from 1046 to 1140 to 1241. I am sure it is a matter of syntax and formatting, but I cannot figure it out. Please help. Below is my query:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS V,
(SELECT CONCAT(ROUND(SUM(server.Tech=2)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=2)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS J
(SELECT CONCAT(ROUND(SUM(server.Tech=4)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=4)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server) AS K
mysql-5.7
mysql-5.7
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asked 16 mins ago
user8846971user8846971
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1 Answer
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You have to join the tables on a value. I'm taking a wild guess here, but I think you might be looking for something like this:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
) AS V,
I don't know what your equals sign is doing in your arithmetic operation. Are you trying to pull all Tech where it equals 7? If so, it would be
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
WHERE server.Tech = 7 AND network.Tech=7
) AS V,
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1 Answer
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1 Answer
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active
oldest
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You have to join the tables on a value. I'm taking a wild guess here, but I think you might be looking for something like this:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
) AS V,
I don't know what your equals sign is doing in your arithmetic operation. Are you trying to pull all Tech where it equals 7? If so, it would be
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
WHERE server.Tech = 7 AND network.Tech=7
) AS V,
add a comment |
You have to join the tables on a value. I'm taking a wild guess here, but I think you might be looking for something like this:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
) AS V,
I don't know what your equals sign is doing in your arithmetic operation. Are you trying to pull all Tech where it equals 7? If so, it would be
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
WHERE server.Tech = 7 AND network.Tech=7
) AS V,
add a comment |
You have to join the tables on a value. I'm taking a wild guess here, but I think you might be looking for something like this:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
) AS V,
I don't know what your equals sign is doing in your arithmetic operation. Are you trying to pull all Tech where it equals 7? If so, it would be
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
WHERE server.Tech = 7 AND network.Tech=7
) AS V,
You have to join the tables on a value. I'm taking a wild guess here, but I think you might be looking for something like this:
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech=7)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech=7)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
) AS V,
I don't know what your equals sign is doing in your arithmetic operation. Are you trying to pull all Tech where it equals 7? If so, it would be
SELECT (SELECT CONCAT(ROUND(SUM(server.Tech)/COUNT(server.Tech) * 100,2), '%') AS Server,
CONCAT(ROUND(SUM(network.Tech)/COUNT(network.Tech) * 100,2), '%') AS Network
FROM network join server
On network.Tech = server.Tech
WHERE server.Tech = 7 AND network.Tech=7
) AS V,
answered 1 min ago
Aaron RheamsAaron Rheams
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user8846971 is a new contributor. Be nice, and check out our Code of Conduct.
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