Connecting mySQL table to itself to cross check results?
Not sure that's the best question title, but what I have is this:
db: usernames
id, username, search_via, value_1
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
4, Jim, Bill, 4
5, Brian, Bob, 5
I want to be able to search for the all the users with a search_via of 'Sue', so obviously I have:
SELECT * FROM usernames WHERE search_via = 'Sue':
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
...but then I want, from those results, to also return any usernames which appear as a search_via as well, so in the case it would be 'Brian', as it is a search_via value for Bob, who appeared as a username for 'Sue':
5, Brian, Bob, 5
..so my full results should be:
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
5, Brian, Bob, 5
...because my first search_via name was Sue, and in Sue's results, there was Bob, who was a search_via value for 'Brian'
...I've really struggled explaining this, so I have absolutely zero idea how to go about it?!?
Any ideas?
mysql
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
|
show 1 more comment
Not sure that's the best question title, but what I have is this:
db: usernames
id, username, search_via, value_1
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
4, Jim, Bill, 4
5, Brian, Bob, 5
I want to be able to search for the all the users with a search_via of 'Sue', so obviously I have:
SELECT * FROM usernames WHERE search_via = 'Sue':
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
...but then I want, from those results, to also return any usernames which appear as a search_via as well, so in the case it would be 'Brian', as it is a search_via value for Bob, who appeared as a username for 'Sue':
5, Brian, Bob, 5
..so my full results should be:
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
5, Brian, Bob, 5
...because my first search_via name was Sue, and in Sue's results, there was Bob, who was a search_via value for 'Brian'
...I've really struggled explaining this, so I have absolutely zero idea how to go about it?!?
Any ideas?
mysql
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
1
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Why does n't4, Jim, Bill, 4also appear in output ?Billis also searchable viaSue
– Madhur Bhaiya
Nov 14 '18 at 13:43
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44
|
show 1 more comment
Not sure that's the best question title, but what I have is this:
db: usernames
id, username, search_via, value_1
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
4, Jim, Bill, 4
5, Brian, Bob, 5
I want to be able to search for the all the users with a search_via of 'Sue', so obviously I have:
SELECT * FROM usernames WHERE search_via = 'Sue':
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
...but then I want, from those results, to also return any usernames which appear as a search_via as well, so in the case it would be 'Brian', as it is a search_via value for Bob, who appeared as a username for 'Sue':
5, Brian, Bob, 5
..so my full results should be:
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
5, Brian, Bob, 5
...because my first search_via name was Sue, and in Sue's results, there was Bob, who was a search_via value for 'Brian'
...I've really struggled explaining this, so I have absolutely zero idea how to go about it?!?
Any ideas?
mysql
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
Not sure that's the best question title, but what I have is this:
db: usernames
id, username, search_via, value_1
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
4, Jim, Bill, 4
5, Brian, Bob, 5
I want to be able to search for the all the users with a search_via of 'Sue', so obviously I have:
SELECT * FROM usernames WHERE search_via = 'Sue':
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
...but then I want, from those results, to also return any usernames which appear as a search_via as well, so in the case it would be 'Brian', as it is a search_via value for Bob, who appeared as a username for 'Sue':
5, Brian, Bob, 5
..so my full results should be:
1, Rita, Sue, 1
2, Bob, Sue, 2
3, Bill, Sue, 3
5, Brian, Bob, 5
...because my first search_via name was Sue, and in Sue's results, there was Bob, who was a search_via value for 'Brian'
...I've really struggled explaining this, so I have absolutely zero idea how to go about it?!?
Any ideas?
mysql
mysql
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
asked Nov 14 '18 at 13:30
TheBounderTheBounder
244
244
1
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Why does n't4, Jim, Bill, 4also appear in output ?Billis also searchable viaSue
– Madhur Bhaiya
Nov 14 '18 at 13:43
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44
|
show 1 more comment
1
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Why does n't4, Jim, Bill, 4also appear in output ?Billis also searchable viaSue
– Madhur Bhaiya
Nov 14 '18 at 13:43
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44
1
1
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Why does n't
4, Jim, Bill, 4 also appear in output ? Bill is also searchable via Sue– Madhur Bhaiya
Nov 14 '18 at 13:43
Why does n't
4, Jim, Bill, 4 also appear in output ? Bill is also searchable via Sue– Madhur Bhaiya
Nov 14 '18 at 13:43
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44
|
show 1 more comment
1 Answer
1
active
oldest
votes
To get results upto second level hierarchy only, we can simply use a subquery to fetch all the username values where search_via = 'Sue'. We can then use WHERE search_via IN (..) to get second level hierarchy results as well.
SELECT *
FROM usernames
WHERE search_via = 'Sue' OR
search_via IN (
SELECT username
FROM usernames
WHERE search_via = 'Sue'
)
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53301411%2fconnecting-mysql-table-to-itself-to-cross-check-results%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
To get results upto second level hierarchy only, we can simply use a subquery to fetch all the username values where search_via = 'Sue'. We can then use WHERE search_via IN (..) to get second level hierarchy results as well.
SELECT *
FROM usernames
WHERE search_via = 'Sue' OR
search_via IN (
SELECT username
FROM usernames
WHERE search_via = 'Sue'
)
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
add a comment |
To get results upto second level hierarchy only, we can simply use a subquery to fetch all the username values where search_via = 'Sue'. We can then use WHERE search_via IN (..) to get second level hierarchy results as well.
SELECT *
FROM usernames
WHERE search_via = 'Sue' OR
search_via IN (
SELECT username
FROM usernames
WHERE search_via = 'Sue'
)
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
add a comment |
To get results upto second level hierarchy only, we can simply use a subquery to fetch all the username values where search_via = 'Sue'. We can then use WHERE search_via IN (..) to get second level hierarchy results as well.
SELECT *
FROM usernames
WHERE search_via = 'Sue' OR
search_via IN (
SELECT username
FROM usernames
WHERE search_via = 'Sue'
)
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
To get results upto second level hierarchy only, we can simply use a subquery to fetch all the username values where search_via = 'Sue'. We can then use WHERE search_via IN (..) to get second level hierarchy results as well.
SELECT *
FROM usernames
WHERE search_via = 'Sue' OR
search_via IN (
SELECT username
FROM usernames
WHERE search_via = 'Sue'
)
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
answered Nov 14 '18 at 13:32
Madhur BhaiyaMadhur Bhaiya
19.5k62236
19.5k62236
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
add a comment |
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
You are picking up an extra record (see here). Not that my answer seems correct either :-(
– Tim Biegeleisen
Nov 14 '18 at 13:40
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
@TimBiegeleisen thanks for the fiddle. Actually OP's expected output is wrong. His sample data covers all the usernames actually. Remaining two users are at second level.
– Madhur Bhaiya
Nov 14 '18 at 13:42
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53301411%2fconnecting-mysql-table-to-itself-to-cross-check-results%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Do you want this search result to extend to third level and beyond as well ?
– Madhur Bhaiya
Nov 14 '18 at 13:32
Why does n't
4, Jim, Bill, 4also appear in output ?Billis also searchable viaSue– Madhur Bhaiya
Nov 14 '18 at 13:43
Yes, very much so.
– TheBounder
Nov 14 '18 at 13:44
"Yes, very much so" - Is this answer to my first question ? or the second question ?
– Madhur Bhaiya
Nov 14 '18 at 13:44
You're right, 'Bill' should also appear. My mistake.
– TheBounder
Nov 14 '18 at 13:44