Mixing
MYSQL LEFT JOIN not showing all the data
I am running this SQL LEFT JOIN query in PHP but it's showing one important column which is user_id as null which is not null.
MYSQL QUERY
SELECT
a.*,
b.*,
c.*,
d.*
FROM ts_users_skills a
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
WHERE (a.`skillsTeach` = '$skill_1' OR a.`skillsTeach` = '$skill_2'
OR a.`skillsTeach` = '$skill_3')
GROUP BY a.`user_id`;
JSON output
{
"status": "success",
"nearby_teachers": [
{
"user_id": null,
"skillsTeach": "PHP",
"email": "praveenkumarkp666@gmail.com",
"country_code": "91",
"username": "praveenkum",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-02-25",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "marketing",
"email": "ashok@procusa.in",
"country_code": "91",
"username": "ashok",
"name": "Ashok the founder 2",
"birthday": "1993-11-18",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "html",
"email": "talspo@gmail.com",
"country_code": "91",
"username": "talspo",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-01-01",
"location": "Bhubaneswar"
}
]
}
Thanks in Advance.
mysql json mysqli left-join
add a comment |
I am running this SQL LEFT JOIN query in PHP but it's showing one important column which is user_id as null which is not null.
MYSQL QUERY
SELECT
a.*,
b.*,
c.*,
d.*
FROM ts_users_skills a
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
WHERE (a.`skillsTeach` = '$skill_1' OR a.`skillsTeach` = '$skill_2'
OR a.`skillsTeach` = '$skill_3')
GROUP BY a.`user_id`;
JSON output
{
"status": "success",
"nearby_teachers": [
{
"user_id": null,
"skillsTeach": "PHP",
"email": "praveenkumarkp666@gmail.com",
"country_code": "91",
"username": "praveenkum",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-02-25",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "marketing",
"email": "ashok@procusa.in",
"country_code": "91",
"username": "ashok",
"name": "Ashok the founder 2",
"birthday": "1993-11-18",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "html",
"email": "talspo@gmail.com",
"country_code": "91",
"username": "talspo",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-01-01",
"location": "Bhubaneswar"
}
]
}
Thanks in Advance.
mysql json mysqli left-join
Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Which tableuser_idcomes from? Show the code where you build the result
– Jens
Jan 2 at 14:00
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12
add a comment |
I am running this SQL LEFT JOIN query in PHP but it's showing one important column which is user_id as null which is not null.
MYSQL QUERY
SELECT
a.*,
b.*,
c.*,
d.*
FROM ts_users_skills a
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
WHERE (a.`skillsTeach` = '$skill_1' OR a.`skillsTeach` = '$skill_2'
OR a.`skillsTeach` = '$skill_3')
GROUP BY a.`user_id`;
JSON output
{
"status": "success",
"nearby_teachers": [
{
"user_id": null,
"skillsTeach": "PHP",
"email": "praveenkumarkp666@gmail.com",
"country_code": "91",
"username": "praveenkum",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-02-25",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "marketing",
"email": "ashok@procusa.in",
"country_code": "91",
"username": "ashok",
"name": "Ashok the founder 2",
"birthday": "1993-11-18",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "html",
"email": "talspo@gmail.com",
"country_code": "91",
"username": "talspo",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-01-01",
"location": "Bhubaneswar"
}
]
}
Thanks in Advance.
mysql json mysqli left-join
I am running this SQL LEFT JOIN query in PHP but it's showing one important column which is user_id as null which is not null.
MYSQL QUERY
SELECT
a.*,
b.*,
c.*,
d.*
FROM ts_users_skills a
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
WHERE (a.`skillsTeach` = '$skill_1' OR a.`skillsTeach` = '$skill_2'
OR a.`skillsTeach` = '$skill_3')
GROUP BY a.`user_id`;
JSON output
{
"status": "success",
"nearby_teachers": [
{
"user_id": null,
"skillsTeach": "PHP",
"email": "praveenkumarkp666@gmail.com",
"country_code": "91",
"username": "praveenkum",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-02-25",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "marketing",
"email": "ashok@procusa.in",
"country_code": "91",
"username": "ashok",
"name": "Ashok the founder 2",
"birthday": "1993-11-18",
"location": "Bhubaneswar"
},
{
"user_id": null,
"skillsTeach": "html",
"email": "talspo@gmail.com",
"country_code": "91",
"username": "talspo",
"name": "Procusa Founder",
"gender": "M",
"birthday": "1997-01-01",
"location": "Bhubaneswar"
}
]
}
Thanks in Advance.
mysql json mysqli left-join
mysql json mysqli left-join
asked Jan 2 at 13:56
user8893870
Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Which tableuser_idcomes from? Show the code where you build the result
– Jens
Jan 2 at 14:00
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12
add a comment |
Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Which tableuser_idcomes from? Show the code where you build the result
– Jens
Jan 2 at 14:00
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12
Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Which table
user_id comes from? Show the code where you build the result– Jens
Jan 2 at 14:00
Which table
user_id comes from? Show the code where you build the result– Jens
Jan 2 at 14:00
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12
add a comment |
2 Answers
2
active
oldest
votes
You have several columns user_id in different tables.
In your JSON there is only one user_id field, it is not clear from which table it is.
It is clear tho, that the data is getting lost at this point, the JSON contains a user_id field from a dataset without a user_id field filled.
Try using user_id from a different table than ts_users_skills a.
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
|
show 1 more comment
The problem with your code is that it returns several columns named user_id. Some of those values, that come from LEFT JOINs, may come up as NULL.
When serializing to JSON, where keys are unique, it is likely that the « wrong » field is being chosen, resulting in a NULL value.
To solve this, you want to list precisely the fields to retrieve from each table, instead of using *. Of course, user_id should come from a table that is not LEFT JOINed :
SELECT
a.user_id, a.field1, a.field2
b.field11, b.field12
c.field21, c.field22, c.field23
d.field31
...
Details
You are selecting fields using the * :
SELECT
a.*,
b.*,
c.*,
d.*
...
But your query shows that you have multiple fields named user_id :
...
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
...
answered Jan 2 at 14:06
GMBGMB
18.8k31028
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have several columns user_id in different tables.
In your JSON there is only one user_id field, it is not clear from which table it is.
It is clear tho, that the data is getting lost at this point, the JSON contains a user_id field from a dataset without a user_id field filled.
Try using user_id from a different table than ts_users_skills a.
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
|
show 1 more comment
You have several columns user_id in different tables.
In your JSON there is only one user_id field, it is not clear from which table it is.
It is clear tho, that the data is getting lost at this point, the JSON contains a user_id field from a dataset without a user_id field filled.
Try using user_id from a different table than ts_users_skills a.
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
|
show 1 more comment
You have several columns user_id in different tables.
In your JSON there is only one user_id field, it is not clear from which table it is.
It is clear tho, that the data is getting lost at this point, the JSON contains a user_id field from a dataset without a user_id field filled.
Try using user_id from a different table than ts_users_skills a.
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
You have several columns user_id in different tables.
In your JSON there is only one user_id field, it is not clear from which table it is.
It is clear tho, that the data is getting lost at this point, the JSON contains a user_id field from a dataset without a user_id field filled.
Try using user_id from a different table than ts_users_skills a.
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
answered Jan 2 at 14:01
DanFromGermanyDanFromGermany
19.9k74796
19.9k74796
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
|
show 1 more comment
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
i did not downvote. I just saw your answer.
– user8893870
Jan 2 at 14:05
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Try using user_id from a different table than ts_users_skills a how do you know that OP takes user_id from this table without seeing any code?
– Jens
Jan 2 at 14:09
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Yeah, I got it actually. the geo_records table was the culprit. I removed it and it worked.
– user8893870
Jan 2 at 14:24
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
Thank you @DanFromGermany for your time. I am indebted and grateful to you.
– user8893870
Jan 2 at 14:25
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
@Jens no code? He shows is SQL statement. I am programing PHP/MySQL for 18 years now, I have seen all errors, I have seen the OPs problem before and the problem and the reason is just logical - you have several fields, some can be NULL because you LEFT JOIN - result: you picked the wrong field (from the RIGHT side of the dataset).
– DanFromGermany
Jan 2 at 15:48
|
show 1 more comment
The problem with your code is that it returns several columns named user_id. Some of those values, that come from LEFT JOINs, may come up as NULL.
When serializing to JSON, where keys are unique, it is likely that the « wrong » field is being chosen, resulting in a NULL value.
To solve this, you want to list precisely the fields to retrieve from each table, instead of using *. Of course, user_id should come from a table that is not LEFT JOINed :
SELECT
a.user_id, a.field1, a.field2
b.field11, b.field12
c.field21, c.field22, c.field23
d.field31
...
Details
You are selecting fields using the * :
SELECT
a.*,
b.*,
c.*,
d.*
...
But your query shows that you have multiple fields named user_id :
...
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
...
answered Jan 2 at 14:06
GMBGMB
18.8k31028
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
add a comment |
The problem with your code is that it returns several columns named user_id. Some of those values, that come from LEFT JOINs, may come up as NULL.
When serializing to JSON, where keys are unique, it is likely that the « wrong » field is being chosen, resulting in a NULL value.
To solve this, you want to list precisely the fields to retrieve from each table, instead of using *. Of course, user_id should come from a table that is not LEFT JOINed :
SELECT
a.user_id, a.field1, a.field2
b.field11, b.field12
c.field21, c.field22, c.field23
d.field31
...
Details
You are selecting fields using the * :
SELECT
a.*,
b.*,
c.*,
d.*
...
But your query shows that you have multiple fields named user_id :
...
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
...
answered Jan 2 at 14:06
GMBGMB
18.8k31028
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
add a comment |
The problem with your code is that it returns several columns named user_id. Some of those values, that come from LEFT JOINs, may come up as NULL.
When serializing to JSON, where keys are unique, it is likely that the « wrong » field is being chosen, resulting in a NULL value.
To solve this, you want to list precisely the fields to retrieve from each table, instead of using *. Of course, user_id should come from a table that is not LEFT JOINed :
SELECT
a.user_id, a.field1, a.field2
b.field11, b.field12
c.field21, c.field22, c.field23
d.field31
...
Details
You are selecting fields using the * :
SELECT
a.*,
b.*,
c.*,
d.*
...
But your query shows that you have multiple fields named user_id :
...
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
...
answered Jan 2 at 14:06
GMBGMB
18.8k31028
The problem with your code is that it returns several columns named user_id. Some of those values, that come from LEFT JOINs, may come up as NULL.
When serializing to JSON, where keys are unique, it is likely that the « wrong » field is being chosen, resulting in a NULL value.
To solve this, you want to list precisely the fields to retrieve from each table, instead of using *. Of course, user_id should come from a table that is not LEFT JOINed :
SELECT
a.user_id, a.field1, a.field2
b.field11, b.field12
c.field21, c.field22, c.field23
d.field31
...
Details
You are selecting fields using the * :
SELECT
a.*,
b.*,
c.*,
d.*
...
But your query shows that you have multiple fields named user_id :
...
LEFT JOIN ts_students_log b
ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
ON a.`user_id` = d.`user_id`
...
answered Jan 2 at 14:06
GMBGMB
18.8k31028
edited Jan 2 at 14:11
answered Jan 2 at 14:06
GMBGMB
18.8k31028
answered Jan 2 at 14:06
GMBGMB
18.8k31028
answered Jan 2 at 14:06
GMBGMB
18.8k31028
18.8k31028
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
add a comment |
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
yeah I have multiple user_id. let me try.
– user8893870
Jan 2 at 14:10
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
This is practically the same as my answer but your explanation is way better understandable
– DanFromGermany
Jan 2 at 15:50
add a comment |
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Learn about prepared Statements to prevent SQL injection
– Jens
Jan 2 at 14:00
Which table
user_idcomes from? Show the code where you build the result– Jens
Jan 2 at 14:00
Thanks for your advice on Prepared Statement. I am implementing it.
– user8893870
Jan 2 at 14:02
ts_students_log is the table where user_id is located as a Primary Key but it is present in almost every table as a Foreign key.
– user8893870
Jan 2 at 14:03
Can you please show the php code wher you build the result?
– Jens
Jan 2 at 14:12