how to join or concat two result in single table
I have two sql query
This to find enroll count
select im.instituteName,count(*) AS enrollcnt from institute_csv_users icu Inner JOIN institute_master im ON icu.instituteId=im.id WHERE im.status=1 and icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
and this to find login count
select im.instituteName,count(*) AS logincnt from institute_csv_users icu LEFT JOIN institute_master im ON icu.instituteId=im.id where icu.email!='' AND im.status=1 AND icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
Need result like
| Institutename | enrollcnt | logincnt | Loginpercentage
mysql
asked 7 mins ago
dipali nigadedipali nigade
1
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I have two sql query
This to find enroll count
select im.instituteName,count(*) AS enrollcnt from institute_csv_users icu Inner JOIN institute_master im ON icu.instituteId=im.id WHERE im.status=1 and icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
and this to find login count
select im.instituteName,count(*) AS logincnt from institute_csv_users icu LEFT JOIN institute_master im ON icu.instituteId=im.id where icu.email!='' AND im.status=1 AND icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
Need result like
| Institutename | enrollcnt | logincnt | Loginpercentage
mysql
asked 7 mins ago
dipali nigadedipali nigade
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have two sql query
This to find enroll count
select im.instituteName,count(*) AS enrollcnt from institute_csv_users icu Inner JOIN institute_master im ON icu.instituteId=im.id WHERE im.status=1 and icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
and this to find login count
select im.instituteName,count(*) AS logincnt from institute_csv_users icu LEFT JOIN institute_master im ON icu.instituteId=im.id where icu.email!='' AND im.status=1 AND icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
Need result like
| Institutename | enrollcnt | logincnt | Loginpercentage
mysql
asked 7 mins ago
dipali nigadedipali nigade
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have two sql query
This to find enroll count
select im.instituteName,count(*) AS enrollcnt from institute_csv_users icu Inner JOIN institute_master im ON icu.instituteId=im.id WHERE im.status=1 and icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
and this to find login count
select im.instituteName,count(*) AS logincnt from institute_csv_users icu LEFT JOIN institute_master im ON icu.instituteId=im.id where icu.email!='' AND im.status=1 AND icu.centerId=1 group by im.id order by rl.region_name,im.instituteName ASC
Need result like
| Institutename | enrollcnt | logincnt | Loginpercentage
mysql
mysql
asked 7 mins ago
dipali nigadedipali nigade
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 mins ago
dipali nigadedipali nigade
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 mins ago
dipali nigadedipali nigade
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 mins ago
dipali nigadedipali nigade
1
asked 7 mins ago
dipali nigadedipali nigade
1
1
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
dipali nigade is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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