- 发票表
SELECT id, fname, gtotal, `date` FROM invoice WHERE id = 1;
| id | fname | gtotal | date |
|----|---------|--------|-----------------------|
| 1 | Brandon | 860 | May, 11 2016 00:00:00 |
- invoice_contents 表,
SELECT * FROM invoice_contents WHERE invoice_id = 1;
| id | invoice_id | item | price | quantity | discount | total |
|----|------------|------------|-------|----------|----------|-------|
| 1 | 1 | Dextrose | 10 | 10 | 5 | 95 |
| 2 | 1 | Nescaine | 20 | 30 | 10 | 540 |
| 3 | 1 | Anticavity | 30 | 10 | 25 | 225 |
- 这个 JOIN 查询
SELECT invoice.id, invoice.fname, invoice_contents.item,
invoice_contents.price, invoice_contents.quantit,
invoice_contents.discount, invoice_contents.total,
invoice.gtotal
FROM invoice_contents
INNER JOIN invoice ON invoice_contents.invoice_id=1 AND invoice.id=1;
给出这个结果。
| id | fname | item | price | quantity | discount | total | gtotal |
|----|---------|------------|-------|----------|----------|-------|--------|
| 1 | Brandon | Dextrose | 10 | 10 | 5 | 95 | 860 |
| 1 | Brandon | Nescaine | 20 | 30 | 10 | 540 | 860 |
| 1 | Brandon | Anticavity | 30 | 10 | 25 | 225 | 860 |
我需要这个结果。
| id | fname | item | price | quantity | discount | total | gtotal |
|----|---------|------------|-------|----------|----------|-------|--------|
| 1 | Brandon | Dextrose | 10 | 10 | 5 | 95 | 860 |
| | | Nescaine | 20 | 30 | 10 | 540 | |
| | | Anticavity | 30 | 10 | 25 | 225 | |
我只是 MySQL 的初学者。从今天早上开始,我一直在尝试通过尝试不同的组合来获得这种输出,请帮助我。
最佳答案
@Rex,您的选择是正确的。您应该使用一些脚本制作所需的输出,例如PHP.
关于没有重复值的MySQL连接查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37345342/