MySQL-根据字段值修改查询

我希望根据sample.snop_轴的值（可以是'M'或'F'）更改以下查询：

SELECT *
FROM specimen,
     topography_index,
     morphology,
     FUNCTION
WHERE SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code
  AND morphology.morphology_code = specimen.snop_code
  AND specimen.topography_index = '_ORGAN_'

因此，如果sample.snop_轴==M，则添加

和morphics.morphics_代码=sample.snop_代码

否则，如果sample.snop_轴==F，则添加

AND functions.function_code=sample.snop_code

更新

我已按建议将查询更改为：

SELECT * FROM specimen, topography_index, morphology, functions 
WHERE 
SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code 
AND 
IF(specimen.snop_axis == 'M', morphology.morphology_code = specimen.snop_code, functions.function_code = specimen.snop_code) 
AND 
specimen.topography_index = '_ORGAN_'

但我有一个错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '== 'M', morphology.morphology_code = specimen.snop_code, functions.function_code' at line 1

试试这个

SELECT *
FROM specimen,
     topography_index,
     morphology,
     FUNCTION
WHERE SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code
  AND specimen.topography_index = '_ORGAN_'
  AND if(specimen.snop_axis = 'M', morphology.morphology_code = specimen.snop_code, functions.function_code = specimen.snop_code);

---

您可以像这样简单地使用case语句：-

SELECT * FROM specimen, topography_index, morphology, function WHERE SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code 
AND morphology.morphology_code = specimen.snop_code 
AND specimen.topography_index = '_ORGAN_'
AND CASE WHEN specimen.snop_axis = M THEN morphology.morphology_code = specimen.snop_code
ELSE function.function_code = specimen.snop_code END;

---

这可能会对您有所帮助。

您可以使用逻辑运算AND和OR将它们作为条件子句，如下所示：

SELECT *
FROM specimen,
     topography_index,
     morphology,
     FUNCTION
WHERE SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code
  AND morphology.morphology_code = specimen.snop_code
  AND specimen.topography_index = '_ORGAN_'
  AND (
    (specimen.snop_axis = 'M' AND morphology.morphology_code = specimen.snop_code)
    OR 
    (specimen.snop_axis = 'F' AND functions.function_code = specimen.snop_code)
);

---

谢谢，但我得到了一个错误…请参阅OPSorry中的changed query，这是因为==，在mysql中，检查相等性是=。嗯…应该可以，但它返回了同一样本的2292条记录。snop_代码…表形态中有2292条记录…您正在连接4个表，您正在执行*这意味着它将返回所有4个表中的所有列。指定所需的列以及这些表之间的关系。

SELECT *
FROM specimen,
     topography_index,
     morphology,
     FUNCTION
WHERE SUBSTRING(specimen.topography_index, 2, 2) = topography_index.topography_index_code
AND morphology.morphology_code = specimen.snop_code
AND specimen.topography_index = '_ORGAN_'
AND (if(specimen.snop_axis = 'M', 
     morphology.morphology_code = specimen.snop_code, 
     functions.function_code = specimen.snop_code) 
OR if(specimen.snop_axis = 'F',functions.function_code = specimen.snop_code));