编写修理成功率查询oracle-sql-查询语句

由于缺乏具体的表结构和数据，我无法提供精确的SQL查询语句。但是，以下是一些可能有用的查询示例，可以根据您的具体情况进行修改和调整：

1. 查询所有维修记录中成功修复的记录数和总记录数，计算成功率：

SELECT 
  COUNT(CASE WHEN status = 'success' THEN 1 END) as success_count,
  COUNT(*) as total_count,
  COUNT(CASE WHEN status = 'success' THEN 1 END) / COUNT(*) as success_rate
FROM repair_records;

2. 按月份统计维修成功率：

SELECT 
  TO_CHAR(repair_date, 'YYYY-MM') as month,
  COUNT(CASE WHEN status = 'success' THEN 1 END) / COUNT(*) as success_rate
FROM repair_records
GROUP BY TO_CHAR(repair_date, 'YYYY-MM');

3. 统计每个维修人员的维修成功率：

SELECT 
  technician_name,
  COUNT(CASE WHEN status = 'success' THEN 1 END) / COUNT(*) as success_rate
FROM repair_records
GROUP BY technician_name;

4. 统计每个设备类型的维修成功率：

SELECT 
  device_type,
  COUNT(CASE WHEN status = 'success' THEN 1 END) / COUNT(*) as success_rate
FROM repair_records
GROUP BY device_type;

这些查询只是示例，您需要根据您的具体情况进行修改。特别是要注意表结构和字段名称的正确性，以及数据类型的匹配。