Mybatis解决In查询条件过长的问题
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2024-03-06 21:29:50
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方法1:分次查询,将参数且分割成多个短的查询后合并
代码:
int splitNum =(int) Math.ceil( (float) ids.length/999); //切片数量
List<String> itemIdList = new ArrayList<>(Arrays.asList(ids));
List<List<String>> splitList = averageAssign(itemIdList, splitNum);
for (List<String> list : splitList) {
param.put("itemIds",list);
List<Map<Object, Object>> itemStatisticsList = iProcessExtMapper.getItemStatisticsList(param);
result.addAll(itemStatisticsList);
}
将list分成N等分方法方法:
public static <T> List<List<T>> averageAssign(List<T> source,int n){
List<List<T>> result=new ArrayList<List<T>>();
int remaider=source.size()%n; //(先计算出余数)
int number=source.size()/n; //然后是商
int offset=0;//偏移量
for(int i=0;i<n;i++){
List<T> value=null;
if(remaider>0){
value=source.subList(i*number+offset, (i+1)*number+offset+1);
remaider--;
offset++;
}else{
value=source.subList(i*number+offset, (i+1)*number+offset);
}
result.add(value);
}
return result;
}
方法2:xml文件中编写sql
i.id in
<foreach collection="itemIds" index="index" item="item" open="(" close=")">
<if test="index != 0">
<choose>
<when test="index % 1000 == 999"> ) OR ID IN( </when>
<otherwise>,</otherwise>
</choose>
</if>
#{item}
</foreach>
sql逻辑:
ID IN(ids[0],ids[1]+...+ids[998])OR ID IN (ids[999],ids[1000],...ids[max])