我有一个包含列的数据库:id,pdate,pvalue1,pvalue2。首先,我用光标进行查询:
Cursor c = ourDatabase.query(DATABASE_TABLE, new String[] { "_id","pdate","pvalue1","pvalue2"},
"pdate >= ? AND pdate <= ?", new String[] { datefrom, dateto }, null, null, null);
这给了我一些行,例如,如果pdate = 20120318,那么pvalue1 = 58,pvalue2 = 29。这些是字符串,所以我可以给pvalue2一个值“XX”。我想在给定的datefrom和dateto之间求和pvalue1,并用pdate将它们分组,其中pvalue2 = XX。我的问题是我不能把这个条件放到查询中(因为它的工作,比如“pvalue2 = XX”..),因为我也需要其他数据。
if (c.moveToFirst())
{
do{
if (c.getString(3).equals("XX")){
Log.i("XX", c.getString(1) + " " + c.getString(2)) + " " + c.getString(3));
}
else {
Log.i("NotXX", c.getString(1) + " " + c.getString(2)) + " " + c.getString(3));
}
while (c.moveToNext());
}
}
到目前为止没关系,所以我可以用pvalue2 = XX和NotXX来记录数据,得到这样的结果: (pdate,pvalue1,pvalue2)20120317,48,29; ------; 20120317,21,54; ------- 20120317,11,XX; ----- 20120318,79,71; ------- 20120318,21,XX
我想做什么?
首先:用pdate对sums(pvalue1)进行分组,如果pvalue2是XX或者不是XX则表明它,所以像这样的somethnig: 20120317,NotXX,69(自48 + 21 = 69)-------- 20120317,XX,11 -------- 20120318,NotXX,79 -------- 20120318,XX ,21
在此之后,我想每天从NotXX总和中减去XX总和。我想得到: 20120317,58(自69-11)------- 20120318,58(自79-21以来)
我该怎么做? 非常感谢你提前!
答案 0 :(得分:1)
我的问题是我无法将此条件纳入查询
你可能错了。您可以添加类似的内容(语法可能包含错误)
“选择总和(从DATABASE_TABLE中选择pdate,其中pdata&gt; x和pdate&lt; y)作为总和”
到投影参数,您将该结果作为名为sum的列获得。唯一的问题是投影中没有?
的支持(至少我没有尝试过,但我想它不会起作用)
如果那不是你想要的,那么很可能会采用不同的方式。 SQLite非常强大。
编辑:
这会是你想要的吗?它不是在SQL中完成的,但它会打印出你想要的每一天的总和。
Cursor c = ourDatabase.query(DATABASE_TABLE, new String[] { "_id","pdate","pvalue1","pvalue2"},
"pdate >= ? AND pdate <= ?", new String[] { datefrom, dateto }, null, null, "pdate");
boolean first = true;
if (c != null) {
String currentDate = null;
int sum = 0;
while (c.moveToNext()) {
String date = c.getString(1);
int value1 = c.getInt(2);
String value2 = c.getString(3);
if (!date.equals(currentDate)) {
if (!first) {
Log.d("TAG", "The result for " + currentDate + " is: " + sum);
} else {
Log.d("TAG", "Date has changed, but we don't have data yet.");
}
first = false;
currentDate = date;
sum = 0;
}
if ("XX".equals(value2)) {
Log.d("TAG", "new line: " + date + ", " + value1 + ", " + value2 + " -");
sum -= value1;
} else {
Log.d("TAG", "new line: " + date + ", " + value1 + ", " + value2 + " +");
sum += value1;
}
}
if (!first) {
Log.d("TAG", "The last result: " + currentDate + " is: " + sum);
}
c.close();
}
Edit2:当您希望数据库完成此操作时,这可能会有效。
Cursor c = ourDatabase.rawQuery(
"SELECT pdate, sum(sum2) AS sum1 FROM " +
"(" +
" SELECT pdate, pvalue1, pvalue2, -sum(pvalue1) AS sum2 " +
" FROM " + DATABASE_TABLE +
" WHERE pvalue2='XX' GROUP BY pdate" +
" UNION " +
" SELECT pdate, pvalue1, pvalue2, sum(pvalue1) AS sum2 " +
" FROM " + DATABASE_TABLE +
" WHERE pvalue2!='XX' GROUP BY pdate" +
") " +
" WHERE pdate>=? AND pdate<=? " +
" GROUP BY pdate",
new String[] { datefrom, dateto });
if (c != null) {
while (c.moveToNext()) {
String date = c.getString(0);
int value1 = c.getInt(1);
Log.d("TAG", "The result for " + date + " is: " + value1);
}
c.close();
}