public boolean panduanNullCondition(XSSFRow from, Properties source) { boolean N = true;
boolean Y = true;
ArrayList<Boolean> al = new ArrayList<Boolean>();
ArrayList<Boolean> Al = new ArrayList<Boolean>();
ProcessRow p = new ProcessRow();
String process = source.getProperty("不符合哪几个条件就取空");
String[] processs = process.split("\\|");
for (int q = 0; q < processs.length; q++) {
String[] P = processs[q].split("\\+");
for (int k = 0; k < P.length; k++) {
switch (P[k]) {
case "J1":
al.add(p.panduanJ1(from, source));
break;
case "J2":
al.add(p.panduanJ2(from, source));
break;
case "J3":
al.add(p.panduanJ3(from, source));
break;
case "J4":
al.add(p.panduanJ4(from, source));
break;
case "L1":
al.add(p.panduanL1(from, source));
break;
case "L2":
al.add(p.panduanL2(from, source));
break;
case "L3":
al.add(p.panduanL3(from, source));
break;
case "L4":
al.add(p.panduanL4(from, source));
break;
case "G1":
al.add(p.panduanG1(from, source));
break;
case "G2":
al.add(p.panduanG2(from, source));
break;
case "G3":
al.add(p.panduanG3(from, source));
break;
case "F1":
al.add(p.panduanF1(from, source));
break;
case "F2":
al.add(p.panduanF2(from, source));
break;
}
for (int i = 0; i < al.size(); i++) {
N = N && al.get(i);
Al.add(N);
}
}
}
for (int u = 0; u < Al.size(); u++) {
Y = Y && !Al.get(u);
}
return Y;
}
以下是配置文件:
不符合哪几个条件就取空=G1+F1|G2+F2
我想实现的是G1和F1求&&运算,然后G2和F2求&&运算,然后二者的值再求&&运算。