Vector abc = new Vector();
abc.add("begindate=2012-05-16, makerdate=2012-05-16, expiredate=2012-06-16, saleschannelid=12, programcode=SS100001, maker=10000289, description=特价支架测试, datelastupdated=2012-05-16, ssbiid=100000533");
abc.add("begindate=2012-05-16, makerdate=2012-05-16, expiredate=2012-06-16, saleschannelid=12, programcode=SS100002, maker=10000289, description=支架特价测试, datelastupdated=2012-05-16, ssbiid=100000536");
for(int i=0;i<abc.size();i++){
HashMap hm=(HashMap)abc.get(i); //这个地方就出错了呢？
}
}
------解决方案--------------------
abc.get(i)是一个字符串，怎么转能HashMap?
- Java code
HashMap hm = new HashMap();for(int i=0;i<abc.size();i++){ hm.put(i, abc.get(i));}
------解决方案--------------------
字符串就是字符串，强制转化为其他类型都会报错
取出来后按照逗号分割成每个键值对，然后按照等号分割成key和value，然后将这些存储到Map里面
- Java code
Vector<String> abc = new Vector<String>(); abc.add("begindate=2012-05-16, makerdate=2012-05-16, expiredate=2012-06-16, saleschannelid=12, programcode=SS100001, maker=10000289, description=特价支架测试, datelastupdated=2012-05-16, ssbiid=100000533"); abc.add("begindate=2012-05-16, makerdate=2012-05-16, expiredate=2012-06-16, saleschannelid=12, programcode=SS100002, maker=10000289, description=支架特价测试, datelastupdated=2012-05-16, ssbiid=100000536"); Vector<Map<String,String>> efg = new Vector<Map<String,String>>(); for(int i=0;i<abc.size();i++) { String[] kvs = abc.get(i).split(","); HashMap<String, String> hm=new HashMap<String, String>(); for(String kv:kvs) { kv = kv.trim(); String key = kv.split("=")[0]; String value = kv.split("=")[1]; hm.put(key, value); } efg.add(hm); }