MySQL报错解决--Parameter index out of range (1 ＞ number of parameters, which is 0)

分析:需要0个参数.实际1个,一般是写like语句时的错误
解决办法:

//错误写法为: "and rname like '%?%' ";
// list1.add(rname);

//修改为:
 sql += "and rname like ? ";
 //百分号作为值传递过去就解决啦!
 list1.add("%"+rname+"%");