Item SUPERIOR's value can't get as string!

var dptid = doc.getItemValueAsString("所属部门");
var sql1="select * from obpm.t_department where ID='"+dptid+"'";
var doc1 = queryBySQL (sql1);

if (doc1 != null && doc1.size() >0 ) {        
        for ( var iter = doc1.iterator();iter.hasNext();) {
        var data1 = iter.next();
                var quyu = data1.getItemValueAsString("SUPERIOR");     
    }
}
上面红色字体错误，怎么解决

您好,您查询的是系统字段.并且您是通过queryBySQL方法来查询的,那么您需要在查询的时候给字段一个"item_"开头的别名,
var sql1="select SUPERIOR as item_SUPERIOR from obpm.t_department where ID='"+dptid+"'";

才能通过getItemValueAsString去获取.
或者您换成数据源的方式去查询,然后通过get方法去取值.
有问题,我们再沟通.

用你上面的语句之后，tomcat提示如下错误，不知道什么原因：

INFO AbstractDocStaticTblDAO:1268 - MY SQL:  SELECT * FROM (select SUPERIOR as item_SUPERIOR from obpm.t_department where ID='
11e1-73fc-6457bd4f-bc51-cd56ce611492') table_0 WHERE table_0.DOMAINID ='11e1-73fb-0eab2cb2-bc51-cd56ce611492'
java.lang.Exception: JavaScriptRunning:[ManualNode(11e1-7263-27558470-894c-a52f2e6687da1338341908018).岗位审核.auditorListScript]: Wrapped c
om.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'table_0.DOMAINID' in 'where clause' (baselib#1241)

您好,不好意思,您在查询语句中添加domainid,将domainid也查询即可
var sql1="select SUPERIOR as item_SUPERIOR,domainid from obpm.t_department where ID='"+dptid+"'";

这个已经解决，谢谢

感谢楼主告知结果!