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本帖最后由 bess 于 2012-12-10 14:00 编辑
Q:查询表单数据异常,怎么解决?
Field:rs_month_details_majorname-值脚本出错 <br>JavaScriptRunning:[Form.rs_month_details.Field(11e0-7b79-ed98d3c4-bfaa-d912e9a69ded).rs_month_details_majorname.OptionsScript]: Wrapped com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'table_0.DOMAINID' in 'where clause' (baselib#1238)
源代码如下:
值脚本里面没有写东西,我在选项脚本里加入了下面的
var sql="select m.id as item_id,m.item_rs_month_majorprojects_name as name from test_obpm.tlk_rs_month_majorprojects m";
var opts = $TOOLS.createOptions();
var query = queryBySQL(sql);
var value = "";
if(query!=null)
{
for(var iter = query.iterator();iter.hasNext();){
var doc= iter.next();
value= doc.getItemValueAsString("name");
opts.add(value,value);
}
}
opts;
A:rs_month_details_majorname这个控件的值脚本错了,查找表单时,必须有domainid
"select id, name item_name, '"+getWebUser().getDomainid()+"' domainid from obpm.t_user"
然后对应name列使用iscript
getItemValue("name")
1. var sql="select id as id,name as rs_month_majorprojects_name, '"+getWebUser().getDomainid()+"' domainid from tlk_rs_month_majorprojects ";
2. var opts = $TOOLS.createOptions();
3. var query = queryBySQL(sql);
4. var value = "";
5. if(query!=null)
6. {
7. for(var iter = query.iterator();iter.hasNext();){
8. var doc= iter.next();
9. value= doc.getItemValueAsString("rs_month_majorprojects_name");
10. opts.add(id,value);
11. }
12. }
13. opts;
摘自jariy提问,原帖子链接:http://www.teemlink.com/bbs/view ... 26amp%3Btypeid%3D36 |
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