submitJob提交的函数中用parseExpr不能解析函数

运行以下代码，直接run_factor()可以，但submitJob报错：Syntax Error: [line #1] Cannot recognize the token factor0000

def factor0000(){
	return table(1..100 as id);
}
def factor0001(){
	return table(101..200 as id);
}
def run_factor(){
	factor="";
	for(i in 0..1){
		sql0 = "select * from factor0000()"
		sql1 = strReplace(sql0 , "factor0000" , "factor" + lpad(string(i) , 4 , "0"))
		print(sql1);
		res= parseExpr(sql1).eval();
		factor+=string(res);
	}
	return factor;
}
submitJob("test000" , "test000" , run_factor)
getRecentJobs()

请问怎么将运行的函数名称使用字符串传进去，在里面动态调用呢？

0 条评论

默认排序时间排序

1 个回答

wale 2023-11-17 10:08

submitJob+parseExpr问题，可以参考以下例子：

def factor0000(){
	return table(1..100 as id);
}
def factor0001(){
	return table(101..200 as id);
}

funlist=dict(STRING,ANY)
f=exec name from defs() where name like "factor%"
for (i in f){
        funlist[i]=funcByName(i)
        }
 
def run_factor(dd){
        factor=0
        for(i in 0..1){
//                sql0="select * from factor0000()"
//                sql1=strReplace(sql0,"factor0000","factor"+lpad(string(i),4,"0"))
//                factor=parseExpr(sql1,dd).eval()
                fname="factor"+lpad(string(i),4,"0")
                factor=factor+makeCall(dd[fname]).eval()
        }
        return factor 
}
jodid=submitJob("test000" , "test000" , run_factor{funlist})
getJobReturn(jodid,1)

submitJob提交的函数中用parseExpr不能解析函数

1 个回答

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