如何使用 Progress 4GL 列出字段中的所有行

Question

Progress 数据库中的地址字段显然是多行。所以如果我做一些简单的事情，比如：

for each customer where customer-number = "123" no-lock.

display customer-number
        customer-name
        customer-address.

我得到以下结果：

123  John Smith  123 Easy St.
                 c/o Jane Smith
                 Apartment C12

我想要以下结果：

 123  John Smith  123 Easy St.  c/o Jane Smith  Apartment C12

我试过 Entry - 但如果元素不存在，则会出现错误。我正在尝试导出到 CSV 并希望每一行成为另一列，如果一列不存在则它是空白的。

任何帮助将不胜感激！

Answer 1

如果分隔符是一个换行符，那么这样的事情应该有效：

define variable n as integer no-undo.
define variable i as integer no-undo.

define variable address as character no-undo.

output to value( "customer.csv").
for each customer no-lock:
  address = "".
  n = num-entries( customer-address, "~n" ).
  do i = 1 to n:
    address = address + entry( i, customer-address, "~n" ) + " ".
  end.
  export delimiter "," customer-number customer-name trim( address ).  
end.
output close.

或者，如果您只需要更改客户地址字段中的分隔符：

output to value( "customer.csv").
for each customer no-lock:
  export delimiter "," customer-number customer-name replace( customer-address, "~n", "," ).  
end.
output close.

（而且，实际上，我的原始代码实际上只是将 ~n 替换为 space，因此您也可以这样做...）

如何使用 Progress 4GL 列出字段中的所有行

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