我正在COBOL中编写一个程序,它从表中读取数据并将其转储到CSV文件中。我的工作定义列定义为(例如)FACILITYCODE。以下是专栏的详细信息:
工作存储:WS-FACILITYCODE PIC X(04) VALUE "XYZ".
计划定义:MOVE WS-FACILITYCODE TO ABI-FACILITYCODE.
编写程序时,CSV标题显示为“Facilitycode”。我想将其更改为“FACILITY_CODE”
用于创建CSV PERFORM 800-WRITECSV-XYZINV.
SQL世界中可比的例子是'AS'。
请提出任何建议。
谢谢!
答案 0 :(得分:2)
正如大家所说,我们需要知道PERFORM 800-WRITECSV-XYZINV.
做了什么......
但我在类似的解决方案中使用了写入CSV文件
*...
*acces to you table and it fills the variables WS-A WS-B and WS-C
*build heather
STRING "A" ","
"B" ","
"C" DELIMITED BY SIZE
INTO WS-LINE
END-STRING
* write header
WRITE FILE FROM WS-LINE
...
然后对于表格中的每个文件,您可以:
INITIALIZE WS-LINE
*build line whit the values
STRING WS-A ","
WS-B ","
WS-C DELIMITED BY "SIZE" <-- it will print all spaces that each variable have
INTO WS-LINE
END-STRING
* write line
WRITE FILE FROM WS-LINE
...