Question

我的列有多个值，如A0045，A00065。我想转换它9945,9965。需要删除所有0和字符值并在该值之前添加99 ..请帮助..

Answer 1

这可以通过多种方式完成。这是一种方式（可能不是最好的）。由于我没有数据库，因此我创建了临时表。

def temp-table tt 
    field val as char.

create tt.
tt.val = "A0045".

create tt.
tt.val = "A00065".

for each tt:
    run filter_zero(input-output val).
    val = replace(val,"A","99").
    DISP val.
end.

procedure filter_zero:
    define input-output parameter val  as character.

    // remvoe all zeroes 
    repeat while val matches("*0*"): 
        val = replace(val,"0","").
    end.

end procedure.

Answer 2

下面的代码也删除了大写字母（A-Z）和小写字母（a-z）以及“0”：

DEF TEMP-TABLE test
    FIELD str1 AS CHAR.

DEF VAR newStr AS CHAR NO-UNDO.
DEF VAR i AS INT NO-UNDO.

CREATE test.
ASSIGN test.str1 = "A0045".

CREATE test.
ASSIGN test.str1 = "A00065".

FOR EACH test:
    DO i = 65 TO 90: /* A - Z */
        IF SUBSTR(test.str1, 1, 1) EQ CHR(i) THEN
        DO:
            test.str1 = REPLACE(test.str1, CHR(i), "").       
        END.
    END.

    DO i = 97 TO 122: /* a - z */
        IF SUBSTR(test.str1, 1, 1) EQ CHR(i) THEN
        DO:
            test.str1 = REPLACE(test.str1, CHR(i), "").       
        END.
    END.

    /* Removes all 0 and add 99 before the value */
    test.str1 = REPLACE(test.str1, "0", "").
    ASSIGN test.str1 = "99" + test.str1.
    DISP test.
END.

Answer 3

将变量单词定义为字符no-undo。

将变量i定义为整数no-undo。

指定word =＆＃34; A00065＆＃34;。

/ *******删除所有零******** /

word = replace（word，substring（word，index（word，＆＃34; 0＆＃34;），（r-index（word，＆＃34; 0＆＃34;） - 1）），＆＃ 34;＆＃34;。）

我= 65到90：如果substring（word，1,1）= chr（i）那么做：

  word = replace(word,substring(word,1,1),"99").
  leave.

端。

显示字词。

Answer 4

public class UriAttribute: ValidationAttribute
{
    protected override ValidationResult IsValid(object value, ValidationContext validationContext)
    {
        Uri uri;
        bool valid = Uri.TryCreate(Convert.ToString(value), UriKind.Absolute, out uri);

        if (!valid)
        {
            return new ValidationResult(ErrorMessage);
        }
        return ValidationResult.Success;
    }
}

Answer 5

如果这对您有用，请告诉我，未安装Progress因此无法编译和测试。

/ * A0045 - ＆gt; 9945 A00065 - ＆gt; 9965 * /

DEFINE VARIABLE v_RawData AS CHARACTER NO-UNDO.
DEFINE VARIABLE v_InpData AS CHARACTER NO-UNDO.
DEFINE VARIABLE i         AS INTEGER NO-UNDO.
DEFINE VARIABLE j         AS INTEGER NO-UNDO.

ASSIGN v_RawData = "A0045,A00065".

DO i =1 TO NUM-ENTRIES(v_RawData):
    ASSIGN v_InpData = ENTRY(i,v_RawData).

    DO j = 1 TO LENGTH(v_InpData):
        IF ASC(SUBSTRING(v_InpData,j,1)) > 48 AND 
           ASC(SUBSTRING(v_InpData,j,1)) < 58 THEN
            DO:
                LEAVE.
            END.
    END.
    MESSAGE REPLACE(v_InpData,SUBSTRING(v_InpData,1,j - 1),"99").
END.

如何将A00073值转换为9973正在进行4gl

5 个答案: