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SSIS派生列处理CSV:将字符串转换为带5位小数的浮点数

Fixing Your SSIS Derived Column Expression

The problem with your current expression is twofold:

  1. It only checks for NULL values using ISNULL(), but the empty rows in your CSV are likely empty strings (not NULL), so those aren't being handled. This leads to conversion errors when trying to turn an empty string into a float.
  2. Converting directly to DT_R8 (double) doesn't enforce a fixed 5-decimal format if you need the output as a formatted string.

Here are two tailored solutions based on your desired output type:

If you want to store the value as a numeric type with exactly 5 decimal places (preserving precision for things like currency or measurements), use DT_NUMERIC with explicit precision and scale:

(DT_NUMERIC, 10, 5)(ISNULL(Col1) || TRIM(Col1) == "" ? "0.00000" : REPLACE(Col1, ",", "."))

Breakdown:

  • ISNULL(Col1) || TRIM(Col1) == "": Catches both NULL values and empty/whitespace-only strings from your CSV.
  • ? "0.00000" : REPLACE(Col1, ",", "."): Replaces invalid entries with a default 5-decimal zero, and swaps commas to dots for valid values to match standard decimal formatting.
  • (DT_NUMERIC, 10, 5): Converts the cleaned string to a numeric type with 10 total digits and 5 decimal places, ensuring consistent precision.

Option 2: Output as a Formatted String (e.g., "3.00000")

If you need the result to be a string with exactly 5 decimal places (for display or text-based storage), use STR() to format the converted number:

LTRIM(STR((DT_R8)(ISNULL(Col1) || TRIM(Col1) == "" ? "0.00000" : REPLACE(Col1, ",", ".")), 10, 5))

Breakdown:

  • The inner expression converts the cleaned value to a double (DT_R8).
  • STR(..., 10, 5) formats the number into a string with 10 total characters and 5 decimal places.
  • LTRIM() removes any leading spaces added automatically by the STR() function for alignment.

Testing with Your Sample Data:

For your CSV example:

  • "3,00000" → becomes 3.00000 (numeric) or "3.00000" (string)
  • Empty rows → become 0.00000 (numeric) or "0.00000" (string)
  • "6.56565" → remains 6.56565 (numeric) or "6.56565" (string)

内容的提问来源于stack exchange,提问作者Aercheon

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最近更新时间:2026.05.14 08:53:07