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From viirya <...@git.apache.org>
Subject [GitHub] spark pull request #15398: [SPARK-17647][SQL] Fix backslash escaping in 'LIK...
Date Wed, 12 Oct 2016 03:35:53 GMT
Github user viirya commented on a diff in the pull request:

    https://github.com/apache/spark/pull/15398#discussion_r82931395
  
    --- Diff: sql/catalyst/src/main/scala/org/apache/spark/sql/catalyst/util/StringUtils.scala
---
    @@ -25,26 +25,25 @@ object StringUtils {
     
       // replace the _ with .{1} exactly match 1 time of any character
       // replace the % with .*, match 0 or more times with any character
    -  def escapeLikeRegex(v: String): String = {
    -    if (!v.isEmpty) {
    -      "(?s)" + (' ' +: v.init).zip(v).flatMap {
    -        case (prev, '\\') => ""
    -        case ('\\', c) =>
    -          c match {
    -            case '_' => "_"
    -            case '%' => "%"
    -            case _ => Pattern.quote("\\" + c)
    -          }
    -        case (prev, c) =>
    -          c match {
    -            case '_' => "."
    -            case '%' => ".*"
    -            case _ => Pattern.quote(Character.toString(c))
    -          }
    -      }.mkString
    -    } else {
    -      v
    +  def escapeLikeRegex(str: String): String = {
    +    val builder = new StringBuilder()
    +    var escaping = false
    +    for (next <- str) {
    +      if (escaping) {
    +        builder ++= Pattern.quote(Character.toString(next))
    --- End diff --
    
    `\Q\\E\Qa\E` is correct. But doesn't it become `\Qa\E` in this change?
    
    For `\\a`, the prefixing `\\` will go the next branch and enable `escaping`. Then the
next char `a` will be quoted here. So it becomes `\Qa\E`. BTW, before this change, it will
be `\Q\a\E`. 



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