Wildcard is not used now by default. User has to set it explicitly if needed #2277
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ssuvorov-fls
changed the title
| private String encodeSearchString(String searchString) { | ||
| // encode search string without encoding wildcards | ||
| if (searchString == null) { | ||
| return null; | ||
| } | ||
| char wildcard = '*'; | ||
| List<Integer> wildcardIndexes = IntStream.range(0, searchString.length()) | ||
| .filter(i -> searchString.charAt(i) == wildcard).boxed() | ||
| .collect(Collectors.toList()); | ||
| StringBuffer buff = new StringBuffer(); | ||
| if (wildcardIndexes.isEmpty()) { | ||
| buff.append(LdapEncoder.filterEncode(searchString)); | ||
| } else { | ||
| int previousWildcardIndex = -1; | ||
| for (int i = 0; i <= wildcardIndexes.size() - 1; i++) { | ||
| int wildcardIndex = wildcardIndexes.get(i); | ||
| if (wildcardIndex > previousWildcardIndex + 1 ) { | ||
| String substring = searchString.substring(previousWildcardIndex + 1, wildcardIndex); | ||
| buff.append(LdapEncoder.filterEncode(substring)); | ||
| } | ||
| buff.append(wildcard); | ||
| // last wildcard in the string but not trailing | ||
| if (i == wildcardIndexes.size() - 1 && wildcardIndex < searchString.length() - 1) { | ||
| String substring = searchString.substring(wildcardIndex + 1, searchString.length()); | ||
| buff.append(LdapEncoder.filterEncode(substring)); | ||
| } | ||
| previousWildcardIndex = wildcardIndex; | ||
| } | ||
| } | ||
|
|
||
| return buff.toString(); | ||
| } |
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Couldn't this all be replaced with:
List<String> tokens = StringUtils.split(searchString, wildcard);
List<String> encodedTokens = tokens.replaceAll(LdapEncoder::filterEncode);
return StringUtils.join(encodedTokens, wildcard);
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Hi
I've checked it and it doesn't work as expected
For * it returns empty string
For user* it returns user
For user*cn* it returns user*cn
My code will return correct string without any limitations of wildcard position
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Understood, I didn't think through the corner cases, but your implementation is pretty complicated and I think I can modify the suggested with a few tweaks to hit the corner cases:
if (searchString == null || searchString.isEmpty() || wildCard.equals(searchString)) return searchString; // nothing to encode
List<String> tokens = StringUtils.split(searchString, wildcard);
List<String> encodedTokens = tokens.replaceAll(LdapEncoder::filterEncode);
String encodedSearchString = StringUtils.join(encodedTokens, wildcard) + searchString.endsWith(wildcard) ? wildcard : "";
return encodedSearchString;
I should note, my implementation will handle cases where wild card length is > 1. Not sure if that is the case with LDAP filtering, but it's good to be flexible.
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I agree
Your code is simpler
I made additional check for case when wildcard is at the first position
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Thanks for fixing that, it's hard doing this without having a compiler in front of me :) Yes, you captured the case where it might be like *user*, and the splitting of that just becomes user (which is the term we need to encode) but I lost the leading-wildcard that we need to check at the end when we re-assemble. Thank you for making this change.
fixes #1790