delete redis keys NOT matching a pattern using Java
I want to delete redis keys that DO NOT match a particular pattern.
Redis data looks like this -
prefix9_key1
prefix9_key2
prefix11_key1
prefix11_key2
prefix2_key1
prefix2_key2
These prefix values are randomly generated numbers (example 1123, 3422, 9879).
I know what my current prefix is (let's say it is prefix11). But there is no way to know what were the earlier prefixes being used. I want to delete all keys that DO NOT use current prefix value (prefix2* and prefix9*).
I read many posts mentioning deleting keys matching a pattern. I want to delete keys that do not match a particular pattern.
redis
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
add a comment |
I want to delete redis keys that DO NOT match a particular pattern.
Redis data looks like this -
prefix9_key1
prefix9_key2
prefix11_key1
prefix11_key2
prefix2_key1
prefix2_key2
These prefix values are randomly generated numbers (example 1123, 3422, 9879).
I know what my current prefix is (let's say it is prefix11). But there is no way to know what were the earlier prefixes being used. I want to delete all keys that DO NOT use current prefix value (prefix2* and prefix9*).
I read many posts mentioning deleting keys matching a pattern. I want to delete keys that do not match a particular pattern.
redis
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
add a comment |
I want to delete redis keys that DO NOT match a particular pattern.
Redis data looks like this -
prefix9_key1
prefix9_key2
prefix11_key1
prefix11_key2
prefix2_key1
prefix2_key2
These prefix values are randomly generated numbers (example 1123, 3422, 9879).
I know what my current prefix is (let's say it is prefix11). But there is no way to know what were the earlier prefixes being used. I want to delete all keys that DO NOT use current prefix value (prefix2* and prefix9*).
I read many posts mentioning deleting keys matching a pattern. I want to delete keys that do not match a particular pattern.
redis
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
I want to delete redis keys that DO NOT match a particular pattern.
Redis data looks like this -
prefix9_key1
prefix9_key2
prefix11_key1
prefix11_key2
prefix2_key1
prefix2_key2
These prefix values are randomly generated numbers (example 1123, 3422, 9879).
I know what my current prefix is (let's say it is prefix11). But there is no way to know what were the earlier prefixes being used. I want to delete all keys that DO NOT use current prefix value (prefix2* and prefix9*).
I read many posts mentioning deleting keys matching a pattern. I want to delete keys that do not match a particular pattern.
redis
redis
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
asked Jan 2 at 6:45
Vibhav Singh RohillaVibhav Singh Rohilla
334213
334213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
From the docs:
h[^e]llo matches hallo, hbllo, ... but not hello
But, the specific example you gave is a bit tricky. Here is how you can get the desired filtering:
=>redis-cli keys *
1) "pre1_234"
2) "pre3_234"
3) "pre11_asv"
4) "pre2_234"
=>redis-cli keys 'pre[^1]*' | redis-cli keys 'pre?[^1]*'
1) "pre1_234"
2) "pre3_234"
3) "pre2_234"
To me redis-cli keys 'pre[^1][^1]'* should have given the desired output. But it's not working. The pattern matcher short circuits at the first [^1] and hence pre1_234 is not part of the output. The vagaries of regex :)
You can use lua script for atomic deletion in one call to Redis. See this.
answered Jan 2 at 16:25
rainhackerrainhacker
1638
add a comment |
You need a lua UDF for this (below example uses jedis as redis client),
String DELETE_SCRIPT = "local keys = redis.call('keys', '%s')" +
" for i,k in ipairs(keys) do" +
" local res = redis.call('del', k)" +
" end";
Your pattern will be a string input to the method with the prefix like prefix[^11]*
public void deleteOthers (String pattern) {
String luaScript = String.format(DELETE_SCRIPT, pattern);
jedis.eval(luaScript);
}
Call to this method would be something like,
deleteOthers("prefix[^11]*");
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
From the docs:
h[^e]llo matches hallo, hbllo, ... but not hello
But, the specific example you gave is a bit tricky. Here is how you can get the desired filtering:
=>redis-cli keys *
1) "pre1_234"
2) "pre3_234"
3) "pre11_asv"
4) "pre2_234"
=>redis-cli keys 'pre[^1]*' | redis-cli keys 'pre?[^1]*'
1) "pre1_234"
2) "pre3_234"
3) "pre2_234"
To me redis-cli keys 'pre[^1][^1]'* should have given the desired output. But it's not working. The pattern matcher short circuits at the first [^1] and hence pre1_234 is not part of the output. The vagaries of regex :)
You can use lua script for atomic deletion in one call to Redis. See this.
answered Jan 2 at 16:25
rainhackerrainhacker
1638
add a comment |
From the docs:
h[^e]llo matches hallo, hbllo, ... but not hello
But, the specific example you gave is a bit tricky. Here is how you can get the desired filtering:
=>redis-cli keys *
1) "pre1_234"
2) "pre3_234"
3) "pre11_asv"
4) "pre2_234"
=>redis-cli keys 'pre[^1]*' | redis-cli keys 'pre?[^1]*'
1) "pre1_234"
2) "pre3_234"
3) "pre2_234"
To me redis-cli keys 'pre[^1][^1]'* should have given the desired output. But it's not working. The pattern matcher short circuits at the first [^1] and hence pre1_234 is not part of the output. The vagaries of regex :)
You can use lua script for atomic deletion in one call to Redis. See this.
answered Jan 2 at 16:25
rainhackerrainhacker
1638
add a comment |
From the docs:
h[^e]llo matches hallo, hbllo, ... but not hello
But, the specific example you gave is a bit tricky. Here is how you can get the desired filtering:
=>redis-cli keys *
1) "pre1_234"
2) "pre3_234"
3) "pre11_asv"
4) "pre2_234"
=>redis-cli keys 'pre[^1]*' | redis-cli keys 'pre?[^1]*'
1) "pre1_234"
2) "pre3_234"
3) "pre2_234"
To me redis-cli keys 'pre[^1][^1]'* should have given the desired output. But it's not working. The pattern matcher short circuits at the first [^1] and hence pre1_234 is not part of the output. The vagaries of regex :)
You can use lua script for atomic deletion in one call to Redis. See this.
answered Jan 2 at 16:25
rainhackerrainhacker
1638
From the docs:
h[^e]llo matches hallo, hbllo, ... but not hello
But, the specific example you gave is a bit tricky. Here is how you can get the desired filtering:
=>redis-cli keys *
1) "pre1_234"
2) "pre3_234"
3) "pre11_asv"
4) "pre2_234"
=>redis-cli keys 'pre[^1]*' | redis-cli keys 'pre?[^1]*'
1) "pre1_234"
2) "pre3_234"
3) "pre2_234"
To me redis-cli keys 'pre[^1][^1]'* should have given the desired output. But it's not working. The pattern matcher short circuits at the first [^1] and hence pre1_234 is not part of the output. The vagaries of regex :)
You can use lua script for atomic deletion in one call to Redis. See this.
answered Jan 2 at 16:25
rainhackerrainhacker
1638
edited Jan 2 at 16:32
answered Jan 2 at 16:25
rainhackerrainhacker
1638
answered Jan 2 at 16:25
rainhackerrainhacker
1638
answered Jan 2 at 16:25
rainhackerrainhacker
1638
1638
add a comment |
add a comment |
You need a lua UDF for this (below example uses jedis as redis client),
String DELETE_SCRIPT = "local keys = redis.call('keys', '%s')" +
" for i,k in ipairs(keys) do" +
" local res = redis.call('del', k)" +
" end";
Your pattern will be a string input to the method with the prefix like prefix[^11]*
public void deleteOthers (String pattern) {
String luaScript = String.format(DELETE_SCRIPT, pattern);
jedis.eval(luaScript);
}
Call to this method would be something like,
deleteOthers("prefix[^11]*");
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
add a comment |
You need a lua UDF for this (below example uses jedis as redis client),
String DELETE_SCRIPT = "local keys = redis.call('keys', '%s')" +
" for i,k in ipairs(keys) do" +
" local res = redis.call('del', k)" +
" end";
Your pattern will be a string input to the method with the prefix like prefix[^11]*
public void deleteOthers (String pattern) {
String luaScript = String.format(DELETE_SCRIPT, pattern);
jedis.eval(luaScript);
}
Call to this method would be something like,
deleteOthers("prefix[^11]*");
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
add a comment |
You need a lua UDF for this (below example uses jedis as redis client),
String DELETE_SCRIPT = "local keys = redis.call('keys', '%s')" +
" for i,k in ipairs(keys) do" +
" local res = redis.call('del', k)" +
" end";
Your pattern will be a string input to the method with the prefix like prefix[^11]*
public void deleteOthers (String pattern) {
String luaScript = String.format(DELETE_SCRIPT, pattern);
jedis.eval(luaScript);
}
Call to this method would be something like,
deleteOthers("prefix[^11]*");
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
You need a lua UDF for this (below example uses jedis as redis client),
String DELETE_SCRIPT = "local keys = redis.call('keys', '%s')" +
" for i,k in ipairs(keys) do" +
" local res = redis.call('del', k)" +
" end";
Your pattern will be a string input to the method with the prefix like prefix[^11]*
public void deleteOthers (String pattern) {
String luaScript = String.format(DELETE_SCRIPT, pattern);
jedis.eval(luaScript);
}
Call to this method would be something like,
deleteOthers("prefix[^11]*");
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
answered Jan 2 at 17:26
swayamrainaswayamraina
668711
668711
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
add a comment |
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
prefix[^11]* will not work. see my answer above. Both prefix[^11]* and prefix[^1][^1]* will ignore keys like 'prefix1_key4'. Also, Redis follows glob-style pattern. In which, "[abc] matches one character given in the bracket" en.wikipedia.org/wiki/Glob_(programming)
– rainhacker
Jan 2 at 20:35
add a comment |
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