Regex Too Complicated - replaceAll - large strings preprocessed before deserialize
Given
- A JSON string with Apex reserved words (like
"currency":"USD"and otherwise undeserializable-into-Date type (like"someDate":"0001-01-01T00:00:00Z")
An apex method that uses
String s = s.replaceAll('"currency":','"currencyX":')
.replaceAll('"0001\-01\-01T00:00:00Z"','null');
before deserializing into a custom Apex type with property String currencyX and Date someDate
When
The incoming Apex string is very large (> 1MB)
Then
Uncatchable error: Regex Too Complicated
What to do?
apex
asked 11 hours ago
cropredycropredy
36k441123
add a comment |
Given
- A JSON string with Apex reserved words (like
"currency":"USD"and otherwise undeserializable-into-Date type (like"someDate":"0001-01-01T00:00:00Z")
An apex method that uses
String s = s.replaceAll('"currency":','"currencyX":')
.replaceAll('"0001\-01\-01T00:00:00Z"','null');
before deserializing into a custom Apex type with property String currencyX and Date someDate
When
The incoming Apex string is very large (> 1MB)
Then
Uncatchable error: Regex Too Complicated
What to do?
apex
asked 11 hours ago
cropredycropredy
36k441123
add a comment |
Given
- A JSON string with Apex reserved words (like
"currency":"USD"and otherwise undeserializable-into-Date type (like"someDate":"0001-01-01T00:00:00Z")
An apex method that uses
String s = s.replaceAll('"currency":','"currencyX":')
.replaceAll('"0001\-01\-01T00:00:00Z"','null');
before deserializing into a custom Apex type with property String currencyX and Date someDate
When
The incoming Apex string is very large (> 1MB)
Then
Uncatchable error: Regex Too Complicated
What to do?
apex
asked 11 hours ago
cropredycropredy
36k441123
Given
- A JSON string with Apex reserved words (like
"currency":"USD"and otherwise undeserializable-into-Date type (like"someDate":"0001-01-01T00:00:00Z")
An apex method that uses
String s = s.replaceAll('"currency":','"currencyX":')
.replaceAll('"0001\-01\-01T00:00:00Z"','null');
before deserializing into a custom Apex type with property String currencyX and Date someDate
When
The incoming Apex string is very large (> 1MB)
Then
Uncatchable error: Regex Too Complicated
What to do?
apex
apex
asked 11 hours ago
cropredycropredy
36k441123
asked 11 hours ago
cropredycropredy
36k441123
asked 11 hours ago
cropredycropredy
36k441123
asked 11 hours ago
cropredycropredy
36k441123
asked 11 hours ago
cropredycropredy
36k441123
36k441123
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This is basically teaching an old dog (me) new tricks
A cursory reading of the Apex String class documentation shows two methods
replace(target,replacement)replaceAll(regExp,replacement)
The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)
But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement
Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.
So, if you have large strings and you are just doing simple text substitution, avoid replaceAll
String s = s.replace('"currency":','"currencyX":')
.replace('"0001-01-01T00:00:00Z"','null');
answered 11 hours ago
cropredycropredy
36k441123
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is basically teaching an old dog (me) new tricks
A cursory reading of the Apex String class documentation shows two methods
replace(target,replacement)replaceAll(regExp,replacement)
The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)
But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement
Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.
So, if you have large strings and you are just doing simple text substitution, avoid replaceAll
String s = s.replace('"currency":','"currencyX":')
.replace('"0001-01-01T00:00:00Z"','null');
answered 11 hours ago
cropredycropredy
36k441123
add a comment |
This is basically teaching an old dog (me) new tricks
A cursory reading of the Apex String class documentation shows two methods
replace(target,replacement)replaceAll(regExp,replacement)
The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)
But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement
Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.
So, if you have large strings and you are just doing simple text substitution, avoid replaceAll
String s = s.replace('"currency":','"currencyX":')
.replace('"0001-01-01T00:00:00Z"','null');
answered 11 hours ago
cropredycropredy
36k441123
add a comment |
This is basically teaching an old dog (me) new tricks
A cursory reading of the Apex String class documentation shows two methods
replace(target,replacement)replaceAll(regExp,replacement)
The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)
But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement
Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.
So, if you have large strings and you are just doing simple text substitution, avoid replaceAll
String s = s.replace('"currency":','"currencyX":')
.replace('"0001-01-01T00:00:00Z"','null');
answered 11 hours ago
cropredycropredy
36k441123
This is basically teaching an old dog (me) new tricks
A cursory reading of the Apex String class documentation shows two methods
replace(target,replacement)replaceAll(regExp,replacement)
The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)
But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement
Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.
So, if you have large strings and you are just doing simple text substitution, avoid replaceAll
String s = s.replace('"currency":','"currencyX":')
.replace('"0001-01-01T00:00:00Z"','null');
answered 11 hours ago
cropredycropredy
36k441123
answered 11 hours ago
cropredycropredy
36k441123
answered 11 hours ago
cropredycropredy
36k441123
answered 11 hours ago
cropredycropredy
36k441123
36k441123
add a comment |
add a comment |
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