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Regex Too Complicated - replaceAll - large strings preprocessed before deserialize
How do you deserialize json properties that are reserved words in apex?APEX ENUM Serialize to and Deserialize from JSONregex too complicated error when using split function for less than 1M charactersCatch Regex too complicated when splitting a string in salesforceError - Heap Size Too LargeWhat is the correct JSON/Apex Strategy?Serialize JSON attribute that is a reserved keyword in ApexRegex too complicated when splitting a stringHow to deserialize into interface type when concrete type is unknownRead through small CSV file returns Regex too complicated
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
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
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
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 1 hour 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');
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');
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');
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');
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 1 hour ago
cropredycropredy
36k441123
36k441123
add a comment |
add a comment |
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