I got to deal here with a problem, caused by a dirty design. I get a list of string and want to parse attributes out of it. Unfortunately, I can't change the source, where these String were created.
Example:
String s = "type=INFO, languageCode=EN-GB, url=http://www.stackoverflow.com, ref=1, info=Text, that may contain all kind of chars., deactivated=false"
Now I want to extract the attributes type, languageCode, url, ref, info and deactivated.
The problem here is the field info, whose text is not limited by quote mark. Also commas may occur in this field, so I can't use the comma at the end of the string, to find out where is ends.
Additional, those strings not always contain all attributes. type, info and deactivated are always present, the rest is optional.
Any suggestions how I can solve this problem?
One possible solution is to search for = characters in the input and then take the single word immediately before it as the field name - it seems that all your field names are single words (no whitespace). If that's the case, you can then take everything after the = until the next field name (accounting for separating ,) as the value.
This assumes that the value cannot contain =.
Edit:
As a possible way to handle embedded =, you can see if the word in front of it is one your known field names - if not, you can possibly treat the = as an embedded character rather than an operator. This, however, assumes that you have a fixed set of known fields (some of which may not always appear). This assumption may be eased if you know that the field names are case-sensitive.
Assuming that order of elements is fixed you could write solution using regex like this one
String s = "type=INFO, languageCode=EN-GB, url=http://www.stackoverflow.com, ref=1, info=Text, that may contain all kind of chars., deactivated=false";
String regex = //type, info and deactivated are always present
"type=(?<type>.*?)"
+ "(?:, languageCode=(?<languageCode>.*?))?"//optional group
+ "(?:, url=(?<url>.*?))?"//optional group
+ "(?:, ref=(?<rel>.*?))?"//optional group
+ ", info=(?<info>.*?)"
+ ", deactivated=(?<deactivated>.*?)";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(s);
if(m.matches()){
System.out.println("type -> "+m.group("type"));
System.out.println("languageCode -> "+m.group("languageCode"));
System.out.println("url -> "+m.group("url"));
System.out.println("rel -> "+m.group("rel"));
System.out.println("info -> "+m.group("info"));
System.out.println("deactivated -> "+m.group("deactivated"));
}
Output:
type -> INFO
languageCode -> EN-GB
url -> http://www.stackoverflow.com
rel -> 1
info -> Text, that may contain all kind of chars.
deactivated -> false
EDIT: Version2 regex searching for oneOfPossibleKeys=value where value ends with:
, oneOfPossibleKeys=
or has end of string after it (represented by $).
Code:
String s = "type=INFO, languageCode=EN-GB, url=http://www.stackoverflow.com, ref=1, info=Text, that may contain all kind of chars., deactivated=false";
String[] possibleKeys = {"type","languageCode","url","ref","info","deactivated"};
String keysStrRegex = String.join("|", possibleKeys);
//above will contain type|languageCode|url|ref|info|deactivated
String regex = "(?<key>\\b(?:"+keysStrRegex+")\\b)=(?<value>.*?(?=, (?:"+keysStrRegex+")=|$))";
// (?<key>\b(?:type|languageCode|url|ref|info|deactivated)\b)
// =
// (?<value>.*?(?=, (?:type|languageCode|url|ref|info|deactivated)=|$))System.out.println(regex);
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(s);
while(m.find()){
System.out.println(m.group("key")+" -> "+m.group("value"));
}
Output:
type -> INFO
languageCode -> EN-GB
url -> http://www.stackoverflow.com
ref -> 1
info -> Text, that may contain all kind of chars.
deactivated -> false
You could use a regular expression, capturing all the "fixed" groups and using whatever remains for info. This should even work if the info part contains , or = characters. Here's some quick example (using Python, but that should not be a problem...).
>>> p = r"(type=[A-Z]+), (languageCode=[-A-Z]+), (url=[^,]+), (ref=\d), (info=.+?), (deactivated=(?:true|false))"
>>> s = "type=INFO, languageCode=EN-GB, url=http://www.stackoverflow.com, ref=1, info=Text, that may contain all kind of chars, even deactivated=true., deactivated=false"
>>> re.search(p, s).groups()
('type=INFO',
'languageCode=EN-GB',
'url=http://www.stackoverflow.com',
'ref=1',
'info=Text, that may contain all kind of chars, even deactivated=true.',
'deactivated=false')
If any of those elements are optional, you can put a ? after those groups, and make the comma optional. If the order can be different, then it's more complicated. In this case, instead of using one RegEx to capture everything at once, use several RegExes to capture the individual attributes and then remove (replace with '') those in the string before matching the next attribute. Finally, match info.
On further consideration, given that those attributes could have any order, it may be more promising to capture just everything spanning from one keyword to the next, regardless of its actual content, very similar to Pshemo's solution:
keys = "type|languageCode|url|ref|info|deactivated"
p = r"({0})=(.+?)(?=\, (?:{0})=|$)".format(keys)
matches = re.findall(p, s)
But this, too, might fail in some very obscure cases, e.g. if the info attribute contains something like ', ref=foo', including the comma. However, there seems to be no way around those ambiguities. If you had a string like info=in this string, ref=1, and in another, ref=2, ref=1, does it contain one ref attribute, or three, or none at all?
Related
My String is like this.
{\\\"692950841314120\\\":[{\\\"type\\\":\\\"ads_management\\\",\\\"call_count\\\":3,\\\"total_cputime\\\":1,\\\"total_time\\\":5,\\\"estimated_time_to_regain_access\\\":0}]}
Since the key here is a variable value I am trying to replace this 692950841314120(or the values which I get from sever) with a constant like ID. My main goal is to parse this as POJO. I have tried using..
string.replaceAll("^[0-9]{15}$","ID")
but due to Slashes I think i am not able to get the desired value. Is there any better way to do this. I know I can do below Code but I don't want any ID123 if I added extra value and distort any other info in JSON.
string.replaceAll("[0-9]{15}","ID")
Strictly speaking, if you have a valid JSON string, you should parse it using something like GSON, rather than using regex. That being said, if you must use regex, you could try removing the starting and ending anchors:
string.replaceAll("[0-9]{15}", "ID")
Or maybe use double quotes instead:
string.replaceAll("\"[0-9]{15}\"", "ID")
It is safer to assume the value is inisde \" and \":.
You can then use
.replaceAll("(\\\\\")[0-9]{15}(\\\\\":)", "$1ID$2")
The regex is (\\")[0-9]{15}(\\":) and it means:
(\\") - match and capture \" substring into Group 1
[0-9]{15} - fifteen digits
(\\":) - Group 2: a \": substring.
The $1 and $2 are placeholders holding the Group 1 and 2 values.
You should use "A word boundary" \b.
Try this.
public static void main(String[] args) {
String input = "{\\\"692950841314120\\\":"
+ "[{\\\"type\\\":\\\"12345678901234567890\\\","
+ "\\\"call_count\\\":3,"
+ "\\\"total_cputime\\\":1,"
+ "\\\"total_time\\\":5,"
+ "\\\"estimated_time_to_regain_access\\\":0}]}";
System.out.println(input.replaceAll("\\b[0-9]{15}\\b", "ID"));
}
output:
{\"ID\":[{\"type\":\"12345678901234567890\",\"call_count\":3,\"total_cputime\":1,\"total_time\":5,\"estimated_time_to_regain_access\":0}]}
I'm trying to make a string reconstruction with parameters based on a regex pattern.
I'm looking for something like this:
String regex = "(?<string>[a-z\-]+)?(/number/(?<number>\\d+))?"
case 1:
Map<String,String> parameters = new HashMap<>();
parameters.put("string","test");
parameters.put("number","1");
generator.generateString(regex, parameters) // returns "test/number/1"
case 2:
Map<String,String> parameters = new HashMap<>();
parameters.put("string","test");
generator.generateString(regex, parameters) // returns "test"
because this group (/number/(?<number>\\d+))? is optional, it would still match
Maybe there's a library for it, just note that I need something that is working with regex named groups (introduced in Java7)
I could write it myself, but at least a library that could detect groups in the pattern would be a big help.
How a library like that could work:
Pattern2 r = Pattern2.compile("(?<string>[a-z\-]+)?(/number/(?<number>\\d+))?");
To find start and end character of the "string" named group
r.start("string"); //0
r.end("string"); //19
or to find start and end of a group by a number of a group
r.start(2); //20
r.end(2); //44
I'm trying to click on element using below code and its working fine if variable does not have special characters.
Values that passed in variable are:
Product Passed
Test Message -> Failed
Double Units 13/2" -> failed
I cannot go with partial text, because sometimes only difference between two element is special character.
String Answertoselect = ElementName.replaceAll("'", "\\\\'");
String selectanswer = Answertoselect.replaceAll(" ", "${nbsp}");
I want to extract particular fields from a url of a facebookpage. Iam not able to extract since link format is not static.eg:if I gave the below examples as input it should give the o/p as what we desire
1)https://www.facebook.com/pages/Ice-cream/109301862430120?rf=102173023157556
o/p -109301862430120
What about this type of link
can anyone help me
So in short, you want to get name after last / and (if there is any) before ? mark.
You can do it with using URI and File classes like
String data = "https://www.facebook.com/pages/Anti-Christian-sentiment/149675731889496?ref=br_tf";
System.out.println(new File(new URI(data).getRawPath()).getName());
Output: 149675731889496
If you need to use regex then you can use
([^/?]+)(\\?|$)
and just read content of group 1 (the one in first pair of parenthesis).
If you don't want to use groups, and make regex match only digit part (without including ? in match) then you can use look around mechanisms like look-ahead (?=...). Regex you would have to use would look like
[^/?]+(?=\\?|$)
Code example:
String data = "https://www.facebook.com/pages/Anti-Christian-sentiment/149675731889496?ref=br_tf";
Pattern p = Pattern.compile("([^/?]+)(\\?|$)");
Matcher m = p.matcher(data);
if (m.find()){
System.out.println(m.group(1));
}
Output:
149675731889496
Given the string
Content ID [9283745997] Content ID [9283005997] There can be text in between Content ID [9283745953] Content ID [9283741197] Content ID [928374500] There can be valid text here which should not be removed.
I want to remove the text starting Content ID followed by [9283745997] any numbers can be present between square brackets. Eventually I want the result string to be
There can be text in between There can be valid text here which should not be removed.
Could anyone please provide a valid regex to capture this recurring text but the numerals within square brackets are unique?
I appreciate your help!
My soulution to this was :
Pattern p = Pattern.compile("(Content ID \\[\\d*\\] )");
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer();
while(m.find()) {
m.appendReplacement(sb, "");
}
m.appendTail(sb);
System.out.println(sb);
So basically you are trying to remove each of Content ID [one or more digits].
To do this you can use replaceAll("regex","replacement") method of String class. As replacement you can use empty String "".
Only problem that stays is what regex should you use.
to match Content ID just write it normally as "Content ID "
to match [ or ] you will have to add \ before each of them because they are regex metacharacters and you need to escape them (in Java you will need to write \ as "\\")
to represent one digit (character from range 0-9) regex uses \d (again in Java you will need to write \ as "\\" which will result in "\\d")
to say "one or more of previously described element" just add + after definition of such element. For example if you want to match one or more letters a you can write it as a+.
Now you should be able to create correct regex. If you will have some questions feel free to ask them in comments.
Try this one:
(Content ID \[[0-9]+\])
You can test it here: http://regexpal.com/
I would use the regex
Content ID \[\d+\] ?
Implement it like this:
str.replaceAll("Content ID \\[\\d+\\] ?", "");
You can find an explanation and demonstration here: http://regex101.com/r/qD5rJ6