I have a String with single quote. I want to replace the single quote with 2 single quotes.
I tried using
String s="Kathleen D'Souza";
s.replaceAll("'","''");
s.replaceAll("\'","\'\'");
s.replace("'","''");
s.replace("\'","\'\'");
But the single quote is not getting replaced with 2 single quotes.
reassign the replaced string to s
String s="Kathleen D'Souza";
s = s.replaceAll("'","''");
Please try
s= "test ' test";
`s.replaceAll("'","\"");` => test " test
`s.replaceAll("'","''");` => test '' test
Strings are immutable. Assign the result of replaceAll to your String:
s = s.replaceAll("'","''");
String s="Kathleen D'Souza";
s= s.replace("'", "''");
Try String#replace(). It will replace all occurrence of single ' with double ''.
Note, with the given solutions successive single quotes will be doubled, so Kathleen D''Souza turns into Kathleen D''''Souza. (I've seen users outsmart themselves like this.) If that is something you are concerned about, you can match successive single quotes with:
s = s.replaceAll("''*","''");
Related
I am looking for regex which can help me replace strings like
source=abc/task=cde/env=it --> source='abc'/task='cde'/env='it'
To be more precise, I want to replace a string which starts with = and ends with either / or end of the string with ''
Tried code like this
"source=abc/task=cde/env=it".replaceAll("=(.*?)/","'$1'")
But that results in
source'abc'task'cde'env=it
Using lookahead and look behind:
(?<==)([^/]*)((?=/)|$)
Lookbehind allows you to specify what comes before your match. In this case an equals: (?<==).
The main match in my regex looks for any non-slash character, zero or more times: ([^/]*)
Lookahead allows you to specify what comes after your match. In this case, a slash: (?=/).
The $ matches the end of the line, so that the last item in your test data becomes quoted. ((?=/)|$) combines with this with the lookahead, meaning "either a slash comes after the match or this is the end of the line".
Here it is in action in a test.
#Test
public void test_quote_items() {
String regex = "(?<==)([^/]*)((?=/)|$)";
String actual = "source=abc/task=cde/env=it".replaceAll(regex,"'$1'");
String expected = "source='abc'/task='cde'/env='it'";
assertEquals(expected, actual);
}
Try
String input = "source=abc/task=cde/env=it".replaceAll("=(.*?)(/|$)","='$1'/");
The problems I found are that you are not replacing the =
and also the / is not there for the end of String, that also needs to be replaced when found.
output
source='abc'/task='cde'/env='it'/
If you don't want the last '/', that is trivial to remove isn't it.
My String is like this.
{\\\"692950841314120\\\":[{\\\"type\\\":\\\"ads_management\\\",\\\"call_count\\\":3,\\\"total_cputime\\\":1,\\\"total_time\\\":5,\\\"estimated_time_to_regain_access\\\":0}]}
Since the key here is a variable value I am trying to replace this 692950841314120(or the values which I get from sever) with a constant like ID. My main goal is to parse this as POJO. I have tried using..
string.replaceAll("^[0-9]{15}$","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}]}
We have a String as below.
\config\test\[name="sample"]\identifier["2"]\age["3"]
I need to remove the quotes surrounding the numbers. For example, the above string after replacement should look like below.
\config\test\[name="sample"]\identifier[2]\age[3]
Currently I'm trying with the regex as below
String.replaceAll("\"\\\\d\"", "");
This is replacing the numbers also. Please help to find out a regex for this.
You can use replaceAll with this regex \"(\d+)\" so you can replace the matching of \"(\d+)\" with the capturing group (\d+) :
String str = "\\config\\test\\[name=\"sample\"]\\identifier[\"2\"]\\age[\"3\"]";
str = str.replaceAll("\"(\\d+)\"", "$1");
//----------------------^____^------^^
Output
\config\test\[name="sample"]\identifier[2]\age[3]
regex demo
Take a look about Capturing Groups
We can try doing a blanket replacement of the following pattern:
\["(\d+)"\]
And replacing it with this:
\[$1\]
Note that we specifically target quoted numbers only appearing in square brackets. This minimizes the risk of accidentally doing an unintended replacement.
Code:
String input = "\\config\\test\\[name=\"sample\"]\\identifier[\"2\"]\\age[\"3\"]";
input = input.replaceAll("\\[\"(\\d+)\"\\]", "[$1]");
System.out.println(input);
Output:
\config\test\[name="sample"]\identifier[2]\age[3]
Demo here:
Rextester
You can use:
(?:"(?=\d)|(?<=\d)")
and replace it with nothing == ( "" )
fast test:
echo '\config\test\[name="sample"]\identifier["2"]\age["3"]' | perl -lpe 's/(?:"(?=\d)|(?<=\d)")//g'
the output:
\config\test\[name="sample"]\identifier[2]\age[3]
test2:
echo 'identifier["123"]\age["456"]' | perl -lpe 's/(?:"(?=\d)|(?<=\d)")//g'
the output:
identifier[123]\age[456]
NOTE
if you have only a single double quote " it works fine; otherwise you should add quantifier + for both beginning and end "
test3:
echo '"""""1234234"""""' | perl -lpe 's/(?:"+(?=\d)|(?<=\d)"+)//g'
the output:
1234234
I want to to input this link in to the string.
String url=www.test.com;
String link=<a href=url>contact info</a>
How can I write this ?
You will need to do:
String url = "www.test.com";
You can use \ character to indicate that we want to include a special character, and that the next character should be treated differently. \" indicates a double quote character and not the termination of the string.
String link = "contact info";
A character preceded by a backslash is an escape sequence and has special meaning to the compiler. The following table shows the Java escape sequences:
Java Escape Sequences:
For More information check this link
First, let's assume you have:
String url = "www.test.com";
(Note the quotes around the string.)
To create your link string, you'd do this:
String link = "contact info";
// Note ---------------^^-----------^^
To put a " inside a string literal, you put a backslash in front of it. This is called "escaping" the quote.
First have the url value within quotes ,then concat the value in the link string.
String url="www.test.com";
String link="contact info";
I want to split the string
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
to
name
employeeno
dob
joindate
I wrote the following java code for this but it is printing only name other matches are not printing.
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
Pattern pattern = Pattern.compile("\\[.+\\]+?,?\\s*" );
String[] split = pattern.split(fields);
for (String string : split) {
System.out.println(string);
}
What am I doing wrong here?
Thank you
This part:
\\[.+\\]
matches the first [, the .+ then gobbles up the entire string (if no line breaks are in the string) and then the \\] will match the last ].
You need to make the .+ reluctant by placing a ? after it:
Pattern pattern = Pattern.compile("\\[.+?\\]+?,?\\s*");
And shouldn't \\]+? just be \\] ?
The error is that you are matching greedily. You can change it to a non-greedy match:
Pattern.compile("\\[.+?\\],?\\s*")
^
There's an online regular expression tester at http://gskinner.com/RegExr/?2sa45 that will help you a lot when you try to understand regular expressions and how they are applied to a given input.
WOuld it be better to use Negated Character Classes to match the square brackets? \[(\w+\s)+\w+[^\]]\]
You could also see a good example how does using a negated character class work internally (without backtracking)?