I have a piece of code where I want to remove the quotation marks around property names.
// Current format
var user = {
'name': 'A',
'loggedIn': true
}
// Desired format
var user = {
name: 'A',
loggedIn: true
}
I've managed to find all the places I wish to change with this regular expression:
'(.+)'\:
Now I want to remove the quotation marks in those strings. I tried to enter (.+)\: into the "replace with" field, but it did not work. Is there some way to do what I want to do with this tool?
Find in Path documentation explains how to use the references:
if you specify the search pattern through a regular expression, use the $n format in back references (to refer to a previously found and saved pattern).
$1 will contain whatever is matched by the parenthesis, so your replacement string would look like $1:.
See also Regular Expression Syntax Reference.
Your regex matches with your desire strings, but you missed using captured groups! $1 returns first group and second and 3rd comes with $2 and $3 and ... .
Additional words:
You can back-referencing by \1 in your find regex to avoid repeating capture groups codes.
I suggest use this regex instead of your own in general cases:
^\s*(['|"])(.*?)\1\s?:
and replace by $2: to extract string between '/".
Related
I have multiple occurance of src={icons.ICON_NAME_HERE} in my code, that I would like to change to name="ICON_NAME_HERE".
Is it possible to do it with regular expressions, so I can keep whatever is in code as ICON_NAME_HERE?
To clarify:
I have for example src={icons.upload} and src={icons.download}, I want to do replace all with one regexp, so those gets converted to name="upload" and name="download"
Try searching on the following pattern:
src=\{icons\.([^}]+)\}
And then replace with your replacement:
name="$1"
In case you are wondering, the quantity in parentheses in the search pattern is captured during the regex search. Then, we can access that captured group using $1 in the replacement. In this case, the captured group should just be the name of the icon.
I would like to use regular expression for replacing a certain pattern in the Kettle. For example, AAAA >5< BBBB, I want to replace this with AAAA 555 BBBB. I know how to find the pattern, but I am not sure how to replace that with new string. The one thing I have to keep is that I have to find pattern together ><, not separately like > or < because there is another pattern <5>.
You can use the "Replace in String" step in a transformation.
Set use RegEx to "Y", type your regex on the Search box, with capturing groups if necessary, and the replacement string in the replacement box, referring to capture groups as $1, $2, ...
It'll replace all occurrences of the regex in the original string.
If the Out Stream field is ommitted, it'll overwrite the In stream field.
If you want the pattern >\d< replaced by a triple of the found digit, you can use Replace-In-String in regex mode:
Search: (.*)(>(\d)<)(.*)
Replace: $1$3$3$3$4
If you want all such patterns treated the same:
Search: (>(\d)<)
Replace: $2$2$2
EDIT due to your improved requirement
Since you intend to convert your "simple" markup to a more HTML-like markup, you better use a User-Defined-Java-Expression. Also, you must avoid to reintroduce simple markup when replacing repeatedly.
I'm trying to clean a huge geoJson datafile. I need to change the format of "text" field from
"text": "(2:Placename,Placename)"
to
"text": "Placename".
In Sublime text I managed to write a regular expression which enabled me to select and remove the first part leaving something like this:
"text": "Placename)"
With following regexp I can select the text above, but I need to narrow it down to the last character:
text\": \".*?\)
No matter what I can't figure out how to select the ")" character in the end of Placename string in the whole file and remove it. Note that the "Placename" here can be any place name, like New York, London etc.
I tried to build an expression where first part finds the text field, then ignores n-amount of characters until it finds the ")" character.
After experimenting and Googling I couldn't find a solution here.
You can capture the value of the second placemark field with the following regexp:
/"text": "+\(\d+:[^,]+,(.*?)\)/
Which will capture "Placename" in $1
More info on capturing parenthesis: http://www.regular-expressions.info/brackets.html
The trick is to use the inverted character classes and to escape any parentheses you want to match.
HTH
I do not know if you are using a Unix system, but probably sed can do much of the work for you. It can interpret regular expressions, capture groups, and substitute by other groups of characters. I have tried an example with sed and the following sed command worked for me:
echo "\"text\": \"(2:Placename,Placename)\"" | sed -r 's/(\"text\": )\"\([[:digit:]]:[^0-9]+,([^0-9]+)\)\"/\1\"\2\"/g'
-r allows sed to interpret regular expressions. I am using parentheses to capture groups that I will use later in the substitution (e.g., a group for "text", and a group for the second placename). In the substitution part of sed, you can use groups by using \n where n is the group number that you want to used. This expression should help you to achieve your desired result.
The regular expression which I have provided will select the string 72719.
Regular expression:
(?<=bdfg34f;\d{4};)\d{0,9}
Text sample:
vfhnsirf;5234;72159;2;668912;28032009;4;
bdfg34f;8467;72719;7;6637912;05072009;7;
b5g342sirf;234;72119;4;774582;20102009;3;
How can I rewrite the expression to select that string even when the number 8467; is changed to 84677; or 846777; ? Is it possible?
First, when asking a regex question, you should always specify which language you are using.
Assuming that the language you are using does not support variable length lookbehind (and most don't), here is a solution which will work. Your original expression uses a fixed-length lookbehind to match the pattern preceding the value you want. But now this preceding text may be of variable length so you can't use a look behind. This is no problem. Simply match the preceding text normally and capture the portion that you want to keep in a capture group. Here is a tested PHP code snippet which grabs all values from a string, capturing each value into capture group $1:
$re = '/^bdfg34f;\d{4,};(\d{0,9})/m';
if (preg_match_all($re, $text, $matches)) {
$values = $matches[1];
}
The changes are:
Removed the lookbehind group.
Added a start of line anchor and set multi-line mode.
Changed the \d{4} "exactly four" to \d{4,} "four or more".
Added a capture group for the desired value.
Here's how I usually describe "fields" in a regex:
[^;]+;[^;]+;([^;]+);
This means "stuff that isn't semi-colon, followed by a semicolon", which describes each field. Do that twice. Then the third time, select it.
You may have to tweak the syntax for whatever language you are doing this regex in.
Also, if this is just a data file on disk and you are using GNU tools, there's a much easier way to do this:
cat file | cut -d";" -f 3
to match the first number with a minimum of 4 digits
(?<=bdfg34f;\d{4,};)\d{0,9}
and to match the first number with 1 or more length
(?<=bdfg34f;\d+;)\d{0,9}
or to match the first number only if the length is between 4 and 6
(?<=bdfg34f;\d{4,6};)\d{0,9}
This is a simple text parsing problem that probably doesn't mandate the use of regular expressions.
You could take the input line by line and split on ';', i.e. (in php, I have no idea what you're doing)
foreach (explode("\n", $string) as $line) {
$bits = explode(";", $line);
echo $bits[3]; // third column
}
If this is indeed in a file and you happen to be using PHP, using fgetcsv would be much better though.
Anyway, context is missing, but the bottom line is I don't think you should be using regular expressions for this.
How can I create a regular expression that will grab delimited text from a string? For example, given a string like
text ###token1### text text ###token2### text text
I want a regex that will pull out ###token1###. Yes, I do want the delimiter as well. By adding another group, I can get both:
(###(.+?)###)
/###(.+?)###/
if you want the ###'s then you need
/(###.+?###)/
the ? means non greedy, if you didn't have the ?, then it would grab too much.
e.g. '###token1### text text ###token2###' would all get grabbed.
My initial answer had a * instead of a +. * means 0 or more. + means 1 or more. * was wrong because that would allow ###### as a valid thing to find.
For playing around with regular expressions. I highly recommend http://www.weitz.de/regex-coach/ for windows. You can type in the string you want and your regular expression and see what it's actually doing.
Your selected text will be stored in \1 or $1 depending on where you are using your regular expression.
In Perl, you actually want something like this:
$text = 'text ###token1### text text ###token2### text text';
while($text =~ m/###(.+?)###/g) {
print $1, "\n";
}
Which will give you each token in turn within the while loop. The (.*?) ensures that you get the shortest bit between the delimiters, preventing it from thinking the token is 'token1### text text ###token2'.
Or, if you just want to save them, not loop immediately:
#tokens = $text =~ m/###(.+?)###/g;
Assuming you want to match ###token2### as well...
/###.+###/
Use () and \x. A naive example that assumes the text within the tokens is always delimited by #:
text (#+.+#+) text text (#+.+#+) text text
The stuff in the () can then be grabbed by using \1 and \2 (\1 for the first set, \2 for the second in the replacement expression (assuming you're doing a search/replace in an editor). For example, the replacement expression could be:
token1: \1, token2: \2
For the above example, that should produce:
token1: ###token1###, token2: ###token2###
If you're using a regexp library in a program, you'd presumably call a function to get at the contents first and second token, which you've indicated with the ()s around them.
Well when you are using delimiters such as this basically you just grab the first one then anything that does not match the ending delimiter followed by the ending delimiter. A special caution should be that in cases as the example above [^#] would not work as checking to ensure the end delimiter is not there since a singe # would cause the regex to fail (ie. "###foo#bar###). In the case above the regex to parse it would be the following assuming empty tokens are allowed (if not, change * to +):
###([^#]|#[^#]|##[^#])*###