I want to replace
$this->input->post("product_name");
with
$post_data["product_name"];
I want to use notepad++ regex, but I couldn't find proper solution
In find --> $this->input->post("[\*w\]");
In replace --> $post_data["$1"];
but its not working
The $this->input->post("[\*w\]"); pattern does not work because:
$ is a special char matching the end of a line, you need to use \$ to match it as a literal char
[\*w'\] is a malformed pattern as there is no matching unescaped ] for the [ that opens a character class. Also, w just matches w, not any letter, digit or underscore, \w does that.
You may use
Find What: \$this->input->post\("(\w*)"\);
Replace With: $post_data["$1"];
If there can be any char inside double quotes use .*? instead of \w*:
Find What: \$this->input->post\("(.*?)"\);
Regulex graph:
NPP test:
Use this pattern to match desired text \$this->input->post\(("[^"]+")\);
And replace it with pattern \$post_data\[\1\]
Explanation:
\$this->input->post - matach $this->input->post literally
\(("[^"]+")\); - match (literally, then match double quates and everything between them with "[^"]+" and store inside first capturing group, then match ); literally
To replace
$this->input->post("product_name");
by
$post_data["product_name"];
do replace, with regex activated
this->input->post\("(.*)"\);
by
post_data\["\1"\];
The \x with x a number, corresponds to the x-th match catched with the parenthesis. Here we catch any character inside this->input->post(XXXX);
Don't forget to escape special character with \.
Your special characters were []()
Related
I'm using Java Pattern class to find a string "keyword" which is at the beginning of the string or after a character that is in a list of characters. For example, the list of characters is ' ' and '<', then:
match:
"keyword..."
"...<keyword..."
"... keyword..."
not match:
"...akeyword..."
I've tried all these:
"[^ <]keyword"
"[ <^]keyword"
"[\\^ <]keyword" note:for a Java/C# string backslash need to be escaped
This question is similar Match only at string start or after whitespace but with only basic skills of Regex I can't adopt it to this problem. I'v tried:
"(?<!\\S<)keyword"
"(?<!([\\S<]))keyword"
And this seems to be a very basic problem, there may be a very easy and clear way.
This should work (^|[< ])keyword
(...|...) has ^ and [< ], stating either it should be start of string of be after char(<) or char( )
You could use an alternation | in a non capturing group (?:^|[ <]) to assert either the start of the string ^ or match a space or < in a character class and use a capturing group for keyword.
(?:^|[ <])(keyword)\b
Regex demo
Or you could use a positive lookbehind (?<=...) and match only keyword
(?<=^|[< ])keyword\b
Regex demo
(^keyword |[< ^]keyword)
Write in the square brackets the character you need.
i want to use regex for replace a string in a file in powershell, this is the regex:
=.*\\app\\client\\.*\\
I applied this regex on that string:
HOME= C:\app\client\Administrateur\product
And i want this result:
= C:\app\client\Administrateur
But I have this result:
= C:\app\client\Administrateur\
How to say to regex i want to stop before the next backslash ?
Your pattern =.*\\app\\client\\.*\\ will match the last occurrence of \app\client\ and will then match until the last occurrence of the forward slash.
To match what comes after app\client\ but not include the last backslash you could use a negated character class matching not a backslash:
=.*\\app\\client\\[^\\]*
Regex demo
If the .* part at the start can not contain a backslash, this would be another option to prevent needless backtracking because the .* would first match until the end of the string:
=[^\\]*\\app\\client\\[^\\]*
Regex demo
I am using Notepad++ to remove some unwanted strings from the end of a pattern and this for the life of me has got me.
I have the following sets of strings:
myApp.ComboPlaceHolderLabel,
myApp.GridTitleLabel);
myApp.SummaryLabel + '</b></div>');
myApp.NoneLabel + ')') + '</label></div>';
I would like to leave just myApp.[variable] and get rid of, e.g. ,, );, + '...', etc.
Using Notepad++, I can match the strings themselves using ^myApp.[a-zA-Z0-9].*?\b (it's a bit messy, but it works for what I need).
But in reality, I need negate that regex, to match everything at the end, so I can replace it with a blank.
You don't need to go for negation. Just put your regex within capturing groups and add an extra .*$ at the last. $ matches the end of a line. All the matched characters(whole line) are replaced by the characters which are present inside the first captured group. .
matches any character, so you need to escape the dot to match a literal dot.
^(myApp\.[a-zA-Z0-9].*?\b).*$
Replacement string:
\1
DEMO
OR
Match only the following characters and then replace it with an empty string.
\b[,); +]+.*$
DEMO
I think this works equally as well:
^(myApp.\w+).*$
Replacement string:
\1
From difference between \w and \b regular expression meta characters:
\w stands for "word character", usually [A-Za-z0-9_]. Notice the inclusion of the underscore and digits.
(^.*?\.[a-zA-Z]+)(.*)$
Use this.Replace by
$1
See demo.
http://regex101.com/r/lU7jH1/5
I'd like to replace this
y[100] with this Ith(y,100) in Sublime3.
I've got the regular expression \by[\d+] in find what and Ith(y,$1) in replace with, but it doesn't work. It finds what to replace correctly but just replaces it with Ith(y, )
You need to pick the \d+ inside () to capture in group($1)
\by[(\d+)]
Also, you may need to to escape the [ and ] characters here.
\by\[(\d+)\]
You have to capture the data you want to backreference: to do that you have to use unescaped parenthesis (...), ie capturing groups. Also, [...] are character classes, a special character in regex that you need to escape.
Try replacing
\by\[(\d+)\]
with
Ith(y, \1)
You need to escape the special characters and capture the number in a capturing group group. Try this regex:
\by\[(\d+)]\b
And replace with
Ith(y, $1)
Online explanation and demonstration: http://regex101.com/r/nX3yJ9
I'm so close to understanding regex. I'm a bit stumped, I thought i understood lazy and greedy.
Here is my current regex: <g_n><!\[CDATA\[([^]]+)(?=]]><\/g_n>)
My current regex makes:
<g_n><![CDATA[xxxxxxxxxx]]></g_n>
match to:
<g_n><![CDATA[xxxxxxxxxx
But I want to make it match like this:
xxxxxxxxxx
You want
<g_n><!\[CDATA\[(.*?)]]></g_n>
then if you want to replace it use
\1
in the replacement box
Your matching the whole string, the brackets around the .*? match all of that and put it in the \1 variable
So the match will be all of the string with \1 referring to what you want
To change the xxxxx
Regex :
(<g_n><![CDATA[)(?:.*?)(]]></g_n>)
Replacement
\1WHAT YOU WANT TO CHANGE TO\2
It looks like you need to add escape slashes to the two closing square brackets, as they are literals from the string you're parsing.
<g_n><!\[CDATA\[.*+?\]\]><\/g_n>
^ ^
Any square brackets not being escaped by backslashes will be treated as regex operational brackets, which in this case won't catch the input string.
EDIT, I think the +? is redundant.
\[.*\]\]> ...
should suffice, since .* means any character, any amount of times.
Tested with notepad++ 6.3.2:
find: (<g_n><!\[CDATA\[)([^]]+)(?=]]></g_n>)
replace: $1WhatYouWant
You can replace + by * in the pattern to match void CDATA:
<g_n><![CDATA[]]></g_n>