I am getting data from api call which contains double quotes. for example data = '{"firstName":""John""}'
how to parse this data into json.
expected-output:result = JSON.parse(data) and result.firstname should give output as "John" not John
As #Cid points out, that is invalid JSON.
You will need to sanitize it first:-
var json = data.replace(/""/g, '"');
var x = JSON.parse(json);
If you want to keep the inner quotes, you'll need to use something like this:-
var json = data.replace(/(\".*\":)\"\"(.*)\"\"/g, '$1 "\\"$2\\""');
var x = JSON.parse(json);
However, you may need to fiddle with the regex if it conflicts with other parameters.
You can review the regex above at https://regex101.com/ to get an explanation of how the regex matches:-
/(\".*\":)\"\"(.*)\"\"/g
1st Capturing Group (\".*\":)
\" matches the character " literally (case sensitive)
.* matches any character (except for line terminators)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\" matches the character " literally (case sensitive)
: matches the character : literally (case sensitive)
\" matches the character " literally (case sensitive)
\" matches the character " literally (case sensitive)
2nd Capturing Group (.*)
.* matches any character (except for line terminators)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\" matches the character " literally (case sensitive)
\" matches the character " literally (case sensitive)
Global pattern flags
g modifier: global. All matches (don't return after first match)
The $1 and $2 in the replacement text correspond to the capture groups in the regex. See String.prototype.replace() for details.
Try With this
var json = '{"firstName":""John""}'; //Let's say you got this
json = json.replace(/\"([^(\")"]+)\":/g,"$1:"); //This will remove all the quotes
json;
Related
Is there a way in Presto or SQL to remove the last backslash only when it is preceded by a character and keep it otherwise?
I am using regexp_replace in presto. So for example if x = '/' The expression should return '/'
and if x = 'beta/alpha/' it should return 'beta/alpha'
I am using select regexp_replace ([expression], '[\/]$', '').
This returns an empty string when there is only a backslash and removes the backslash from the end of the string if the expression has some characters before the backslash.
You can use
regexp_replace([expression], '([^/])/$', '$1')
-- or
regexp_replace([expression], '(?<=[^/])/$', '')
See the regex demo.
Details
([^/])/$ - matches and captures any char but / into Group 1 (with the ([^/]) pattern, the $1 in the replacement pattern is a replacement backreference that refers to the Group 1 value), then matches a / at the end of string ($)
(?<=[^/])/$ matches a / at the end of a string only when the char immediately on the left is not a / char (and not the start of a string).
I just want to format a decimal number for output to a simple CSV formatted file.
I feel like I'm stupid, but I can't find a way to do it without leading zeroes or spaces, of course I can simply trim the leading spaces, but there has to be a proper way to just format like I that, isn't there?
Example
define variable test as decimal.
define variable testString as character.
test = 12.3456.
testString = string(test, '>>>>>9.99').
message '"' + testString + '"' view-as alert-box. /* " 12.35" */
I tried using >>>>>9.99 and zzzzz9.99 for the number format, but both format the string with leading spaces. I actually have no idea what the difference is between using > and z.
The SUBSTITUTE() function will do what you describe wanting:
define variable c as character no-undo.
c = substitute( "&1", 1.23 ).
display "[" + c + "]".
(Toss in a TRUNCATE( 1.2345, 2 ) if you really only want 2 decimal places.)
Actually, this also works:
string( truncate( 1.2345, 2 )).
If you are creating a CSV file you might want to think about using EXPORT. EXPORT format removes leading spaces and omits decorations like ",". The SUBSTITUTE() function basically uses EXPORT format to make its substitutions. The STRING() function uses EXPORT format when no other format is specified.
The EXPORT statement will format your data for you. Here is an example:
DEFINE VARIABLE test AS DECIMAL NO-UNDO.
DEFINE VARIABLE testRound AS DECIMAL NO-UNDO.
DEFINE VARIABLE testString AS CHARACTER NO-UNDO.
test = 12.3456.
testRound = ROUND(test, 2).
testString = STRING(test).
OUTPUT TO VALUE("test.csv").
EXPORT DELIMITER "," test testRound testString.
OUTPUT CLOSE.
Here is the output:
12.3456,12.35,"12.3456"
The EXPORT statement's default delimiter is a space so you have to specify a comma for your CSV file. Since the test and testRound variables are decimals, they are not in quotes in the output. testString is character so it is in quotes.
In a wordpress theme, I am using the "posts_where" filter to add search to the "excerpt" field. It is working excepted when there is a single quote in the search string, leading to a SQL synthax error.
It seems to be a bug in the preg_replace function off the posts_where filter.
For example, for the string "o'kine" , the $where string received in the posts_where filter is :
"AND (((cn_posts.post_title LIKE '%o\'kine%') OR (cn_posts.post_content LIKE '%o\'kine%')))"
Then this is my preg_replace to add the post_excerpt field :
$where = preg_replace(
"/post_title\s+LIKE\s*(\'[^\']+\')/",
"post_title LIKE $1) OR (post_excerpt LIKE $1", $where );
And the value of the $where after :
"AND (((cn_posts.post_title LIKE '%o\') OR (post_excerpt LIKE '%o\'kine%') OR (cn_posts.post_content LIKE '%o\'kine%')))"
See the '%o\' part that is causing the SQL synthax error.
The expected result would be :
"AND (((cn_posts.post_title LIKE '%o\'kine%') OR (post_excerpt LIKE '%o\'kine%') OR (cn_posts.post_content LIKE '%o\'kine%')))"
The bug is clearly in my regular expression, more precisely in my capturing parentheses. I do not know how to deal with the possibility of zero or more single quote in my search string?
EDIT : With Casimir et Hippolyte answer, this is the working filter with single quote in the search string :
function cn_search_where( $where ) {
$where = preg_replace(
"/post_title\s+LIKE\s*('[^'\\\\]*+(?s:\\\\.[^'\\\\]*)*+')/",
"post_title LIKE $1) OR (post_excerpt LIKE $1", $where );
return $where;
}
The subpattern to match a quoted string with eventual escaped quotes (or other characters) is:
'[^'\\]*+(?s:\\.[^'\\]*)*+'
(note that to figure a literal backslash in a regex pattern, it must be escaped since the backslash is a special character)
So in a php string (backslashes need to be escaped one more time):
$pattern = "~'[^'\\\\]*+(?s:\\\\.[^'\\\\]*)*+'~";
With this information, I think you can build the pattern yourself.
details:
' # a literal single quote
[^'\\]*+ # zero or more characters that are not a single quote or a backslash
(?s: # open a non-capture group with the s modifier (the dot can match newlines)
\\. # an escaped character
[^'\\]*
)*+ # repeat the group zero or more times
'
Am using opencsv 2.3 and it does not appear to be dealing with escape characters as I expect. I need to be able to handle an escaped separator in a CSV file that does not use quoting characters.
Sample test code:
CSVReader reader = new CSVReader(new FileReader("D:/Temp/test.csv"), ',', '"', '\\');
String[] nextLine;
while ((nextLine = reader.readNext()) != null) {
for (String string : nextLine) {
System.out.println("Field [" + string + "].");
}
}
and the csv file:
first field,second\,field
and the output:
Field [first field].
Field [second].
Field [field].
Note that if I change the csv to
first field,"second\,field"
then I get the output I am after:
Field [first field].
Field [second,field].
However, in my case I do not have the option of modifying the source CSV.
Unfortunately it looks like opencsv does not support escaping of separator characters unless they're in quotes. The following method (taken from opencsv's source) is called when an escape character is encountered.
protected boolean isNextCharacterEscapable(String nextLine, boolean inQuotes, int i) {
return inQuotes // we are in quotes, therefore there can be escaped quotes in here.
&& nextLine.length() > (i + 1) // there is indeed another character to check.
&& (nextLine.charAt(i + 1) == quotechar || nextLine.charAt(i + 1) == this.escape);
}
As you can see, this method only returns true if the character following the escape character is a quote character or another escape character. You could patch the library to this, but in its current form, it won't let you do what you're trying to do.
I am using Laravel 4 and have set up the following query:
if(Input::get('keyword')) {
$keyword = Input::get('keyword');
$search = DB::connection()->getPdo()->quote($keyword);
$query->whereRaw("MATCH(resources.name, resources.description, resources.website, resources.additional_info) AGAINST(? IN BOOLEAN MODE)",
array($search)
);
}
This query runs fine under normal use, however, if the user enters a string such as ++, an error is thrown. Looking at the MySQl docs, there are some keywords, such as + and - which have specific purposes. Is there a function which will escape these types of special characters from a string so it can be used in a fulltext search like above without throwing any errors?
Here is an example of an error which is thrown:
{"error":{"type":"Illuminate\\Database\\QueryException","message":"SQLSTATE[42000]: Syntax error or access violation: 1064 syntax error, unexpected '+' (SQL: select * from `resources` where `duplicate` = 0 and MATCH(resources.name, resources.description, resources.website, resources.additional_info) AGAINST('c++' IN BOOLEAN MODE))","file":"\/var\/www\/html\/[...]\/vendor\/laravel\/framework\/src\/Illuminate\/Database\/Connection.php","line":555}}
Solutions I've tried:
$search = str_ireplace(['+', '-'], ' ', $keyword);
$search = filter_var($keyword, FILTER_SANITIZE_STRING);
$search = DB::connection()->getPdo()->quote($keyword);
I'm assuming I will need to use regex. What's the best approach here?
Only the words and operators have meaning in Boolean search mode. Operators are: +, -, > <, ( ), ~, *, ", #distance. After some research I found what word characters are: Upper case, Lower case letters, Numeral (digit) and _. I think you can use one of two approaches:
Replace all non word characters with spaces (I prefer this approach). This can be accomplished with regex:
$search = preg_replace('/[^\p{L}\p{N}_]+/u', ' ', $keyword);
Replace characters-operators with spaces:
$search = preg_replace('/[+\-><\(\)~*\"#]+/', ' ', $keyword);
Only words are indexed by full text search engine and can be searched. Non word characters isn't indexed, so it does not make sense to leave them in the search string.
References:
Boolean Full-Text Searches
Fine-Tuning MySQL Full-Text Search (see: "Character Set Modifications")
PHP: preg_replace
PHP: Unicode character properties
PHP: Possible modifiers in regex patterns
While the answer from Rimas is technically correct, it will suit you only if you do not want users to use the MATCH operators, because it will strip them all completely. For example, I do want to allow use of all of them except #distance in search forms on my site, thus I've come up with this:
#Trim first
$newValue = preg_replace('/^\p{Z}+|\p{Z}+$/u', '', string);
#Remove all symbols except allowed operators and space. #distance is not included, since it's unlikely a human will be using it through UI form
$newValue = preg_replace('/[^\p{L}\p{N}_+\-<>~()"* ]/u', '', $newValue);
#Remove all operators, that can only precede a text and that are not preceded by either beginning of string or space
$newValue = preg_replace('/(?<!^| )[+\-<>~]/u', '', $newValue);
#Remove all double quotes and asterisks, that are not preceded by either beginning of string, letter, number or space
$newValue = preg_replace('/(?<![\p{L}\p{N}_ ]|^)[*"]/u', '', $newValue);
#Remove all double quotes and asterisks, that are inside text
$newValue = preg_replace('/([\p{L}\p{N}_])([*"])([\p{L}\p{N}_])/u', '', $newValue);
#Remove all opening parenthesis which are not preceded by beginning of string or space
$newValue = preg_replace('/(?<!^| )\(/u', '', $newValue);
#Remove all closing parenthesis which are not preceded by beginning of string or space or are not followed by end of string or space
$newValue = preg_replace('/(?<![\p{L}\p{N}_])\)|\)(?! |$)/u', '', $newValue);
#Remove all double quotes if the count is not even
if (substr_count($newValue, '"') % 2 !== 0) {
$newValue = preg_replace('/"/u', '', $newValue);
}
#Remove all parenthesis if count of closing does not match count of opening ones
if (substr_count($newValue, '(') !== substr_count($newValue, ')')) {
$newValue = preg_replace('/[()]/u', '', $newValue);
}
Unfortunately I was not able to figure out a way to do this in 1 regex, thus doing multiple runs. It's also possible, that I am missing some edge cases, as well. Any additions or corrections are appreciated: either here or create an issue for https://github.com/Simbiat/database where I implement this.