I'm currently learning Angular I'm having an issue with transforming a certain long string.
So, this is my div with a lot of lines, generated by ng-repeat:
http://pastebin.com/raw/bJqqUvpY
Yeah, it's pretty nasty, I know. What i want to do is to remove all the ng attributes and other stuff that I don't need from that string, because I'm going to pass this data to PHP via ajax.
Here's what I tried to do (angular):
$scope.updateHtml = function() {
var testVal = angular.element('#sendHtml').val();
testVal = testVal.replace(/ng-class="{'selected':selection.indexOf($index) != -1}"/g,"");
$scope.sendHtml.html = testVal;
};
But it doesn't work. Perhaps it's because of the quotation marks inside of the phrase, or is it?
This, for instance, works with a replacement of a single letter:
$scope.updateHtml = function() {
var testVal = angular.element('#sendHtml').val();
testVal = 'abcabcabc';
testVal = testVal.replace(/b/g,"");
$scope.sendHtml.html = testVal;
};
Then $scope.sendHtml.html is equal to 'acacac' like it should.
Could this be solved with another kind of RegEx?
Escape dot ., parenthesis (), and dollar $ signs.
testVal = testVal.replace(/ng-class="{'selected':selection\.indexOf\(\$index\) != -1}"/g,"");
Demo: http://regexr.com/3enmj
On the website you can examine all characters that should be escaped, by opening menu Reference - Escaped Characters.
If you need to remove all the ng-* directives as well as the html comments generated you can try the following regex.
\sng-[a-z]*="(.*?)"|(<!--(.*?)-->)
You remove a whitespace \s followed by ng- and any number of characters [a-z]* followed by double quotes and what's inside "(.*?)" as well as the html comments <!--(.*?)--> and what's inside.
Can probably be improved but it works for cleaning your input.
Related
My goal is to take a markdown text and create the necessary bold/italic/underline html tags.
Looked around for answers, got some inspiration but I'm still stuck.
I have the following typescript code, the regex matches the expression including the double asterisk:
var text = 'My **bold\n\n** text.\n'
var bold = /(?=\*\*)((.|\n)*)(?<=\*\*)/gm
var html = text.replace(bold, '<strong>$1</strong>');
console.log(html)
Now the result of this is : My <\strong>** bold\n\n **<\strong> text.
Everything is great aside from the leftover double asterisk.
I also tried to remove them in a later 'replace' statement, but this creates further issues.
How can I ensure they are removed properly?
With your pattern (?=\*\*)((.|\n)*)(?<=\*\*) you assert (not match) with (?=\*\*) that there is ** directly to the right.
Then directly after that, you capture the ** using ((.|\n)*) so then it becomes part of the match.
Then at the end you assert again with (?<=\*\*) that there is ** directly to the left, but ((.|\n)*) has already matched it.
This way so you will end up with all the ** in the match.
You don't need lookarounds at all, as you are already using a capture group.
In Javascript you could match the ** on the left and right and capture any character in a capture group:
\*\*([^]*?)\*\*
Regex demo
But I would suggest using a dedicated parser to parse markdown instead of using a regex.
Just make another call to replaceAll removing the ** with and empty string.
var text = 'My **bold\n\n** text.\n'
var bold = /(?=\*\*)((.|\n)*)(?<=\*\*)/gm
var html = text.replace(bold, '<strong>$1</strong>');
html = html.replaceAll(/\*\*/gm,'');
console.log(html)
I need to fix a bug in AngularJS application, which has many forms to submit data-
Every Text box in forms is accepting whitespaces(both leading and trailing) and saving them into the database. So in order to fix this I used ng-trim="true", it worked and data is getting saved correctly in the back-end.
Problem: Even after using ng-trim when I click on save/update, the form UI shows the text with white-spaces not the trimmed data. It shows correct data only when I refresh the page.
Can anyone guide me.. what will be the approach to fix this?
P.S. - I'm new to both JavaScript and Angular!
Thanks
Using trim() method works fine, but is used in newer browsers.
function removeWhitespaceUsingTrimMethod {
var str = " This is whitespace string for testing purpose ";
var wsr = str.trim();
alert(wsr);
}
Output:
This is whitespace string for testing purpose
From Docs:
(method) String.trim(): string
Removes the leading and trailing white space and line terminator
characters from a string.
Using replace() method – works in all browsers
Syntax:
testStr.replace(rgExp, replaceText);
str.replace(/^\s+|\s+$/g, '');
function removeWhitespaceUsingReplaceMethod {
var str = " This is whitespace string for testing purpose ";
var wsr = str.replace(/^\s+|\s+$/g, '');
alert( wsr);
}
Output:
This is whitespace string for testing purpose
Use string = string.trim() where string is the variable holding your string value.
When using reactive forms
this.form.controls['email'].valueChanges
.subscribe(x=>{
if(x.includes(' ')){
console.log('contains spaces')
this.form.controls['email'].setValue(x.trim())
}else{
console.log('does not contain spaces')
}
})
I'm trying to replace multiple occurrences of a string and nothing seems to be working for me. In my browser or even when testing online. Where am I going wrong?
str = '[{name}] is happy today as data-name="[{name}]" won the match today. [{name}] made 100 runs.';
str = str.replace('/[{name}]/gi','John');
console.log(str);
http://jsfiddle.net/SXTd4/
I got that example from here, and that too wont work.
You must not quote regexes, the correct notation would be:
str = str.replace(/\[{name}\]/gi,'John');
Also, you have to escape the [], because otherwise the content inside is treated as character class.
Updating your fiddle accordingly makes it work.
There are two ways declaring regexes:
// literal notation - the preferred option
var re = /regex here/;
// via constructor
var re = new Regexp('regex here');
You should not put your regex in quotes and you need to escape []
Simply use
str = str.replace(/\[{name}\]/gi,'John');
DEMO
While there are plenty of regex answers here is another way:
str = str.split('[{name}]').join('John');
The characters [ ] { } should be escaped in your regular expression.
I have a project, in which some JavaScript var is evaluated. Because the string needs to be escaped (single quotes only), I have written the exact same code in a test function. I have the following bit of pretty simple JavaScript code:
function testEscape() {
var strResult = "";
var strInputString = "fsdsd'4565sd";
// Here, the string needs to be escaped for single quotes for the eval
// to work as is. The following does NOT work! Help!
strInputString.replace(/'/g, "''");
var strTest = "strResult = '" + strInputString + "';";
eval(strTest);
alert(strResult);
}
And I want to alert it, saying: fsdsd'4565sd.
The thing is that .replace() does not modify the string itself, so you should write something like:
strInputString = strInputString.replace(...
It also seems like you're not doing character escaping correctly. The following worked for me:
strInputString = strInputString.replace(/'/g, "\\'");
Best to use JSON.stringify() to cover all your bases, like backslashes and other special characters. Here's your original function with that in place instead of modifying strInputString:
function testEscape() {
var strResult = "";
var strInputString = "fsdsd'4565sd";
var strTest = "strResult = " + JSON.stringify(strInputString) + ";";
eval(strTest);
alert(strResult);
}
(This way your strInputString could be something like \\\'\"'"''\\abc'\ and it will still work fine.)
Note that it adds its own surrounding double-quotes, so you don't need to include single quotes anymore.
I agree that this var formattedString = string.replace(/'/g, "\\'"); works very well, but since I used this part of code in PHP with the framework Prado (you can register the js script in a PHP class) I needed this sample working inside double quotes.
The solution that worked for me is that you need to put three \ and escape the double quotes.
"var string = \"l'avancement\";
var formattedString = string.replace(/'/g, \"\\\'\");"
I answer that question since I had trouble finding that three \ was the work around.
Only this worked for me:
searchKeyword.replace(/'/g, "\\\'");//searchKeyword contains "d'av"
So, the result variable will contain "d\'av".
I don't know why with the RegEx didn't work, maybe because of the JS framework that I'm using (Backbone.js)
That worked for me.
string address=senderAddress.Replace("'", "\\'");
There are two ways to escaping the single quote in JavaScript.
1- Use double-quote or backticks to enclose the string.
Example: "fsdsd'4565sd" or `fsdsd'4565sd`.
2- Use backslash before any special character, In our case is the single quote
Example:strInputString = strInputString.replace(/ ' /g, " \\' ");
Note: use a double backslash.
Both methods work for me.
var str ="fsdsd'4565sd";
str.replace(/'/g,"'")
This worked for me. Kindly try this
The regular expression in the following code also handles the possibility of escaped single quotes in the string - it will only prepend backslashes to single quotes that are not already escaped:
strInputString = strInputString.replace(/(?<!\\)'/g, "\\'");
Demo: https://regex101.com/r/L1lF7J/1
Compatibility
The regex above uses negative lookbehind, which is widely supported but if using an older Javascript version, this clunkier regex (which uses a capturing group backreference instead) will also do the job:
strInputString = strInputString.replace(/(^|[^\\])'/g, "$1\\'");
Demo: https://regex101.com/r/9niyYw/1
strInputString = strInputString.replace(/'/g, "''");
I've a string done like this: "http://something.org/dom/My_happy_dog_%28is%29cool!"
How can I remove all the initial domain, the multiple underscore and the percentage stuff?
For now I'm just doing some multiple replace, like
str = str.replace("http://something.org/dom/","");
str = str.replace("_%28"," ");
and go on, but it's really ugly.. any help?
Thanks!
EDIT:
the exact input would be "My happy dog is cool!" so I would like to get rid of the initial address and remove the underscores and percentage and put the spaces in the right place!
The problem is that trying to put a regex on Chrome "something goes wrong". Is it a problem of Chrome or my regex?
I'd suggest:
var str = "http://something.org/dom/My_happy_dog_%28is%29cool!";
str.substring(str.lastIndexOf('/')+1).replace(/(_)|(%\d{2,})/g,' ');
JS Fiddle demo.
The reason I took this approach is that RegEx is fairly expensive, and is often tricky to fine tune to the point where edge-cases become less troublesome; so I opted to use simple string manipulation to reduce the RegEx work.
Effectively the above creates a substring of the given str variable, from the index point of the lastIndexOf('/') (which does exactly what you'd expect) and adding 1 to that so the substring is from the point after the / not before it.
The regex: (_) matches the underscores, the | just serves as an or operator and the (%\d{2,}) serves to match digit characters that occur twice in succession and follow a % sign.
The parentheses surrounding each part of the regex around the |, serve to identify matching groups, which are used to identify what parts should be replaced by the ' ' (single-space) string in the second of the arguments passed to replace().
References:
lastIndexOf().
replace().
substring().
You can use unescape to decode the percentages:
str = unescape("http://something.org/dom/My_happy_dog_%28is%29cool!")
str = str.replace("http://something.org/dom/","");
Maybe you could use a regular expression to pull out what you need, rather than getting rid of what you don't want. What is it you are trying to keep?
You can also chain them together as in:
str.replace("http://something.org/dom/", "").replace("something else", "");
You haven't defined the problem very exactly. To get rid of all stretches of characters ending in %<digit><digit> you'd say
var re = /.*%\d\d/g;
var str = str.replace(re, "");
ok, if you want to replace all that stuff I think that you would need something like this:
/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g
test
var string = "http://something.org/dom/My_happy_dog_%28is%29cool!";
string = string.replace(/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g,"");