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Convert string to json object giving error

I have a string which needs to be converted to JSONObject, I added the dependency, but I'm getting error, which I'm not able to figure out. I have the following dependency:
<!-- https://mvnrepository.com/artifact/org.json/json -->
<dependency>
<groupId>org.json</groupId>
<artifactId>json</artifactId>
<version>20220924</version>
</dependency>
String s ="{name=Alex, sex=male}";
JSONObject obj = new JSONObject(s);
System.out.println(obj.get("name"));
I'm getting an exception:
org.json.JSONException: Expected a ':' after a key at line 5

The JSON you've provided is not valid because separator colon : should be used a separator between a Key and a Value (not an equals sign).
A quote from the standard The JavaScript Object Notation (JSON) Data Interchange Format:
4. Objects
An object structure is represented as a pair of curly brackets
surrounding zero or more name/value pairs (or members). A name is a
string. A single colon comes after each name, separating the name
from the value. A single comma separates a value from a following
name. The names within an object SHOULD be unique.
You can preprocess you JSON before parsing it using String.replace()
String s ="{name=Alex, sex=male}";
JSONObject obj = new JSONObject(s.replace('=', ':'));
System.out.println(obj.get("name"));
Output:
Alex
Also, note that Org.json (as well as some other libraries like Gson) would take care of fixing missing double quotes ". But it would not be the case with libraries like Jackson.

The string you're assigning to the variable s is not valid JSON. Property names and properties should be separated by : instead of =, and double quotes should be used around strings and property names.
So the string in your example should be like this (with \ being used to escape the quote characters within the string quotes):
String s = "{\"name\":\"Alex\",\"sex\":\"male\"}";

You should use : instead of =
String s = """{"name":"Alex","sex":"male"}"""; (Since Java 13 preview feature)

read unique char: 'あ' from json file in java

I am reading a JSON file in Java using this code:
String data = Files.readFile(jsonFile)
.trim()
.replaceAll("[^\\x00-\\x7F]", "")
.replaceAll("[\\p{Cntrl}&&[^\r\n\t]]", "")
.replaceAll("\\p{C}", "");
In my JSON file, there is a unique char: 'あ' (12354) that is interpreted to: "" (nothing) when reading the file.
How can I make this char show up in my variable "data"?
Due to answers I've got, I understand that the data is cleaned from high ASCII characters by adding replaceAll("[^\\x00-\\x7F]", ""). But what can I do if I want all high ASCII characters to be cleaned except this one 'あ'?

The character you want is the unicode character HIRAGANA LETTER A and has code U+3042.
You can simply add it to the list of valid characters:
...
.replaceAll("[^\\x00-\\x7F\\u3042]", "")
...

Java regex lookbehind

I want to match a string that has "json" (occurs more than 2 times) and without string "from" between two "json".
For example(what I want the string match or not):
select json,json from XXX -> Yes
select json from json XXXX -> No
select json,XXXX,json from json XXX -> Yes
Why the third is matching because I just want two "json" string occurs without "from" inside between it.
After learning regex lookbehind, I'm write the regex like this:
select.*json.*?(?<!from)json.*from.*
I'm using regex lookbehind to except the from string.
But after test, I find this regex match the string "select get_json_object from get_json_object" too.
What wrong for my regex? Any suggestion is appreciated.

You need to use tempered greedy token for achieving this. Use this regex,
\bjson\b(?:(?!\bfrom\b).)+\bjson\b
This expression (?:(?!\bfrom\b).)+ will match any text that does not contain from as a whole word inside it.
Regex Demo
For matching the whole line, you can use,
^.*\bjson\b(?:(?!\bfrom\b).)+\bjson\b.*$
Like you wanted in your post, this regex will match the line as long as it finds a string where a from does not appear between two jsons
Regex Demo with full line match
Edit:
Why OP's regex select.*json.*?(?<!from)json.*from.* didn't work as expected
Your regex starts matching with select and then .* matches as much as possible, while making sure it finds json ahead followed by some optional characters and then again expects to find a json string then .* matches again some characters then expects to find a from and finally using .* zero or more optional characters.
Let's take an example string that should match.
select json from json json XXXX
It has two json string without from in between so it should match but it doesn't, because in your regex, the order or presence of json and from is fixed which is json then again json then from which is not the case in this string.
Here is a Java code demo
List<String> list = Arrays.asList("select json,json from XXX","select json from json XXXX","select json,json from json XXX","select json from json json XXXX");
list.forEach(x -> {
System.out.println(x + " --> " + x.matches(".*\\bjson\\b(?:(?!\\bfrom\\b).)+\\bjson\\b.*"));
});
Prints,
select json,json from XXX --> true
select json from json XXXX --> false
select json,json from json XXX --> true
select json from json json XXXX --> true

Escape XML Characters for Attribute values Java

I have an XML represented in String. I need to replace all the special characters in the Attribute values with the Escape Characters.
For Ex:
I want to convert 1st one to the second one as following.
<r1 c1=\"01\" c168=\"<A_ATTR><Updates A_VALUE="959" /><Current A_VALUE="100" /></A_ATTR>\"/>
<r1 c1=\"01\" c168=\"<A_ATTR><Updates A_VALUE="959" /><Current A_VALUE="100" /></A_ATTR>\"/>
This questions is similar to the below one : But I need to escape the attribute values. Please advise.
Escape xml characters within nodes of string xml in java

Use string replace function to replace the required character by the encoding. Example below
if your xml string is s then
s = s.replace("<", "<");
s = s.replace(">", ">");

Insert " in correct form in a String

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";