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What are the differences between JSON and JavaScript object? [duplicate]
...or example:
var o = { if: "foo" }; // SyntaxError in ES3
While, using a string literal as a property name (quoting the property name) gives no problems:
var o = { "if": "foo" };
So for "compatibility" (and easy eval'ing maybe?) the quotes are mandatory.
The data types in JSON are also restri...
How in node to split string by newline ('\n')?
How in node to split string by newline ('\n') ?
I have simple string like var a = "test.js\nagain.js" and I need to get ["test.js", "again.js"] .
I tried
...
prevent property from being serialized in web API
...ialization:
public class Foo
{
public int Id { get; set; }
public string Name { get; set; }
[JsonIgnore]
public List<Something> Somethings { get; set; }
}
But, this way does not support XML format. So, in case your application has to support XML format more (or only support...
Why does the PHP json_encode function convert UTF-8 strings to hexadecimal entities?
...ll get this error: "Warning: json_encode() expects parameter 2 to be long, string given in ...". See CertaiN's answer below for 5.3 solution.
– Octavian Naicu
Oct 1 '14 at 10:03
...
Seeking useful Eclipse Java code templates [closed]
...
try {
in = new BufferedReader(new FileReader(${fileName}));
String line;
while ((line = in.readLine()) != null) {
${process}
}
}
catch (FileNotFoundException e) {
logger.error(e) ;
}
catch (IOException e) {
logger.error(e) ;
} finally { ...
How do I convert a org.w3c.dom.Document object to a String?
I want to convert a org.w3c.dom.Document object to a String. I'm using Java 6 and am open to using any (completely free) technology that is up to the task. I tried the solution from this thread -- Is there a more elegant way to convert an XML Document to a String in Java than this code? , where t...
How to maintain a Unique List in Java?
...used, which one shall I use? I just need to have maintain a list of unique strings. So basically even when an existing string is added it should actually get added.
– user1804287
Nov 6 '12 at 21:25
...
JSON.net: how to deserialize without using the default constructor?
...hat you want Json.Net to call.
[JsonConstructor]
public Result(int? code, string format, Dictionary<string, string> details = null)
{
...
}
It is important that the constructor parameter names match the corresponding property names of the JSON object (ignoring case) for this to work cor...
Different names of JSON property during serialization and deserialization
... ObjectMapper();
System.out.println("Serialization: " + mapper.writeValueAsString(c));
Coordinates r = mapper.readValue("{\"red\":25}",Coordinates.class);
System.out.println("Deserialization: " + r.getR());
Result:
Serialization: {"r":5}
Deserialization: 25
...
if arguments is equal to this string, define a variable like this string
... I ran into issues with this when the variable on the left was an empty string. Fix was if [ "$1" = "country" ]; then.
– andrewb
Mar 20 '15 at 1:42
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