问题
Is it possible to get the raw version of a template string in iojs ?
var s = `foo${1+1}bar`
console.log(s); // foo2bar
In the previous example I would like to get the string: foo${1+1}bar
edit1: My need is to detect whether a template string depends on its context of if is is just a 'constant' string that may contain CR and LF
回答1:
Is it possible to get the raw version of a template string in iojs ?
No it is not. It's not possible to get the raw representation of the literal, just like there is no way to get the "raw" literal in these cases:
var foo = {[1+1]: 42};
var bar = 1e10;
var baz = "\"42\"";
Note that the term "template string" is misleading (as it may indicate that you could somehow get the raw value of the string (which is also not the case as shown above)). The correct term is "template literal".
My need is to detect whether a template string depends on its context of if is is just a 'constant' string that may contain CR and LF
Seems like a job for a static analysis tool. E.g. you can use recast to parse the source code and traverse all template literals.
For example, the AST representation of `foo${1+1}bar` is:
If such an AST node as an empty expression property, then you know that the value is constant.
There is a way to determine whether a template literal is "static" or "dynamic" at runtime, but that involves changing the behavior of the code.
You can use tagged templates. Tagged templates are functions that get passed the static and dynamic portions of a template literal.
Example:
function foo(template, ...expressions) {
console.log(template, expressions);
}
foo`foo${1+1}bar` // logs (["foo", "bar"], [2]) but returns `undefined`
I.e. if foo gets passed only a single argument, the template literal does not contain expressions. However, foo would also have to interpolate the static parts with the dynamic parts and return the result (not shown in the above example).
来源：https://stackoverflow.com/questions/31293928/how-to-get-the-raw-version-of-a-template-string-in-iojs