javascript - Try catch ex.stack get first line only - Stack Overflow

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How do I get the first line only of a console.log(ex.stack)>

For example I only want this:

TypeError: Object #<Object> has no method 'debug'

Out of this:

TypeError: Object #<Object> has no method 'debug'
    at eval at <anonymous> (unknown source)
    at eval (native)
    at Object._evaluateOn (unknown source)
    at Object._evaluateAndWrap (unknown source)
    at Object.evaluate (unknown source)

How do I get the first line only of a console.log(ex.stack)>

For example I only want this:

TypeError: Object #<Object> has no method 'debug'

Out of this:

TypeError: Object #<Object> has no method 'debug'
    at eval at <anonymous> (unknown source)
    at eval (native)
    at Object._evaluateOn (unknown source)
    at Object._evaluateAndWrap (unknown source)
    at Object.evaluate (unknown source)

• javascript
• google-chrome-devtools
• console.log

Share Improve this question Follow asked Jun 14, 2016 at 15:18 HarryHarry 55k7676 gold badges186186 silver badges270270 bronze badges

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1 Answer 1

Sorted by: Reset to default 10

If you want the error message, just grab it directly. There's no need to parse it out from the full stack trace:

var Object = {};
try {
  Object.debug();
} catch(ex) {
  console.log(ex.message);
}

If that's not possible for whatever the reason, the stack trace appears to be nothing but a string:

console.log(typeof ex.stack);

string

... so pick your favourite string manipulation technique:

var Object = {};
try {
  Object.debug();
} catch(ex) {
  console.log(ex.stack.split("\n", 1).join(""));
}

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