How do I wait for an exec process to finish in Jest?
I have the following test, but I can't seem to get Jest to wait for my exec call to finish running:
var exec = require('child_process').exec
test('render', async () =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out)
expect(...some file to be created)
);
)
What should I do to make jest wait for the exec callback to be called?
node.js jestjs
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
add a comment |
I have the following test, but I can't seem to get Jest to wait for my exec call to finish running:
var exec = require('child_process').exec
test('render', async () =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out)
expect(...some file to be created)
);
)
What should I do to make jest wait for the exec callback to be called?
node.js jestjs
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
add a comment |
I have the following test, but I can't seem to get Jest to wait for my exec call to finish running:
var exec = require('child_process').exec
test('render', async () =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out)
expect(...some file to be created)
);
)
What should I do to make jest wait for the exec callback to be called?
node.js jestjs
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
I have the following test, but I can't seem to get Jest to wait for my exec call to finish running:
var exec = require('child_process').exec
test('render', async () =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out)
expect(...some file to be created)
);
)
What should I do to make jest wait for the exec callback to be called?
node.js jestjs
node.js jestjs
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
edited Nov 12 '18 at 20:05
skyboyer
3,33611128
3,33611128
asked Nov 12 '18 at 19:15
Toli
2,05141935
asked Nov 12 '18 at 19:15
Toli
2,05141935
asked Nov 12 '18 at 19:15
Toli
2,05141935
2,05141935
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You need to call done() after your assertion.
test('render', async (done) =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out);
expect(...some file to be created);
done();
);
)
Which will mark the test to complete.
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
add a comment |
Another possibility I found that works with async
const util = require('util');
const exec = util.promisify(require('child_process').exec);
test('render', async () =>
const stdout, stderr = await exec('./render.local.sh');
console.log(stdout, stderr);
)
answered Nov 12 '18 at 19:48
Toli
2,05141935
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to call done() after your assertion.
test('render', async (done) =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out);
expect(...some file to be created);
done();
);
)
Which will mark the test to complete.
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
add a comment |
You need to call done() after your assertion.
test('render', async (done) =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out);
expect(...some file to be created);
done();
);
)
Which will mark the test to complete.
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
add a comment |
You need to call done() after your assertion.
test('render', async (done) =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out);
expect(...some file to be created);
done();
);
)
Which will mark the test to complete.
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
You need to call done() after your assertion.
test('render', async (done) =>
await exec('./render.local.sh', (err, out) =>
console.log(err, out);
expect(...some file to be created);
done();
);
)
Which will mark the test to complete.
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
answered Nov 12 '18 at 19:40
Arun Kumar M S
477
477
add a comment |
add a comment |
Another possibility I found that works with async
const util = require('util');
const exec = util.promisify(require('child_process').exec);
test('render', async () =>
const stdout, stderr = await exec('./render.local.sh');
console.log(stdout, stderr);
)
answered Nov 12 '18 at 19:48
Toli
2,05141935
add a comment |
Another possibility I found that works with async
const util = require('util');
const exec = util.promisify(require('child_process').exec);
test('render', async () =>
const stdout, stderr = await exec('./render.local.sh');
console.log(stdout, stderr);
)
answered Nov 12 '18 at 19:48
Toli
2,05141935
add a comment |
Another possibility I found that works with async
const util = require('util');
const exec = util.promisify(require('child_process').exec);
test('render', async () =>
const stdout, stderr = await exec('./render.local.sh');
console.log(stdout, stderr);
)
answered Nov 12 '18 at 19:48
Toli
2,05141935
Another possibility I found that works with async
const util = require('util');
const exec = util.promisify(require('child_process').exec);
test('render', async () =>
const stdout, stderr = await exec('./render.local.sh');
console.log(stdout, stderr);
)
answered Nov 12 '18 at 19:48
Toli
2,05141935
answered Nov 12 '18 at 19:48
Toli
2,05141935
answered Nov 12 '18 at 19:48
Toli
2,05141935
answered Nov 12 '18 at 19:48
Toli
2,05141935
2,05141935
add a comment |
add a comment |
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