How to display the current state of an object in console.log?
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P粉868586032 2023-08-22 13:37:22
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<p>In Safari without any add-ons (and indeed in most other browsers), <code>console.log</code> will show the final state of the object during execution, not <code> ;console.log</code>The status when called. </p> <p>I have to clone the object in order to output it via <code>console.log</code> to get the status of the object at that line of code. </p> <p><strong>Example:</strong></p> <pre class="brush:php;toolbar:false;">var test = {a: true} console.log(test); // {a: false} test.a = false; console.log(test); // {a: false}</pre> <p><br /></p>
P粉868586032
P粉868586032

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P粉473363527

If I want to see its status at the time of recording, I usually convert it to a JSON string.

console.log(JSON.stringify(a));
P粉464113078

I think you are looking for console.dir().

console.log() cannot achieve the function you want, because it prints a reference to the object, and it has changed when you open it. console.dirThe attribute directory of the object will be printed when called.

The JSON idea below is a good one; you could even go ahead and parse the JSON string and get a browsable object, like .dir() would give you:

console.log(JSON.parse(JSON.stringify(obj)));

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