return false or true while iterating an array of object

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How to short circuit Array.forEach like calling break? 24 answers

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如果在回调中返回真值，可以使用Array#some停止迭代的Array#some 。

var array = [{ name: 'paul', age:20 }, { name: 'john', age:30 }, { name: 'albert', age:40 }],
    find = function(array, name) {
        return array.some(function(object) {
            return object.name === name;
        });
    };

console.log(find(array, 'paul'));  // true
console.log(find(array, 'maria')); // false

---

You are returning in the forEach(function(i) {}) , which is only returning in the inside function function(i) {} , that does not help return from the outer function find() . Also, your logic with return false; seems also problematic. Simply use normal for loops would be fine.

var array = [
    {name:'paul',age:20},
    {name:'john',age:30},
    {name:'albert',age:40}
];

var find = function(arr, name) {
  for (let i of arr) {
    if(i.name === name){
      console.log('found ' + name);
      return true;
    }
  }
  console.log(name + ' not there');
  return false;
};

find(array, 'paul');
find(array, 'maria');

---

停止使用forEach方法，并尝试使用它：

var array = [
    {name:'paul',age:20},
    {name:'john',age:30},
    {name:'albert',age:40}
];

var find = function(arr){
	var returnValue = false;
    
    for (var i = 0; i <= arr.length; i++) {
      if(arr[i].name === 'john') {
      	ret = true;
      	break;
      }
    }

    return returnValue;
};
find(array);

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