blocks|key|1367834|text|在students数组中，每个项都是一个数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1367835|​|1367836|const+students+=+[
++['Toto',+4,+17],
++['Titi',+11,+12],
++['Tata',+12,+14]
];


for+(let+i+=+0;+i+<+students.length;+i%2B%2B)+{
++const+student+=+students[i];
++let+result+=+0;
++for+(let+j+=+0;+j+<+student.length;+j%2B%2B)+{
++++if+(typeof(student[j])+===+"number")+{
++++++result+%2B=+student[j];
++}
++
++}
++console.log("Student+"+%2B+students[i][0]+%2B+"+has+the+result+"+%2B+result);
}|code-block|syntax|javascript|1367837|1367838|entityMap^0|1|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

<p>Inside your <code>students</code> array, every item is an array.</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const students = [
  ['Toto', 4, 17],
  ['Titi', 11, 12],
  ['Tata', 12, 14]
];


for (let i = 0; i &lt; students.length; i++) {
  const student = students[i];
  let result = 0;
  for (let j = 0; j &lt; student.length; j++) {
    if (typeof(student[j]) === "number") {
      result += student[j];
  }
  
  }
  console.log("Student " + students[i][0] + " has the result " + result);
}</code></pre>
</div>
</div>
</p>


blocks|key|818344|text|您只需在这里使用与Array.prototype.reduce一起的析构：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|818345|​|818346|const+students+=+[['Toto',+4,+17],+['Titi',+11,+12],+['Tata',+12,+14]];

for+(const+student+of+students)+{
++const+[+name,+...summands+]+=+student;
++console.log(`${name}+scored+${summands.reduce((acc,val)=>acc%2Bval,0)}`)
}|code-block|syntax|javascript|818347|818348|对于更简洁的版本，您可以在For+-definition中直接重构：|818349|818350|const+students+=+[['Toto',+4,+17],+['Titi',+11,+12],+['Tata',+12,+14]];

for+(const+[+name,+...summands+]+of+students)+{
++console.log(`${name}+scored+${summands.reduce((acc,val)=>acc%2Bval,0)}`)
}|818351|818352|这也适用于长度不同的总结表：|818353|818354|const+students+=+[['Toto'],+['Titi',+1,2,3,4,5,6],+['Tata',+12,+14]];

for+(const+[+name,+...summands+]+of+students)+{
++console.log(`${name}+scored+${summands.reduce((acc,val)=>acc%2Bval,0)}`)
}|818355|818356|entityMap^0|9|M|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|17|8|@]|D|@]|E|$]]|$1|P|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|S|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1B|8|@]|D|@]|E|$]]|$1|V|3|G|5|6|7|1C|8|@]|D|@]|E|$]]|$1|W|3|X|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Y|3|G|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|10|$]]

<p>You can simply use destructuring alongside <code>Array.prototype.reduce</code> here:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const students = [['Toto', 4, 17], ['Titi', 11, 12], ['Tata', 12, 14]];

for (const student of students) {
  const [ name, ...summands ] = student;
  console.log(`${name} scored ${summands.reduce((acc,val)=&gt;acc+val,0)}`)
}</code></pre>
</div>
</div>
</p>
<p>For an even more concise version, you can destructure right in the for-definition:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const students = [['Toto', 4, 17], ['Titi', 11, 12], ['Tata', 12, 14]];

for (const [ name, ...summands ] of students) {
  console.log(`${name} scored ${summands.reduce((acc,val)=&gt;acc+val,0)}`)
}</code></pre>
</div>
</div>
</p>
<p>This will also work for lists of summands varying in length:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const students = [['Toto'], ['Titi', 1,2,3,4,5,6], ['Tata', 12, 14]];

for (const [ name, ...summands ] of students) {
  console.log(`${name} scored ${summands.reduce((acc,val)=&gt;acc+val,0)}`)
}</code></pre>
</div>
</div>
</p>


blocks|key|2213309|text|你的循环迭代学生，而不是他们的成绩:你需要两个循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2213310|此外，您的日志记录知道每个学生都由一个数组表示，该数组的第一个元素是学生名称的字符串，因此，当您添加循环来迭代学生成绩时，跳过第一个条目：|2213311|for+(var+i+=+0;+i+<+students.length;+i%2B%2B)+{
++var+result+=+0;
++for+(var+j+=+1;+j+<+students[i].length;+j%2B%2B)+{
++++result+%2B=+students[i][j];
++}
++console.log("Student+"+%2B+students[i][0]+%2B+"+has+the+result+"+%2B+result);
++}
}|code-block|syntax|javascript|2213312|注意，我也将您的result移动到每个学生计数，您正在为所有学生计数。|offset|length|style|CODE|2213313|还有其他方法可以做到这一点，但这可能与自动取款机无关。|2213314|entityMap^0|0|0|0|8|6|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<p>Your loop iterates over the students, not their grades: you need two loops.</p>
<p>Furthermore, your logging knows that each student is represented by an array whose first element is a string of the student name, so when you add the loop to iterate over the student grades, skip the first entry:</p>
<pre><code>for (var i = 0; i &lt; students.length; i++) {
  var result = 0;
  for (var j = 1; j &lt; students[i].length; j++) {
    result += students[i][j];
  }
  console.log(&quot;Student &quot; + students[i][0] + &quot; has the result &quot; + result);
  }
}
</code></pre>
<p>Note I've also moved your <code>result</code> to count per student, you were counting for all students.</p>
<p>There are other ways to do this, but that might not be relevant atm.</p>


blocks|key|1367988|text|你可以运行这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1367989|​|1367990|var+students+=+[
++['Toto',+4,+17],
++['Titi',+11,+12],
++['Tata',+12,+14]
];



for+(var+i+=+0;+i+<+students.length;+i%2B%2B)+{
++var+result+=+0;
++for+(var+j+=+0;+j+<=+students[i].length;+j%2B%2B)+{
++++if+(typeof(students[i][j])+===+"number")+{
++++++result+%2B=+students[i][j];
++++}
++}

++console.log("Student+"+%2B+students[i][0]+%2B+"+has+the+result+"+%2B+result);
}|code-block|syntax|javascript|1367991|1367992|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>you can run this</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var students = [
  ['Toto', 4, 17],
  ['Titi', 11, 12],
  ['Tata', 12, 14]
];



for (var i = 0; i &lt; students.length; i++) {
  var result = 0;
  for (var j = 0; j &lt;= students[i].length; j++) {
    if (typeof(students[i][j]) === "number") {
      result += students[i][j];
    }
  }

  console.log("Student " + students[i][0] + " has the result " + result);
}</code></pre>
</div>
</div>
</p>


blocks|key|1368007|text|您可以使用for循环执行此操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1368008|​|1368009|var+students+=+[
++['Toto',+4,+17],
++['Titi',+11,+12],
++['Tata',+12,+14]
];

var+result+=+0;

for+(var+i+=+0;+i+<+students.length;+i%2B%2B)+{
++console.log("Student+"+%2B+students[i][0]+%2B+"+has+the+result+"+%2B+(students[i][1]+%2B+students[i][2]));
}|code-block|syntax|javascript|1368010|1368011|或者用for..of和Array.prototype.reduce()|offset|length|style|CODE|1368012|1368013|var+students+=+[
++++['Toto',+4,+17],
++++['Titi',+11,+12],
++++['Tata',+12,+14]
];

for(let+[name,+...values]+of+students)+{
+const+sum+=+values.reduce((acc,+curr)+=>+acc+%2B=+curr,+0);
+console.log("Student+"+%2B+name+%2B+"+has+the+result+"+%2B+sum);
}|1368014|1368015|entityMap^0|0|0|0|0|3|7|B|O|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Z|8|@$L|10|M|11|N|O]|$L|12|M|13|N|O]]|9|@]|A|$]]|$1|P|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|S|3|C|5|6|7|16|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|U|$]]

<p>You can do this with for loop</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var students = [
  ['Toto', 4, 17],
  ['Titi', 11, 12],
  ['Tata', 12, 14]
];

var result = 0;

for (var i = 0; i &lt; students.length; i++) {
  console.log("Student " + students[i][0] + " has the result " + (students[i][1] + students[i][2]));
}</code></pre>
</div>
</div>
</p>
<p>or this with <code>for..of</code> and <code>Array.prototype.reduce()</code></p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var students = [
    ['Toto', 4, 17],
    ['Titi', 11, 12],
    ['Tata', 12, 14]
];

for(let [name, ...values] of students) {
 const sum = values.reduce((acc, curr) =&gt; acc += curr, 0);
 console.log("Student " + name + " has the result " + sum);
}</code></pre>
</div>
</div>
</p>


blocks|key|2213349|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2213350|+var+students+=+[
++++++++++++['Toto',+4,+17],
++++++++++++['Titi',+11,+12],
++++++++++++['Tata',+12,+14]
++++++++];

++++++++let+result+=+0;
++++++++students.forEach(item+=>+{
++++++++++++item.forEach(realItem+=>+{
++++++++++++++++if+(typeof+realItem+===+'number'){
++++++++++++++++++++result+%2B=+realItem
++++++++++++++++}
++++++++++++})
++++++++})
++++++++console.log(result)|code-block|syntax|javascript|2213351|2213352|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code> var students = [
            ['Toto', 4, 17],
            ['Titi', 11, 12],
            ['Tata', 12, 14]
        ];

        let result = 0;
        students.forEach(item =&gt; {
            item.forEach(realItem =&gt; {
                if (typeof realItem === 'number'){
                    result += realItem
                }
            })
        })
        console.log(result)</code></pre>
</div>
</div>
</p>


<p>I have an array with 3 indices for each student:</p>
<pre><code>var students = [
    ['Toto', 4, 17],
    ['Titi', 11, 12],
    ['Tata', 12, 14]
];
</code></pre>
<p>I have to calculate the sum for each student. Except that, I must avoid the string.</p>
<p>My result obtained for the student <code>Toto</code> is <code>0</code> instead of <code>21</code>.</p>
<p>I don't understand where the problem?</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var students = [
  ['Toto', 4, 17],
  ['Titi', 11, 12],
  ['Tata', 12, 14]
];

var result = 0;

for (var i = 0; i &lt; students.length; i++) {
  if (typeof(students[i]) === "number") {
    result += students[i];
  }
  console.log("Student " + students[i][0] + " has the result " + result);
}</code></pre>
</div>
</div>
</p>


Calculate the sum of an array by removing a string?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个数组，每个学生有3个索引：var students = [    ['Toto', 4, 17],    ['Titi', 11, 12],    ['Tata', 12, 14]];我必须计算每个学生的总数。除此之外，我必须避开绳子。我为学生Toto获得的结果是0而不是21。我不明白问题出在哪里？​var s...

问通过删除字符串计算数组的和？
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问通过删除字符串计算数组的和？EN