blocks|key|612173|text|当您执行.splice()时，数组将被重新索引，这意味着当一个索引被删除时，您将跳过一个索引，并且缓存的.length已过时。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|612174|要修复它，您需要在.splice()之后递减i，或者简单地反向迭代……|612175|var+i+=+Auction.auctions.length
while+(i--)+{
++++...
++++if+(...)+{+
++++++++Auction.auctions.splice(i,+1);
++++}+
}|code-block|syntax|javascript|612176|这样，重新索引不会影响迭代中的下一项，因为索引只影响从当前点到数组末尾的项，并且迭代中的下一项低于当前点。|612177|entityMap^0|4|9|1G|7|0|9|9|M|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

<p>The array is being re-indexed when you do a <code>.splice()</code>, which means you'll skip over an index when one is removed, and your cached <code>.length</code> is obsolete.</p>

<p>To fix it, you'd either need to decrement <code>i</code> after a <code>.splice()</code>, or simply iterate in reverse...</p>

<pre><code>var i = Auction.auctions.length
while (i--) {
    ...
    if (...) { 
        Auction.auctions.splice(i, 1);
    } 
}
</code></pre>

<p>This way the re-indexing doesn't affect the next item in the iteration, since the indexing affects only the items from the current point to the end of the Array, and the next item in the iteration is lower than the current point.</p>


blocks|key|612216|text|这是一个非常常见的问题。解决方案是向后循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|612217|for+(var+i+=+Auction.auctions.length+-+1;+i+>=+0;+i--)+{
++++Auction.auctions[i].seconds--;
++++if+(Auction.auctions[i].seconds+<+0)+{+
++++++++Auction.auctions.splice(i,+1);
++++}
}|code-block|syntax|javascript|612218|如果你从最后弹出它们，这并不重要，因为当你向后返回时，索引将被保留。|612219|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<p>This is a pretty common issue. The solution is to loop backwards:</p>

<pre><code>for (var i = Auction.auctions.length - 1; i &gt;= 0; i--) {
    Auction.auctions[i].seconds--;
    if (Auction.auctions[i].seconds &lt; 0) { 
        Auction.auctions.splice(i, 1);
    }
}
</code></pre>

<p>It doesn't matter if you're popping them off of the end because the indices will be preserved as you go backwards.</p>


blocks|key|612064|text|每次循环时都要重新计算长度，而不是一开始就重新计算，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|612065|for+(i+=+0;+i+<+Auction.auctions.length;+i%2B%2B)+{
++++++auction+=+Auction.auctions[i];
++++++Auction.auctions[i]['seconds']+--;
++++++if+(auction.seconds+<+0)+{+
++++++++++Auction.auctions.splice(i,+1);
++++++++++i--;+//decrement
++++++}
}|code-block|syntax|javascript|612066|这样你就不会越界了。|612067|编辑:在if语句中添加了减量。|612068|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<p>Recalculate the length each time through the loop instead of just at the outset, e.g.:</p>

<pre><code>for (i = 0; i &lt; Auction.auctions.length; i++) {
      auction = Auction.auctions[i];
      Auction.auctions[i]['seconds'] --;
      if (auction.seconds &lt; 0) { 
          Auction.auctions.splice(i, 1);
          i--; //decrement
      }
}
</code></pre>

<p>That way you won't exceed the bounds.</p>

<p>EDIT: added a decrement in the if statement.</p>


blocks|key|612181|text|Auction.auctions+=+Auction.auctions.filter(function(el)+{
++return+--el["seconds"]+>+0;
});|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|612182|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Auction.auctions = Auction.auctions.filter(function(el) {
  return --el["seconds"] &gt; 0;
});
</code></pre>


blocks|key|396444|text|如果你正在使用ES6%2B+-为什么不直接使用Array.filter方法呢？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|396445|Auction.auctions+=+Auction.auctions.filter((auction)+=>+{
++auction['seconds']+--;
++return+(auction.seconds+>+0)
})++|code-block|syntax|javascript|396446|请注意，在过滤器迭代期间修改数组元素仅适用于对象，而不适用于原始值的数组。|396447|entityMap^0|L|C|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

<p>If you are e using ES6+ - why not just use <code>Array.filter</code> method?</p>
<pre><code>Auction.auctions = Auction.auctions.filter((auction) =&gt; {
  auction['seconds'] --;
  return (auction.seconds &gt; 0)
})  
</code></pre>
<p>Note that modifying the array element during filter iteration only works for objects and will not work for array of primitive values.</p>


blocks|key|396528|text|这里是这个简单线性时间问题的简单线性时间解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396529|当我运行这段代码时，当n=1百万时，每次调用filterInPlace()都需要花费.013到.016秒。一个二次解决方案(例如，公认的答案)将需要一百万倍的时间，或者更多。|396530|​|396531|//+Remove+from+array+every+item+such+that+!condition(item).
function+filterInPlace(array,+condition)+{
+++var+iOut+=+0;
+++for+(var+i+=+0;+i+<+array.length;+i%2B%2B)
+++++if+(condition(array[i]))
+++++++array[iOut%2B%2B]+=+array[i];
+++array.length+=+iOut;
}

//+Try+it+out.++A+quadratic+solution+would+take+a+very+long+time.
var+n+=+1*1000*1000;
console.log("constructing+array...");
var+Auction+=+{auctions:+[]};
for+(var+i+=+0;+i+<+n;+%2B%2Bi)+{
++Auction.auctions.push({seconds:1});
++Auction.auctions.push({seconds:2});
++Auction.auctions.push({seconds:0});
}
console.log("array+length+should+be+"%2B(3*n)%2B":+",+Auction.auctions.length)
filterInPlace(Auction.auctions,+function(auction)+{return+--auction.seconds+>=+0;+})
console.log("array+length+should+be+"%2B(2*n)%2B":+",+Auction.auctions.length)
filterInPlace(Auction.auctions,+function(auction)+{return+--auction.seconds+>=+0;+})
console.log("array+length+should+be+"%2Bn%2B":+",+Auction.auctions.length)
filterInPlace(Auction.auctions,+function(auction)+{return+--auction.seconds+>=+0;+})
console.log("array+length+should+be+0:+",+Auction.auctions.length)|code-block|syntax|javascript|396532|396533|请注意，这会就地修改原始数组，而不是创建新的数组；这样做会很有好处，例如，在数组是程序的单一内存瓶颈的情况下；在这种情况下，您不希望创建另一个相同大小的数组，即使是临时的。|396534|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|S|8|@]|9|@]|A|$I|J]]|$1|K|3|E|5|6|7|T|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|U|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|O|$]]

<p>Here is a simple linear time solution to this simple linear time problem.</p>
<p>When I run this snippet, with n = 1 million, each call to filterInPlace() takes .013 to .016 seconds.  A quadratic solution (e.g. the accepted answer) would take a million times that, or so.</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>// Remove from array every item such that !condition(item).
function filterInPlace(array, condition) {
   var iOut = 0;
   for (var i = 0; i &lt; array.length; i++)
     if (condition(array[i]))
       array[iOut++] = array[i];
   array.length = iOut;
}

// Try it out.  A quadratic solution would take a very long time.
var n = 1*1000*1000;
console.log("constructing array...");
var Auction = {auctions: []};
for (var i = 0; i &lt; n; ++i) {
  Auction.auctions.push({seconds:1});
  Auction.auctions.push({seconds:2});
  Auction.auctions.push({seconds:0});
}
console.log("array length should be "+(3*n)+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds &gt;= 0; })
console.log("array length should be "+(2*n)+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds &gt;= 0; })
console.log("array length should be "+n+": ", Auction.auctions.length)
filterInPlace(Auction.auctions, function(auction) {return --auction.seconds &gt;= 0; })
console.log("array length should be 0: ", Auction.auctions.length)</code></pre>
</div>
</div>
</p>
<p>Note that this modifies the original array in place rather than creating a new array; doing it in place like this can be advantageous, e.g. in the case that the array is the program's single memory bottleneck; in that case, you don't want to create another array of the same size, even temporarily.</p>


blocks|key|396341|text|另一个简单的解决方案是一次消化数组元素：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396342|while(Auction.auctions.length){
++++//+From+first+to+last...
++++var+auction+=+Auction.auctions.shift();
++++//+From+last+to+first...
++++var+auction+=+Auction.auctions.pop();

++++//+Do+stuff+with+auction
}|code-block|syntax|javascript|396343|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Another simple solution to digest an array elements once:</p>

<pre><code>while(Auction.auctions.length){
    // From first to last...
    var auction = Auction.auctions.shift();
    // From last to first...
    var auction = Auction.auctions.pop();

    // Do stuff with auction
}
</code></pre>


blocks|key|396296|text|下面是正确使用splice的另一个示例。此示例将从“array”中删除“attribute”。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396297|for+(var+i+=+array.length;+i--;)+{
++++if+(array[i]+===+'attribute')+{
++++++++array.splice(i,+1);
++++}
}|code-block|syntax|javascript|396298|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Here is another example for the proper use of splice. This example is about to remove 'attribute' from 'array'.</p>

<pre><code>for (var i = array.length; i--;) {
    if (array[i] === 'attribute') {
        array.splice(i, 1);
    }
}
</code></pre>


blocks|key|396784|text|普通的for循环对我来说更熟悉，我只需要在每次从数组中删除一项时递减索引|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396785|​|396786|//5+trues+,+5+falses
var+arr1+=+[false,+false,+true,+true,+false,+true,+false,+true,+true,+false];

//remove+falses+from+array
for+(var+i+=+0;+i+<+arr1.length;+i%2B%2B){
++++if+(arr1[i]+===+false){
++++++++arr1.splice(i,+1);
++++++++i--;//+decrement+index+if+item+is+removed
++++}
}
console.log(arr1);//+should+be+5+trues|code-block|syntax|javascript|396787|396788|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>The normal for loop is more familiar for me, I just need to decrement the index each time I remove an item from the 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>//5 trues , 5 falses
var arr1 = [false, false, true, true, false, true, false, true, true, false];

//remove falses from array
for (var i = 0; i &lt; arr1.length; i++){
    if (arr1[i] === false){
        arr1.splice(i, 1);
        i--;// decrement index if item is removed
    }
}
console.log(arr1);// should be 5 trues</code></pre>
</div>
</div>
</p>


blocks|key|396694|text|为什么要在.splice上浪费cpu周期呢？该操作必须一次又一次地遍历整个循环，以删除数组中的元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396695|为什么不在一个循环中使用传统的两个标志呢？|396696|​|396697|const+elements+=+[1,+5,+5,+3,+5,+2,+4];
const+remove+=+5

i+=+0

for(let+j+=+0;+j+<+elements.length;+j%2B%2B){
++if+(elements[j]+!==+remove)+{
++++elements[i]+=+elements[j]
++++i%2B%2B
++}
}
elements.length+=+i|code-block|syntax|javascript|396698|396699|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Q|8|@]|9|@]|A|$I|J]]|$1|K|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>why waste cpu cycles on .splice? that operation has to go through the whole loop again and again to remove an element in an array.</p>
<p>why not just use traditional 2 flags in one 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>const elements = [1, 5, 5, 3, 5, 2, 4];
const remove = 5

i = 0

for(let j = 0; j &lt; elements.length; j++){
  if (elements[j] !== remove) {
    elements[i] = elements[j]
    i++
  }
}
elements.length = i</code></pre>
</div>
</div>
</p>


blocks|key|612255|text|在循环时尝试将数组中继到newArray中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|612256|var+auctions+=+Auction.auctions;
var+auctionIndex;
var+auction;
var+newAuctions+=+[];

for+(
++auctionIndex+=+0;+
++auctionIndex+<+Auction.auctions.length;
++auctionIndex%2B%2B)+{

++auction+=+auctions[auctionIndex];

++if+(auction.seconds+>=+0)+{+
++++newAuctions.push(
++++++auction);
++}++++
}

Auction.auctions+=+newAuctions;|code-block|syntax|javascript|612257|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Try to relay an array into newArray when looping:</p>

<pre><code>var auctions = Auction.auctions;
var auctionIndex;
var auction;
var newAuctions = [];

for (
  auctionIndex = 0; 
  auctionIndex &lt; Auction.auctions.length;
  auctionIndex++) {

  auction = auctions[auctionIndex];

  if (auction.seconds &gt;= 0) { 
    newAuctions.push(
      auction);
  }    
}

Auction.auctions = newAuctions;
</code></pre>


blocks|key|396564|text|两个有效的示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396565|(Example+ONE)
//+Remove+from+Listing+the+Items+Checked+in+Checkbox+for+Delete
let+temp_products_images+=+store.state.c_products.products_images
if+(temp_products_images+!=+null)+{
++++for+(var+l+=+temp_products_images.length;+l--;)+{
++++++++//+'mark'+is+the+checkbox+field
++++++++if+(temp_products_images[l].mark+==+true)+{
++++++++++++store.state.c_products.products_images.splice(l,1);+++++++++//+THIS+WORKS
++++++++++++//+this.$delete(store.state.c_products.products_images,l);++//+THIS+ALSO+WORKS
++++++++}
++++}
}

(Example+TWO)
//+Remove+from+Listing+the+Items+Checked+in+Checkbox+for+Delete
let+temp_products_images+=+store.state.c_products.products_images
if+(temp_products_images+!=+null)+{
++++let+l+=+temp_products_images.length
++++while+(l--)
++++{
++++++++//+'mark'+is+the+checkbox+field
++++++++if+(temp_products_images[l].mark+==+true)+{
++++++++++++store.state.c_products.products_images.splice(l,1);+++++++++//+THIS+WORKS
++++++++++++//+this.$delete(store.state.c_products.products_images,l);++//+THIS+ALSO+WORKS
++++++++}
++++}
}|code-block|syntax|javascript|396566|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Two examples that work:</p>

<pre><code>(Example ONE)
// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    for (var l = temp_products_images.length; l--;) {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}

(Example TWO)
// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    let l = temp_products_images.length
    while (l--)
    {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}
</code></pre>


blocks|key|396582|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|396583|RemoveItems.forEach((i,+j)+=>+{
++++OriginalItems.splice((i+-+j),+1);
});|code-block|syntax|javascript|396584|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Give this a try</p>

<pre class="lang-js prettyprint-override"><code>RemoveItems.forEach((i, j) =&gt; {
    OriginalItems.splice((i - j), 1);
});
</code></pre>


blocks|key|612674|text|删除参数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|612675|++++++++oldJson=[{firstName:'s1',lastName:'v1'},
+++++++++++++++++{firstName:'s2',lastName:'v2'},
+++++++++++++++++{firstName:'s3',lastName:'v3'}]
++++++++
++++++++newJson+=+oldJson.map(({...ele})+=>+{
++++++++++delete+ele.firstName;
++++++++++return+ele;
++++++++++})|code-block|syntax|javascript|612676|它删除和创建新的数组，因为我们在每个对象上使用扩展操作符，所以原始数组对象也不会受到损害|612677|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<h1>Deleting Parameters</h1>
<pre><code>        oldJson=[{firstName:'s1',lastName:'v1'},
                 {firstName:'s2',lastName:'v2'},
                 {firstName:'s3',lastName:'v3'}]
        
        newJson = oldJson.map(({...ele}) =&gt; {
          delete ele.firstName;
          return ele;
          })
</code></pre>
<p>it deletes and and create new array and as we are using spread operator on each objects so the original array objects are also remains unharmed</p>


blocks|key|612352|text|for+(i+=+0,+len+=+Auction.auctions.length;+i+<+len;+i%2B%2B)+{
++++auction+=+Auction.auctions[i];
++++Auction.auctions[i]['seconds']+--;
++++if+(auction.seconds+<+0)+{
++++++++Auction.auctions.splice(i,+1);
++++++++i--;
++++++++len--;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|612353|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>for (i = 0, len = Auction.auctions.length; i &lt; len; i++) {
    auction = Auction.auctions[i];
    Auction.auctions[i]['seconds'] --;
    if (auction.seconds &lt; 0) {
        Auction.auctions.splice(i, 1);
        i--;
        len--;
    }
}
</code></pre>


blocks|key|396386|text|您可以直接查看并使用shift()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|396387|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/shift^0|A|7|A|7|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<p>You can just look through and use <a href="https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/shift" rel="nofollow noreferrer"><code>shift()</code></a></p>


<p>I have the following for loop, and when I use <code>splice()</code> to remove an item, I then get that 'seconds' is undefined.  I could check if it's undefined, but I feel there's probably a more elegant way to do this.  The desire is to simply delete an item and keep on going.  </p>

<pre><code>for (i = 0, len = Auction.auctions.length; i &lt; len; i++) {
    auction = Auction.auctions[i];
    Auction.auctions[i]['seconds'] --;
    if (auction.seconds &lt; 0) { 
        Auction.auctions.splice(i, 1);
    }           
}
</code></pre>


Looping through array and removing items, without breaking for loop

我有下面的for循环，当我使用splice()删除一个项目时，我得到的是'seconds‘是未定义的。我可以检查它是否是未定义的，但我觉得可能有一种更优雅的方法来做到这一点。我们的愿望是简单地删除一项并继续进行。for (i = 0, len = Auction.auctions.length; i < len; i++) {    auction = Auction.auctions[i];  

问循环遍历数组并删除项，而不中断for循环
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回答 16

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问循环遍历数组并删除项，而不中断for循环
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