entityMap|blocks|key|4c70q|text|因为start+with的参数必须是字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6anpc|所以这样做效果会更好|3ga6i|for+(var+i+=+0;+i+<+value.length;+i%2B%2B)+{
++++if+(typeof+value[i].keyword+==+String(undefined)+%7C%7C+value[i].keyword.startsWith("keyword"))
++++++++out.push(value[i]);
}|code-block|syntax|javascript^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|K|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|L|9|@]|A|@]|B|$H|I]]]]

<code>value[i].keyword.startsWith(&quot;keyword&quot;)</code> because the parameter of start with must be a string.
So that will work better this way
<pre class="lang-js prettyprint-override"><code>for (var i = 0; i &lt; value.length; i++) {
 if (typeof value[i].keyword == String(undefined) || value[i].keyword.startsWith(&quot;keyword&quot;))
 out.push(value[i]);
}
</code></pre>

entityMap|blocks|key|2406s|text|我假设有价值|type|unstyled|depth|inlineStyleRanges|entityRanges|data|dpioh|我|3ne5d|.keyword是一个字符串。旧版浏览器不支持String.prototype.startWith。请参见|7hek|浏览器支持|aj9f0|..。|fnrpo|要在较旧的浏览器中使用它，可以使用现有的|dd1tf|多边形填充|d98sj|..。另请参阅答案来自|4thlq|如何检查一个字符串是否“StartsWith”另一个字符串？^0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|S|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|T|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|U|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|V|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|W|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|X|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|Y|9|@]|A|@]|B|$]]|$2|O|4|P|6|7|8|Z|9|@]|A|@]|B|$]]|$2|Q|4|R|6|7|8|10|9|@]|A|@]|B|$]]]]

I assume value[i].keyword is a string. String.prototype.startWith is not supported in older browsers. See <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/startsWith" rel="nofollow noreferrer">browser support</a>.

To use it in older browsers, you can use one of existing <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/startsWith#Polyfill" rel="nofollow noreferrer">polyfills</a>. See also answers from <a href="https://stackoverflow.com/questions/646628/how-to-check-if-a-string-startswith-another-string">How to check if a string “StartsWith” another string?</a>

entityMap|blocks|key|9g7gr|text|有没有一种方法可以让我检查它是否是一个字符串，而不是让它出错？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|da31n|6205j|检查类型？|29fga|var+out+=+values.filter(v+=>+v.keyword+==+null+%7C%7C+typeof+v.keyword+===+"string"+&&+v.keyword.startsWith(+keyword+));|code-block|syntax|javascript|27mmo|或者简单地强制类型|4uhu8|var+out+=+values.filter(v+=>+v.keyword+==+null+%7C%7C+String(v.keyword).startsWith(+keyword+));|2t03o|或者，如果你使用反结构，你甚至可以使用这个：|fchs1|var+out+=+values.filter({keyword:+v})+=>+v+==+null+%7C%7C+String(v).startsWith(+keyword+));|3sghn|我建议您使用Array方法，而不是手动编写循环。|c8ig2|国际海事组织。它能更好地描述你的意图(很适合浏览代码)。|6sfta|如果这是一条热门路径(经常被调用)，JS优化器可以利用知道你只是在过滤代码的优势，并对代码进行优化。也许可以跳过循环期间对Array执行的大量完整性检查。|2r2di|如果这不是热路径，那么在JS中，调用几次函数的性能影响甚至很难衡量。^0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|10|9|@]|A|@]|B|$]]|$2|C|4|-4|6|7|8|11|9|@]|A|@]|B|$]]|$2|D|4|E|6|7|8|12|9|@]|A|@]|B|$]]|$2|F|4|G|6|H|8|13|9|@]|A|@]|B|$I|J]]|$2|K|4|L|6|7|8|14|9|@]|A|@]|B|$]]|$2|M|4|N|6|H|8|15|9|@]|A|@]|B|$I|J]]|$2|O|4|P|6|7|8|16|9|@]|A|@]|B|$]]|$2|Q|4|R|6|H|8|17|9|@]|A|@]|B|$I|J]]|$2|S|4|T|6|7|8|18|9|@]|A|@]|B|$]]|$2|U|4|V|6|7|8|19|9|@]|A|@]|B|$]]|$2|W|4|X|6|7|8|1A|9|@]|A|@]|B|$]]|$2|Y|4|Z|6|7|8|1B|9|@]|A|@]|B|$]]]]

<blockquote>
 Is there a way that I can check if it's a string rather than have it error out?
</blockquote>

checking the type?

<pre><code>var out = values.filter(v =&gt; v.keyword == null || typeof v.keyword === "string" &amp;&amp; v.keyword.startsWith( keyword ));
</code></pre>

or simply enforcing the type

<pre><code>var out = values.filter(v =&gt; v.keyword == null || String(v.keyword).startsWith( keyword ));
</code></pre>

or if you use desctructuring, you can use even this:

<pre><code>var out = values.filter({keyword: v}) =&gt; v == null || String(v).startsWith( keyword ));
</code></pre>

I'd reccomend you to use the Array-methods instead of manually writing loops. 

<ul>
<li>Imo. it is a better description of your intent (good for scanning over code). </li>
<li>If this is a hot path (it's called often) the JS-optimizer can take advantage of knowing, that youre'just filtering, and optimize the code. Maybe by skipping the lots of integrity-checks for the Array it performs during your loop. </li>
<li>And if this is not a hot-path, the performance-impact of calling a function a few times will even be hard to measure, in JS.</li>
</ul>

entityMap|blocks|key|fdbp4|text|找到|type|unstyled|depth|inlineStyleRanges|entityRanges|data|cemub|一篇有用的文章|fabb0|关于这个主题|esk7k|转换为字符串的三种方法是：|92alq|value.toString()|af6ll|"“%2B值|f0147|字符串(值)|e0550|这里要注意的一点是，如果值为空或未定义，则方法#1不起作用。|9ah0h|在我的例子中，方法#2由于某种原因也不起作用，所以最好的选择是String(value)|c023a|var+col+=+"rt_"+%2B+rows[i]["results"][r].ResultTypeID.substring(1);+//did+not+work

var+col+=+"rt_"+%2B+String(rows[i]["results"][r].ResultTypeID).substring(1);|code-block|syntax|javascript^0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|X|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|Y|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|Z|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|10|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|11|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|12|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|13|9|@]|A|@]|B|$]]|$2|O|4|P|6|7|8|14|9|@]|A|@]|B|$]]|$2|Q|4|R|6|7|8|15|9|@]|A|@]|B|$]]|$2|S|4|T|6|U|8|16|9|@]|A|@]|B|$V|W]]]]

Found <a href="http://2ality.com/2012/03/converting-to-string.html" rel="nofollow noreferrer">a useful article</a> on this topic

The three approaches for converting to string are:

<ol>
<li>value.toString()</li>
<li>"" + value</li>
<li>String(value)</li>
</ol>

The point to note here is that approach # 1 doesn’t work if the value is null or undefined.

In my case, approach # 2 for some reason did not work either so the best option would be String(value)

<pre><code>var col = "rt_" + rows[i]["results"][r].ResultTypeID.substring(1); //did not work

var col = "rt_" + String(rows[i]["results"][r].ResultTypeID).substring(1);
</code></pre>

entityMap|blocks|key|ddd30|text|这绝对是一个显而易见的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|21qtg|(这完全是为那些有类似问题的人准备的，但他们并不真正理解)|eaimr|当然，就像很多人说的那样，|dbfdk|它在类型中。|5s426|如果您不知道什么是类型(如果您知道这种高级类型，您可能会知道，但以防万一！)，这很简单。基本类型为|3bu23|/|beb7f|51tjl|c2cd2|1kog4|，以及|cocqo|，任何|dl0ps|类型不为空或数字不等于|7lbqt|导致存在|do4cp|如果通过|a9i1u|靠他们自己。现在我们已经解决了这个问题，是时候谈谈|3upcd|..。如果你运行|1gpr4|，它将输出|26ghg|到控制台，这是由于字符串以字符串开头的方式。如果你运行|875ge|，将以|dv9sk|由于|e5ekm|需要字符串。我们可以看出这是一个|6d374|从错误输出中。所以，上面的问题(已经回答过了)是变量|3ei2b|没有被定义为字符串，而是任何其他变量，并且需要被定义为字符串^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|1H|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|1I|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|1J|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|1K|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|1L|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|1M|9|@]|A|@]|B|$]]|$2|M|4|-4|6|7|8|1N|9|@]|A|@]|B|$]]|$2|N|4|L|6|7|8|1O|9|@]|A|@]|B|$]]|$2|O|4|-4|6|7|8|1P|9|@]|A|@]|B|$]]|$2|P|4|Q|6|7|8|1Q|9|@]|A|@]|B|$]]|$2|R|4|S|6|7|8|1R|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|1S|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|1T|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|1U|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|1V|9|@]|A|@]|B|$]]|$2|11|4|12|6|7|8|1W|9|@]|A|@]|B|$]]|$2|13|4|14|6|7|8|1X|9|@]|A|@]|B|$]]|$2|15|4|16|6|7|8|1Y|9|@]|A|@]|B|$]]|$2|17|4|18|6|7|8|1Z|9|@]|A|@]|B|$]]|$2|19|4|1A|6|7|8|20|9|@]|A|@]|B|$]]|$2|1B|4|1C|6|7|8|21|9|@]|A|@]|B|$]]|$2|1D|4|1E|6|7|8|22|9|@]|A|@]|B|$]]|$2|1F|4|1G|6|7|8|23|9|@]|A|@]|B|$]]]]

<h1>There's a definitely obvious problem.</h1>

(This is entirely for people who have a problem like this one, but they don't really understand)

Of course, just like many others are saying, it's in the type. If you don't know what types are (You likely do if you know this advanced type of stuff, but just in case!), its simple. The primitive types are <code>True</code>/<code>False</code>, <code>Undefined</code>/<code>Null</code>, <code>Numbers</code>, and <code>Strings</code>, any <code>string</code> type that is not empty or numbers not equal to <code>0</code> resulting in being <code>true</code> if put through an <code>if statement</code> on their own. Now that we've got that out of the way, its time to talk about <code>startsWith()</code>. If you run <code>console.log("Hey VSauce, Michael here.".startsWith("Hey"))</code>, it would output <code>true</code> to the console due to how the string starts out with a string. If you run <code>console.log(1234.startsWith(1))</code>, would occur with a <code>TypeError</code> due to <code>startsWith()</code> requiring a string. We can tell this is a <code>TypeError</code> from the error output. So, the problem above (that was already answered) was that the variable <code>keyword</code> wasn't being defined as a string, but rather any other variable, and needed to be defined as a string

I have this code:
<pre class="lang-js prettyprint-override"><code>for (var i = 0; i &lt; value.length; i++) {
 if (typeof value[i].keyword == 'undefined' || value[i].keyword == null || value[i].keyword.startsWith(keyword)) {
 out.push(value[i]);
 }
}
</code></pre>
I am getting an error message saying:
<blockquote>
<pre><code>&gt; TypeError: r[e].startsWith is not a function
&gt; at js-cf2cc68….min.js.gz:85
&gt; at fn (eval at compile (js-cf2cc68….min.js.gz:8), &lt;anonymous&gt;:4:1003)
&gt; at js-cf2cc68….min.js.gz:7
&gt; at p.$digest (js-cf2cc68….min.js.gz:7)
&gt; at p.$apply (js-cf2cc68….min.js.gz:7)
&gt; at HTMLBodyElement.&lt;anonymous&gt; (js-cf2cc68….min.js.gz:9)
</code></pre>
</blockquote>
How is this possible? I think I've accounted for everything.

startsWith is not a function getting called for no reason

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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我有这个代码： for (var i = 0; i < value.length; i++) { if (typeof value[i].keyword == 'undefined' || value[i].keyword == null || value[i].keyword.startsWith(keywo...

问startsWith不是一个无缘无故被调用的函数
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问startsWith不是一个无缘无故被调用的函数EN