jQuery .inArray() 总是正确的？
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jQuery .inArray() always true?
提问by Charles Marsh
I'm trying to use inarray but its always returning true? Any ideas? (all li's are showing)
我正在尝试使用 inarray 但它总是返回 true？有任何想法吗？（所有的 li 都在展示）
$("#select-by-color-list li").hide();
// get the select
var $dd = $('#product-variants-option-0');
if ($dd.length > 0) { // make sure we found the select we were looking for
// save the selected value
var selectedVal = $dd.val();
// get the options and loop through them
var $options = $('option', $dd);
var arrVals = [];
$options.each(function(){
// push each option value and text into an array
arrVals.push({
val: $(this).val(),
text: $(this).text()
});
});
};
//This is where it is returning true...
if($.inArray('Aqua', arrVals)) {
$("#select-by-color-list li#aqua").show();
};
if($.inArray('Army', arrVals)) {
$("#select-by-color-list li#army").show();
};
回答by user113716
You need to do this:
你需要这样做：
if( $.inArray('Aqua', arrVals) > -1 ) {
or this:
或这个：
if( $.inArray('Aqua', arrVals) !== -1 ) {
The $.inArray()methodreturns the 0based index of the item. If there's no item, it returns -1, which the if()statement will consider as true.
该$.inArray()方法返回项目的0基础索引。如果没有项目，则返回-1，if()语句将其视为true。
From the docs:
从文档：
Because JavaScript treats 0 as loosely equal to false (i.e. 0 == false, but 0 !== false), if we're checking for the presence of value within array, we need to check if it's not equal to (or greater than) -1.
因为 JavaScript 将 0 视为大致等于 false（即 0 == false，但 0 !== false），如果我们检查数组中是否存在值，我们需要检查它是否不等于（或大于） ) -1。
EDIT:Instead of pushing both values into the array as an object, just use one or the other, so you have an Array of strings from which you can build a multiple selector.
编辑：不是将两个值作为对象推送到数组中，只需使用一个或另一个，这样您就有了一个字符串数组，您可以从中构建一个多重选择器。
One way is like this:
一种方法是这样的：
// Create an Array from the "value" or "text" of the select options
var arrVals = $.map( $dd[0].options, function( opt, i ){
return opt.value || opt.text;
});
// Build a multiple selector by doing a join() on the Array.
$( "#" + arrVals.join(',#') ).show();
If the Array looks like:
如果数组看起来像：
['Army','Aqua','Bread'];
The resulting selector will look like:
结果选择器将如下所示：
$( "#Army,#Aqua,#Bread" ).show();
回答by Darren Sweeney
ES6 to the rescue! Although not jQuery I thought worth answering for future readers...
ES6 来救援！虽然不是 jQuery，但我认为值得为未来的读者回答......
ES6 introduces .includes()which works as you think $.inArraySHOULD work:
ES6 介绍了.includes()您认为$.inArray应该工作的工作方式：
const myArray = ["a", "b", "c"];
console.log(myArray.includes("a")) /* true */
console.log(myArray.includes("d")) /* false */