Question

我使用以下JavaScript检查组合条目：

var proteins = [
    "charred sirloin steak", 
    "roasted salmon", 
    "grilled chicken", 
    "grilled salmon", 
    "jerk chicken",
    "miami ribs", 
    "charred steak", 
    "tandoori chicken supreme"
];

var sides = [
    "israeli couscous salad",
    "kale caesar", 
    "field o greens",
    "grilled broccoli",
    "greek salad", 
    "coconut curry quinoa salad",
    "vegetable egg fried rice",
    "general tao brussel sprouts",
    "tokyo slaw",
    "mac cheese",
    "potato salad"
];

var comboEntries = document.querySelectorAll('.combo-display .combo-item');
for (var i = 0; i < comboEntries.length; i++) {
  if (proteins.indexOf(comboEntries[i].innerText) == -1 || sides.indexOf(comboEntries[i].innerText) == -1) {
    comboEntries[i].style.color = "orangered";
    console.log(comboEntries[i].innerText + " is not a regular combo item.");
  }
}

但是，这会将所有comboEntries输出为不规则项。如果我将条件反转为!= -1，则可以看到它正常工作，突出显示常规项目，而排除不规则项目。

我在这里做错了什么？一定是我没意识到的简单事情。

编辑：对不起，我之前没有包含HTML，因为这是生成内容的在线厨房显示屏的一部分。这是一个示例：

<div class="combo-display">
<p><strong>Protein 1:</strong> <span id="protein-1" class="combo-item">Custom item</span></p><div class="item-count"> x <span>1</span></div>
<p><strong>Protein 2:</strong> <span id="protein-2" class="combo-item">Custom item</span></p><div class="item-count"> x <span>1</span></div>
<p><strong>Protein 3:</strong> <span id="protein-3" class="combo-item">grilled chicken</span></p><div class="item-count"> x <span class="grilled-chicken-counter">1</span></div>
<p><strong>Side 1:</strong> <span id="side-1" class="combo-item">Custom item</span></p><div class="item-count"> x <span>1</span></div>
<p><strong>Side 2:</strong> <span id="side-2" class="combo-item">kale caesar</span></p><div class="item-count"> x <span class="kale-caesar-counter">1</span></div>
<p><strong>Side 3:</strong> <span id="side-3" class="combo-item">field o greens</span></p><div class="item-count"> x <span class="field-o-greens-counter">1</span></div>
</div>

因此，custom item应该突出显示为不规则项

Answer 1

尽管由于缺少HTML代码而导致您的问题尚不清楚，但是如果我没记错的话，您将尝试突出显示 值，而这两项都不在{{1} }或proteins中都没有。

为此，您应该在sides中使用&&。

这是一个有效的代码段：

if condition

var proteins = [
    "charred sirloin steak", 
    "roasted salmon", 
    "grilled chicken", 
    "grilled salmon", 
    "jerk chicken",
    "miami ribs", 
    "charred steak", 
    "tandoori chicken supreme"
];

var sides = [
    "israeli couscous salad",
    "kale caesar", 
    "field o greens",
    "grilled broccoli",
    "greek salad", 
    "coconut curry quinoa salad",
    "vegetable egg fried rice",
    "general tao brussel sprouts",
    "tokyo slaw",
    "mac cheese",
    "potato salad"
];

var comboEntries = document.querySelectorAll('.combo-display .combo-item');
for (var i = 0; i < comboEntries.length; i++) {
  if (proteins.indexOf(comboEntries[i].innerText) == -1 && sides.indexOf(comboEntries[i].innerText) == -1) {
    comboEntries[i].style.color = "orangered";
    console.log(comboEntries[i].innerText + " is not a regular combo item.");
    }
  }

indexOf检查不规则组合项不起作用？

1 个答案: