小编Leo*_*Leo的帖子

我正在使用此脚本来允许用户更改背景颜色...

document.onclick = function SetFavColor(e) {
    if (e.target.className == 'AvcGbtn') {
        var favColor = e.target.style.backgroundColor;

        localStorage.setItem('color', favColor);
        document.body.style.backgroundColor = favColor;
        console.log(favColor);
    }
};

document.addEventListener('DOMContentLoaded', function GetFavColor() {
    var favColor = document.body.style.backgroundColor;
    var color = localStorage.getItem('color');

    if (color === '') {
        document.body.style.backgroundColor = favColor;
    } else {
      document.body.style.backgroundColor = color;
    }
});

CSS:

body {
  max-width: 100%;
  min-width: 100%;
  height: 100%;
  font-family: normal;
  font-style: normal;
  font-weight: normal;
  background-color: transparent;
}

.AvcGbtn {
  display: inline-block;
  width: 2em;
  height: 2em;
}

HTML:

<span …

所以在我的网站上我使用谷歌自定义搜索引擎,当我搜索示例'youtub'而不是'youtube'it显示为一个建议(显示结果为youtube - 搜索而不是youtub)或(你的意思是:youtube)

所以当我点击文本youtube时,我想将它添加到我的搜索框中,我该怎么做？这是我的代码

searchbox .html

<form action="" method="GET" id="searchform">
<input required="" type="search" name="q" id="searchbox" placeholder="Search..." title="Search..." enableAutoComplete="true" autofocus/>
<button><div class="Search-Button"><img src="MYIMGHERE" class="search-png"/></div></button>
</form>

.js文件

<script type="text/javascript">
(function() {
var cx = '007072537540236505002:fsvzbpwgtgc';
var gcse = document.createElement('script');
gcse.type = 'text/javascript';
gcse.async = true;
gcse.src = 'https://cse.google.com/cse.js?cx=' + cx;
var s = document.getElementsByTagName('script')[0];
s.parentNode.insertBefore(gcse, s);
})();
var urlParams = new URLSearchParams(window.location.search);
document.getElementById('searchbox').value = urlParams.get('q');
</script>

<gcse:searchresults-only linktarget="_self"></gcse:searchresults-only>

我的问题不是很明确,因为我找不到合适的词语!我很抱歉.

当复选框被选中克隆正确的股利,并显示在例如:<div id="favorite"></div>当复选框未选中删除克隆,伴随着localStorage.有人可以帮我解决这个问题吗？

function onClickAvGamesCheckBox() {
  var arr = $('.AvGamesCheckBox').map(function() {
    return this.checked;
  }).get();
  localStorage.setItem("checked", JSON.stringify(arr));
}

$(document).ready(function() {
  var arr = JSON.parse(localStorage.getItem('checked')) || [];
  arr.forEach(function(checked, i) {
    $('.AvGamesCheckBox').eq(i).prop('checked', checked);
  });
  $(".AvGamesCheckBox").click(onClickAvGamesCheckBox);
});

//* Clone script
$(".avclone :checkbox").change(function() {
  var name = $(this).closest("div").attr("name");
  if (this.checked)
    $(".columns[name=" + name + "]").clone().appendTo("#favorite");
  else
    $("#favorite .columns[name=" + name + "]").remove();
});

* {
  box-sizing: border-box;
  padding: 5px;
}

.AvGamesContainer {
  display: block;
  position: relative;
  padding-left: 35px;
  margin-bottom: 12px;
  cursor: pointer; …

如何在Google自定义搜索引擎(而非CSE面板)上设置手动过滤器？

例如,要在新选项卡中打开链接,我正在使用此代码:

<gcse:searchresults-only linktarget="_blank"></gcse:searchresults-only>

现在我想设置"安全模式",并设置每页的结果(例如,显示10个结果).

为了更清楚,我想允许用户从我的网站更改这些过滤器,这就是我试图手动添加它们的原因!谢谢.

复选框默认选中不起作用！我试图修复它，但我找不到这里的错误在哪里？所以在页面加载被选中，在页面加载之后未被选中！？我试过了

<div class="onoffswitch" style="margin: 0 auto;">
<input type="checkbox" class="avacheckbox onoffswitch-checkbox" id="AvButtonAutoGames" checked="checked"/>
<label class="onoffswitch-label" for="AvButtonAutoGames">
    <span class="onoffswitch-inner"></span>
    <span class="onoffswitch-switch"></span>
</label>
</div>

然后

  <div class="onoffswitch" style="margin: 0 auto;">
    <input type="checkbox" class="avacheckbox onoffswitch-checkbox" id="AvButtonAutoGames" checked/>
    <label class="onoffswitch-label" for="AvButtonAutoGames">
        <span class="onoffswitch-inner"></span>
        <span class="onoffswitch-switch"></span>
    </label>
    </div>

我想允许用户在chrome扩展示例文件夹名称（上传）上上传图像，并且没有提交按钮

<form action="/upload">
  <input type="file" name="myimages" accept="image/*">
</form>

显示图片

<span class="AvGbtn" id="AvBgIds" style="background-image: url(Here i want to show upload images url
); background-size: 100% 100%;" oncontextmenu="return false">  
</span>

在这里阅读了很多文章后,我没有找到解决方案,所以我需要帮助才能做到这一点......我的网址就是那些例子:

主页

https://mywebsite.com/
https://mywebsite.com/al/
https://mywebsite.com/it/
https://mywebsite.com/videos/
https://mywebsite.com/al/videos/
https://mywebsite.com/it/videos/
https://mywebsite.com/news/
https://mywebsite.com/al/news/
https://mywebsite.com/it/news/

询问

https://mywebsite.com/search/?q=YouTube
https://mywebsite.com/videos/search/?q=YouTube
https://mywebsite.com/news/search/?q=YouTube

https://mywebsite.com/al/search/?q=YouTube
https://mywebsite.com/al/videos/search/?q=YouTube
https://mywebsite.com/al/news/search/?q=YouTube

https://mywebsite.com/it/search/?q=YouTube
https://mywebsite.com/it/videos/search/?q=YouTube
https://mywebsite.com/it/news/search/?q=YouTube

我的php和html改变了语言

<?php $Ava_Sulg = $_SERVER["REQUEST_URI"];?>

<a class="x" href="/<?php echo $Ava_Sulg;?>">EN</a>
<a class="x" href="/al<?php echo $Ava_Sulg;?>">AL</a>
<a class="x" href="/it<?php echo $Ava_Sulg;?>">IT</a>

所以我允许用户更改他们的语言,什么我想要做的,是他们改变语言的URL可以是上述情况,例如之一,如果他们从改变语言AL to IT和url是https://mywebsite.com/al/videos/search/?q=YouTube与PHP我想这https://mywebsite.com/it/videos/search/?q=YouTube使我想改变这个url only(/al/ to /it/)或者exmaple from IT to EN(/it/ to Nothing)但是我要改变的是在中间,在主页上是不同的,对我来说很难,我怎么能做到这一点是可能的或者没有？如果可能的话,我希望在这里找到解决方案!非常感谢你.

从此 application/ld+json 我想获取示例 1.genre、2.director、3.dateCreated、4.duration、5.uploadDate

 <script type="application/ld+json">{
      "@context": "http://schema.org",
      "@type": "Movie",
      "url": "/title/tt2293640/",
      "name": "Minions",
      "image": "https://m.media-amazon.com/images/M/MV5BMTg2MTMyMzU0M15BMl5BanBnXkFtZTgwOTU3ODk4NTE@._V1_.jpg",
      "genre": [
        "Animation",
        "Action",
        "Adventure",
        "Comedy",
        "Family"
      ],
      "contentRating": "PG",
      "actor": [
        {
          "@type": "Person",
          "url": "/name/nm0000113/",
          "name": "Sandra Bullock"
        },
        {
          "@type": "Person",
          "url": "/name/nm0358316/",
          "name": "Jon Hamm"
        },
        {
          "@type": "Person",
          "url": "/name/nm0000474/",
          "name": "Michael Keaton"
        },
        {
          "@type": "Person",
          "url": "/name/nm1853544/",
          "name": "Pierre Coffin"
        }
      ],
      "director": [
        {
          "@type": "Person",
          "url": "/name/nm0049633/",
          "name": "Kyle Balda"
        },
        {
          "@type": "Person",
          "url": …

我想$LinkOpen根据checkbox元素的当前状态更新PHP变量.avflipswitch.

基于.avflipswitch复选框状态,我想在我的PHP变量$LinkOpen之间切换'_blank','_self'因此我可以将该值推送到我的Google CSE链接目标属性.

到目前为止我尝试了什么:

 $('.avflipswitch').on("change", function (e){ 
   if(this.checked){
     functionOne(<?php $LinkOpen = '_blank';?>);
   }
   else{
     functionTwo(<?php $LinkOpen = '_self';?>);
   }
});

<gcse:searchresults-only linktarget="<?php echo $LinkOpen;?>"></gcse:searchresults-only>