小编pea*_*ove的帖子

有没有更简单、更聪明、更清晰的方法来写这个？

if (array_key_exists("name", $array) AND array_key_exists("age", $array) AND array_key_exists("size", $array) AND array_key_exists("gender", $array) {
    echo "Keys exist!";
}

我想colors通过选项创建具有类名的div .on("change").它运作良好,但只要我改变选项,就会div创建越来越多的选项.但我只有两个colors,所以任何时候我改变选项,我只想看到两个div:"蓝色"和"红色".我试图$('.result').remove();在追加之前使用.但是没有附加任何内容.

$(document).ready(function() {
  $("#select").on("change", function() {
    $(".colors").each(function() {
      var color = $(this).attr('title');
      $(".result").append("<div>" + color + "</div>");
    });
  });
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="select">
  <option class="colors" title="blue">a</option>
  <option class="colors" title="red">b</option>
  <option title="green">c</option>
</select>

<div class="result"></div>

我想用 jQuery 在我的每个选择框周围设置一个红色边框。但它不起作用：

http://jsfiddle.net/jEADR/3795/

<script type="text/javascript" src="//code.jquery.com/jquery-1.8.3.js"></script> 
<script type="text/javascript" src="//cdnjs.cloudflare.com/ajax/libs/select2/3.2/select2.min.js"></script>
 <link rel="stylesheet" type="text/css" href="//cdnjs.cloudflare.com/ajax/libs/select2/3.2/select2.css">

<select class="select" style="width:300px">
        <option value="AL">Alabama</option>
        <option value="Am">Amalapuram</option>
        <option value="An">Anakapalli</option>
        <option value="Ak">Akkayapalem</option>
        <option value="WY">Wyoming</option>
    </select>
    <select class="select" style="width:300px">
        <option value="AL">Alabama</option>
        <option value="Am">Amalapuram</option>
        <option value="An">Anakapalli</option>
        <option value="Ak">Akkayapalem</option>
        <option value="WY">Wyoming</option>
    </select>

<script>
    $(".select").select2();
    $(".select").each(function() {
    $(this).siblings().find(".select2-container").css({"border-color": "red", "border-weight":"5px", "border-style":"solid"});
    });
</script>

我想在我的bootstrap数据表中使用select2选择框.我需要使用完整的select2 js(https://cdnjs.cloudflare.com/ajax/libs/select2/4.0.3/js/select2.full.js)但它不起作用,我想知道为什么.

$(document).ready(function() {
    var table = $('#example').DataTable( {
        scrollY:        "300px",
        scrollX:        true,
        scrollCollapse: true,
        paging:         false,
        fixedColumns:   {
            leftColumns: 1,
            rightColumns: 1
        }
    } );
} );

$("#e1").select2();

td, th{background-color:white;}

<link href="https://netdna.bootstrapcdn.com/bootstrap/3.0.0/css/bootstrap.min.css" rel="stylesheet"/>

<link href="http://cdnjs.cloudflare.com/ajax/libs/select2/3.2/select2.css" rel="stylesheet"/>

<script src="https://code.jquery.com/jquery-1.12.4.js"></script>
<script src="https://cdn.datatables.net/1.10.15/js/jquery.dataTables.min.js"></script>
<script src="https://cdn.datatables.net/fixedcolumns/3.2.2/js/dataTables.fixedColumns.min.js"></script>

<script src="https://netdna.bootstrapcdn.com/bootstrap/3.0.0/js/bootstrap.min.js"></script>


<script src="https://cdnjs.cloudflare.com/ajax/libs/select2/4.0.3/js/select2.full.js"></script>


<table id="example" class="stripe row-border order-column" width="100%" cellspacing="0">
        <thead>
            <tr>
                <th>First name</th>
                <th>Last name</th>
                <th>Position</th>
                <th>Office</th>
                <th>Age</th>
                <th>Start date</th>
                <th>Salary</th>
                <th>Extn.</th>
                <th>E-mail</th>
            </tr>
        </thead>
        <tbody>
            <tr>
                <td><select  id="e1">
        <option value="AL">Alabama</option>
        <option value="Am">Amalapuram</option> …

我在 MAMP 中创建了一个名为“project”的数据库。

在我的.env文件中，我添加了这一行：

DATABASE_URL=mysql://root:root@localhost:3306/project

现在我想跑

php bin/console doctrine:database:create

但我收到一条错误消息：

SQLSTATE[HY000] [2002] 没有那个文件或目录

我的学说配置：

parameters:
    # Adds a fallback DATABASE_URL if the env var is not set.
    # This allows you to run cache:warmup even if your
    # environment variables are not available yet.
    # You should not need to change this value.
    env(DATABASE_URL): ''

doctrine:
    dbal:
        # configure these for your database server
        driver: 'pdo_mysql'
        server_version: '5.7'
        charset: utf8mb4
        default_table_options:
            charset: utf8mb4
            collate: utf8mb4_unicode_ci

        url: '%env(resolve:DATABASE_URL)%'
    orm: …

我正在使用序列化器创建一个 json 字符串：

$table = $this->getDoctrine()->getRepository(Article::class)->findAll();

$serializer = new Serializer(array(new GetSetMethodNormalizer()), array('json' => new JsonEncoder()));
$json_string = $serializer->serialize($table, 'json');

$表的结果：

array(2) { [0]=> object(App\Entity\Article)#5975 (3) { ["id":"App\Entity\Article":private]=> int(1) ["title":"App\Entity\Article":private]=> string(9) "Article 1" ["body":"App\Entity\Article":private]=> string(32) "This is the body for article one" } [1]=> object(App\Entity\Article)#5979 (3) { ["id":"App\Entity\Article":private]=> int(2) ["title":"App\Entity\Article":private]=> string(11) "Article Two" ["body":"App\Entity\Article":private]=> string(32) "This is the body for article two" } } 

这是 $json_string 的结果：

string(145) "[{"id":1,"title":"Article 1","body":"This is the body for article one"},{"id":2,"title":"Article Two","body":"This is the body for …

我不知道,为什么我看不到任何文字:

$(".elipsis").text($(this).text().substr(0, 200)+'...');

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

 <p class="elipsis">Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua. At vero eos et accusam et justo duo dolores et ea rebum. Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit amet. Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam erat, sed diam voluptua. …

我想将来自两个不同列的值呈现到我的jquery数据表中

https://cdn.datatables.net/1.10.16/js/jquery.dataTables.min.js
https://cdn.datatables.net/1.10.16/css/jquery.dataTables.min.css

$(document).ready(function(){

    var table = $('#table').DataTable({

         "ajax": {
                "url": "data.json",
                "dataSrc": "",
            },

         "columnDefs": [
            {
               "render": function (data, type, row) {
                   var result = 'The id is ' + data[0] + ' and the name is ' + data[1];
                        return result;
                    },
                    "targets": 0,
                },      

         ],

        "columns": [
                {
                    "data": "id"
                },
                {
                    "data": "name"
                }
            ]
    });

});

data.json：

[{
    "id": "12",
    "name": "Laura"
}]

但是我的结果是：

The id is 12 and the name is undefined

我在 Atom 编辑器中安装了 atom-beautify 包。

我打开一个文件 index.php

 <ul>
 <li>Coffee</li>
        <li>Tea</li>
  <li>Milk</li>
        </ul> 

然后我发出“美化”命令。

但我的代码仍然是这样的：

 <ul>
 <li>Coffee</li>
        <li>Tea</li>
  <li>Milk</li>
        </ul> 

我希望代码更改为：

 <ul>
   <li>Coffee</li>
   <li>Tea</li>
   <li>Milk</li>
 </ul> 

我在 .env 文件中进行了更改，然后在终端中写入git status：

Changes not staged for commit:
  (use "git add <file>..." to update what will be committed)
  (use "git restore <file>..." to discard changes in working directory)
    modified:   .env

我希望 .env 文件中的更改被 git 忽略，所以我在 .gitignore 中添加了这一行：

/.env

现在当我写的时候git status我得到的结果是：

Changes not staged for commit:
  (use "git add <file>..." to update what will be committed)
  (use "git restore <file>..." to discard changes in working directory)
    modified:   .env
    modified:   .gitignore

我现在能做什么？为什么 gitignore 没有被忽略？