如何接受传递GET或POST值不区分大小写?
喜欢sample.php?OrderBy=asc仍然会和sample.php?orderby=ascor一样sample.php?ORDERBY=asc
有没有办法有效地实现上述目标?
我试图将两个部分放在一起使用inline-block和宽度作为百分比,但它没有填满我的窗口的整个宽度.
到目前为止我所拥有的:
HTML
<section class="left-content">
"Some Code"
</section>
<section class="main-content">
"Some More Code"
</section>
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CSS
.left-content, .right-content {
width: 15%;
min-width: 150px;
padding: 5px;
display: inline-block;
overflow: hidden;
vertical-align: top;
}
.main-content {
width: 85%;
min-width: 712px;
padding: 10px;
display: inline-block;
overflow: hidden;
vertical-align: top;
}
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但除非我在屏幕上计算出小数点的确切百分比,否则它不起作用.有没有人知道如何使用inline-block或我必须使用float?
是否可以在mysql shell中显示如下的查询结果?
mysql> select code, created_at from my_records;
code created_at
1213307927 2013-04-26 09:52:10
8400000000 2013-04-29 23:38:48
8311000001 2013-04-29 23:38:48
3 rows in set (0.00 sec)
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代替
mysql> select code, created_at from my_records;
+------------+---------------------+
| code | created_at |
+------------+---------------------+
| 1213307927 | 2013-04-26 09:52:10 |
| 8400000000 | 2013-04-29 23:38:48 |
| 8311000001 | 2013-04-29 23:38:48 |
+------------+---------------------+
3 rows in set (0.00 sec)
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我问的原因是因为我需要复制输出并将其粘贴到其他工具上,这是一项繁琐的工作.
我试图通过timestampin mysql表进行分区。但是它返回一个错误
CREATE TABLE tblemployeepunch (
fld_id int(11) NOT NULL AUTO_INCREMENT,
fld_date Varchar(15) DEFAULT NULL,
fld_rawpunchdate varchar(25) DEFAULT NULL,
fld_rawpunchtime varchar(25) DEFAULT NULL,
fld_cardno varchar(50) DEFAULT NULL,
fld_reasoncard varchar(20) DEFAULT NULL,
fld_mode varchar(20) DEFAULT NULL,
fld_punchdatetime varchar(50) DEFAULT NULL,
fld_crtdate TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
UNIQUE KEY fld_id (fld_id,fld_crtdate),
KEY in_timesheet (fld_cardno,fld_punchdatetime,fld_mode,fld_rawpunchtime),
KEY in_emppunch (fld_cardno,fld_rawpunchdate,fld_punchdatetime)
)
PARTITION BY HASH(DAYOFYEAR(fld_crtdate))
PARTITIONS 12;
Run Code Online (Sandbox Code Playgroud) 我试图将图像和h1标签对齐在一条线上时遇到问题.我试过显示:内联和内联块它们不起作用,只制作了两个容器.我在该部分添加了100%的宽度,但仍然没有.Float也不起作用,如果确实如此,它会搞砸页面的对齐方式.我究竟做错了什么?有时很难理解为什么它不能按预期工作并需要一些帮助.
HTML
<section>
<img id="me" src="assets/img/pose1.jpg" alt="A photo of me." />
<h1>FOLLOW ME ON...</h1>
</section>
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CSS
section{
display:inline-block;
width:100%;
}
h1{
position:relative; /*position wherever desired on the page*/
top:0;
bottom:0;
right:0;
left:0;
font-weight:bold;
font-size:40px;
color:#FFFFFF;
background-color:#FFB405;
}
Run Code Online (Sandbox Code Playgroud) 我有来自API Web服务的响应,我想获取此对象的值.示例:TimeStamp,TransactionID,SubscriptionResponse
并在HTML代码中显示此值不同的DIV.如何获得这个价值并使用主题?
来自API的回复:
object(SubscriptionResponseContainer)#11 (5) {
["Timestamp":protected]=> string(25) "2015-09-15T04:30:00+04:30"
["TransactionId":protected]=> string(10) "1445756101"
["SubscriptionResponse":protected]=> array(1) {
[0]=> object(SubscriptionResponseItemCollection)#12 (2) {
["Activate":protected]=> NULL
["Renew":protected]=> NULL
}
}
["SubscriptionError":protected]=> array(1) {
[0]=> object(SubscriptionResponseErrorCollection)#13 (2) {
["ActivateError":protected]=> array(1) {
[0]=> object(ResponseItemErrorType)#14 (4) {
["UnitId":protected]=> int(1)
["SubscriberId":protected]=> string(11) "09394354387"
["ErrorCode":protected]=> int(1)
["ErrorMessage":protected]=> string(70) "SubscriberId specified in the activate order was previously activated."
}
}
["RenewError":protected]=> NULL
}
}
["TransactionError":protected]=> NULL
}
Run Code Online (Sandbox Code Playgroud) 我有一个愚蠢的问题,我无法弄清楚.当我刷新我的登录页面时,密码和用户名出现在输入字段中.这是我的代码:
<?php include_once "includes/scripts.php"; ?>
<?php
session_start();
include_once ("includes/connect.php");
if (isset($_SESSION['logged_in'])) {
header('location: admin_cms.php');
}else{
if (isset($_POST['username'], $_POST['password'])){
$username = $_POST['username'];
$password = md5($_POST['password']);
if (empty($username) or empty($password)){
$error = '<p>NOTE: Fields are blank</p>';
}else{
$query = $pdo->prepare("SELECT * FROM users WHERE user_name = ? AND user_password =?");
$query->bindValue(1, $username);
$query->bindValue(2, $password);
$query->execute();
$num = $query->rowCount();
if ($num == 1){
$_SESSION['logged_in'] = true;
header('location: admin_cms.php');
exit();
}else{
$error = "<p>NOTE: The username or password is incorrect</p>";
}
}
}
?> …Run Code Online (Sandbox Code Playgroud) 关于如何将文本放在导航栏上,就像它只向左移动一样,我有点难过.我已经尝试过中心对齐,但它仍然继续坚持到一边.如果有人能告诉我我犯了什么错误,我将不胜感激.
HTML:
<body>
<div id="wrap">
</div>
<ul id="nav">
<li><a href="#">About Us</a></li>
<li><a href="#">Our Products</a></li>
<li><a href="#">FAQs</a></li>
<li><a href="#">Contact</a></li>
<li><a href="#">Login</a></li>
</ul>
<div id="content">
</div>
</body>
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CSS:
body {
margin: 0;
padding: 0;
text-align: justify;
font-family: "Trebuchet MS", Arial, Helvetica, sans-serif;
font-size: 13px;
background-color:#425eb4;
}
*{
margin: 0 auto 0 auto;
}
#wrap {
height:150px;
background:url(images/header1.png) no-repeat center;
text-align:center;
background-color:#FFF;
}
#nav {
width: 100%;
float: left;
padding: 0;
list-style: none;
background-color: #f2f2f2;
border-bottom: 1px solid #ccc;
border-top: 1px solid #ccc; …Run Code Online (Sandbox Code Playgroud) 我有一个包含时间戳字段的数据模型createdAt.
struct Box {
var title: String
var createdAt: Timestamp
var dictionary: [String : Any] {
return [
"title": title,
"createdAt": createdAt
]
}
// ...
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但是,我无法分配Firestore服务器时间戳.swift中云Firestore服务器时间戳的等效数据类型是什么?我在这里错过了一点吗?
let box = Box(title: title, createdAt: FieldValue.serverTimestamp());
Run Code Online (Sandbox Code Playgroud) 嗨朋友我想复制文本div到textarea包括所有换行符请在这里查看小提琴http://jsfiddle.net/SEz5y/
我试过了
脚本
$('.cancel').click(function(){
var fr = $('.fromCont').html();
$('.contentF').val(fr)
})
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HTML
<textarea name="fromCont" rows="5" class="contentF" id="f" placeholder="Add Text"></textarea>
<div class="fromCont" style="">zxczxczc <br> 1 line <br> 2 line <br> 3 line</div>
<a href="#" class="cancel">submitl</a>
Run Code Online (Sandbox Code Playgroud) 我正在尝试使用CREATE DATABASE命令创建一个数据库,但它给了我一个错误.这是我的代码:
$db_usr = mysql_real_escape_string($_POST["email"]);
$con=mysql_connect("localhost","root");
if (! $con)
{
die('Could not connect: ' . mysql_error());
}
else
{
test();
}
function test()
{
$sql = "CREATE DATABASE '$db_usr'";
mysql_query($sql);
}
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它总是返回"未定义的变量"
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