我有以下JSON数据:{"success":"You are welcome"}我json在我的JavaScript代码中命名.
当我想提醒You are welcome我的时候json.success.所以现在我面临的问题是,如果我想提醒,那该怎么办success？有没有办法得到它？
我是laravel的新手,我很享受.在处理社交媒体项目时,我收到此错误:htmlspecialchars() expects parameter 1 to be string, object given (View: C:\wamp64\www\histoirevraie\resources\views\user\profile.blade.php)
我在本网站上查了一些问题,但我没有找到解决问题的问题.
这就是我profile.blade.php的意思:
<ul class="profile-rows">
<li>
<span class="the-label">Last visit: </span>
<span class="the-value mark green">{{ \Carbon\Carbon::createFromFormat('Y-m-d H:i:s', $user->lastVisit)->diffForHumans(\Carbon\Carbon::now())}}</span>
</li>
<li>
<span class="the-label">Member since: </span>
<span class="the-value mark light-gray">{{ $user->created_at->format('F Y') }}</span>
</li>
<li>
<span class="the-label">Profile views: </span>
<span class="the-value mark light-gray">5146</span>
</li>
<li>
<span class="the-label">Living In: </span>
<span class="the-value">{{ $user->town }}</span>
</li>
<li>
<span class="the-label">Website: </span>
<span class="the-value"><a href="{{ url($user->website) }}">{{ $user->website }}</a></span>
</li>
</ul>
有关用户的所有信息均由控制器提供:
public function …我正在制作一个标签系统:
<script>
$(function () {
var links = $('.sidebar-links > div');
links.on('click', function () {
links.removeClass('selected');
$(this).addClass('selected');
});
});
function openCity(evt, cityName) {
// Declare all variables
var i, tabcontent, tablinks;
// Get all elements with class="tabcontent" and hide them
tabcontent = document.getElementsByClassName("tabcontent");
for (i = 0; i < tabcontent.length; i++) {
tabcontent[i].style.display = "none";
}
// Get all elements with class="tablinks" and remove the class "active"
tablinks = document.getElementsByClassName("tablinks");
for (i = 0; i < tabcontent.length; i++) {
tablinks[i].className …我正在使用 PHP，我收到了这条消息
无法修改标头信息 - 标头已由（输出开始于
执行如下代码后：
<?php
if (condition) {
if (condition) {
//Statement
}
$to = $_POST['email'];
$subject = "Registration Confirmation";
$message = '
<html>
<head>
<title>'.$subject.'</title>
</head>
</html>
';
// Always set content-type when sending HTML email
$headers = "MIME-Version: 1.0" . "\r\n";
$headers .= "Content-type:text/html;charset=UTF-8" . "\r\n";
// More headers
$headers .= 'From: <contact@ghanalifestore.com>' . "\r\n";
mail($to,$subject,$message,$headers);
header('Location: reset.php?username='.$username);
exit();
}else {
//statement
}
根据Stackoverflow 的这个答案，我认为错误来自变量中的 HTML 代码$message。我真的不知道如何修改该变量的内容以避免错误。
请帮我解决这个问题。
我是javascript的新手并且非常享受它,但我现在面临的一个问题对我来说有点混乱.
我写下面的代码用类隐藏所有元素male,但它不起作用.当我更换类male由id代码开始工作.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<form class="form-register" method="post" action="">
<div class="form-title-row">
<h1>Create an account</h1>
</div>
<div class="form-row">
<label>
<span>Gender</span>
<select name="register_as" id="dropdown">
<option value="none">Select One</option>
<option value="male">Male</option>
<option value="female">Female</option>
</select>
</label>
</div>
<div class="form-row" class="male">
<label>
<span>First Name</span>
<input type="text" name="firstname">
</label>
</div>
<div class="form-row" class="male">
<label>
<span>Last Name</span>
<input type="text" name="lastname">
</label>
</div>
<div class="form-row">
<label>
<span>Email</span>
<input type="email" name="email">
</label>
</div>
<div class="form-row">
<button type="submit">Register</button>
</div>
<script>
$(document).ready(function(){
$(".male").hide();
$("#dropdown").change(function(){
if ($("#dropdown").val() == …