我的代码PHP有问题,我的数据库是Mongodb:
我收到此错误:
警告:date()期望参数2很长,
我的代码是:
<?php echo date('g:i a, F j', $comment['posted_at']); ?>
我试图name从这块JSON中获取值:
{
"tournament": {
"id": 1605872,
"name": "eBotMatchMakerTest",
"event_id": null,
"participants": [
{
"participant": {
"id": 24899481,
"tournament_id": 1605872,
"name": "Team1",
"seed": 1,
"active": true,
"created_at": "2015-04-18T07:05:48.601-05:00",
"updated_at": "2015-04-18T07:05:48.601-05:00",
"invite_email": null,
"final_rank": null,
"misc": null,
"icon": null,
"on_waiting_list": false,
"invitation_id": null,
"group_id": null,
"checked_in_at": null,
"challonge_username": null,
"challonge_email_address_verified": null,
"removable": true,
"participatable_or_invitation_attached": false,
"confirm_remove": true,
"invitation_pending": false,
"display_name_with_invitation_email_address": "Team1",
"email_hash": null,
"username": null,
"display_name": "Team1",
"attached_participatable_portrait_url": null,
"can_check_in": false,
"checked_in": false,
"reactivatable": false
}
}
],
"matches": [ …我不只是在谈论验证卡号的格式,我正在谈论实际通过authorize.net(例如)发送它以验证卡信息.有没有办法通过authorize.net进行交易？
我已经看到了这个链接,而不是我正在寻找的东西.我想将信用卡信息发送到authorize.net进行验证/验证,而不仅仅是验证格式.
我正在创建一个函数来检查哪个日期在一个充满日期的数据库表中小于当前日期.和过去一样.
我有3个日期来测试函数,以及它们后面的输出:
上个月的
日期:28-04-2015 16:32:00
日期尚未到来:11-06-2015 13:12:00
日期:上一周:04-05-2015 09:45:00
$dateNow = date('d-m-Y H:i:s'); //Current
$deadlineDate = "28-04-2015 16:33:18";
if($deadlineDate < $dateNow){ //If date from last month is smaller then the current date
echo '<tr class="overdue">'; //Overdue class gives that tr an red background color to mark it
echo '<td>'.$deadlineDate.' is smaller then '.$dateNow.'</td>';
}else{
echo '<tr>';
echo '<td>'.$deadlineDate.' is bigger dan '.$dateNow.'</td>';
}
</tr>
输出:
28-04-2015 16:32:00 返回更大
11-06-2015 13:12:00 返回较小 …
其他一切都应该没问题,但我仍然有问题显示else功能.理论上我相信一切都应该有效.我查看了所有语法,一切看起来都很好,一切都已正确关闭.我想要做的是,如果$_GET['cat'] == 1, 2, or 3(这就是为什么我使它成为一个数组)是有效的,那么我希望能够拉出那个特定的类别,如果没有那么我希望它能一起显示所有的类别.很像WordPress的功能.
<?php
include('config.inc.php');
// security check to make sure it's not grabbing anything extra that's not there
$get_cat_id = mysqli_query($mysql_conn, "SELECT `id` FROM `categories`");
$cat_array = array();
while($row = mysql_fetch_assoc($get_cat_id)) {
$cat_array[] = $row['id'];
}
echo $cat_array[];
// checks to see if it's calling for a specific category
if(isset($_GET['cat']) && ($_GET['cat'] == $cat_array)) {
$fetch_post_content = mysqli_query($mysql_conn, "SELECT * FROM `posts` WHERE `category_id` = '".$_GET['cat']."' ORDER BY `id` DESC");
$fetch_category_content = mysqli_query($mysql_conn, "SELECT …我有一个简单的Vars.php页面:
<?php
//Vars.php
$WeekOfDateSelected = date('l, m/d/Y', strtotime($MonthYear));
$NextSundayOfDateSelected = date('l, m/d/Y', strtotime('this Sunday', strtotime($WeekOfDateSelected)));
?>
我有另一个包含Vars.php的PHP并构建一个表:
<html>
<?php
//AnalyticsTest.php
include($_SERVER['DOCUMENT_ROOT']."/~/~/~/~/~/Vars.php");
function WeekTable() {
echo "<table id=\"a\">
<tr>
<th style=\"text-align: center;\"><a href=\"#\">< previous week</a></th>
<th colspan=\"4\" style=\"text-align: center;\"><h2>Week of ";
echo $WeekOfDateSelected;
echo " - ";
echo $NextSundayOfDateSelected;
echo "</th>
<th style=\"text-align: center;\"><a href=\"#\">next week ></a></th>
</tr>
</table>";
}
?>
</html>
基本上,当我WeekTable()正确调用所有输出时,除了PHP变量$WeekOfDateSelected和$NextSundayOfDateSelected输出空白.
我正在尝试按字母顺序对字符串进行排序.我以为我可以将一个字符串分解为一个数组并对其进行排序,但是回声没有返回任何内容.
$schools = "high*low*other*";
$schools = explode("*", $schools);
$schools = sort($schools);
echo $schools[0];
我有$first和$second.他们可以有价值0或1.
if ( $first AND $second ) {
// True
} else {
// False
}
我的心(和谷歌搜索)告诉我,这个结果是真实的,只有当$first == 1和$second == 0,反之亦然.但是当这两个变量都是时,结果都是正确的1.
我不明白它是如何工作的.
我的代码看起来像这样:
$muted = 'true'; //Note this is only for testing
if ($muted == 'false' || $muted == '' || $do_not_text == '' || $do_not_text =='false'){
//do this first thing
}
else{
//do something else
}
我不能让我else跑.搞砸了什么语法？
与此相关,我在我的数据库中存储一个值(这将是$muted我的实际代码中设置的值),它是'true','false'或''.我应该将这些数据类型存储为什么？目前,我正在使用VARCHAR,但我怀疑这是问题的一部分.
我不小心在PHP中做了一个除法运算符,如下所示:
$a = 012; //or 0012 or even 00012
echo $a/4;
echo gettype($a/4);
我得到的输出是:2.5 double.//我不知道为什么输出2.5.
我再试一次:
$a=12;
echo $a/4;
echo gettype($a/4);
这将输出3 Integer.
我已阅读文档https://secure.php.net/manual/en/language.operators.arithmetic.php但仍无法理解.
有人可以帮我吗？
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