php 不是有效的 AllXsd 值
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not a valid AllXsd value
提问by Loreto Gabawa Jr.
I got this from a Soap client request:
我从 Soap 客户端请求中得到了这个：
Exception: SoapFault exception: [soap:Client] Server was unable to read request. ---> There is an error in XML document (2, 273). ---> The string '2010-5-24' is not a valid AllXsd value. in /path/filinet.php:21 Stack trace: #0 [internal function]: SoapClient->__call('SubIdDetailsByO...', Array) #1 /path/filinet.php(21): SoapClient->SubIdDetailsByOfferId(Array) #2 {main}
异常：SoapFault 异常：[soap:Client] 服务器无法读取请求。---> XML 文档中存在错误 (2, 273)。---> 字符串“2010-5-24”不是有效的 AllXsd 值。在 /path/filinet.php:21 堆栈跟踪：#0 [内部函数]：SoapClient->__call('SubIdDetailsByO...', Array) #1 /path/filinet.php(21)：SoapClient->SubIdDetailsByOfferId(数组）#2 {main}
Seems like I am sending an incorrect value, how do I format my value in an AllXsd in php?
好像我发送的值不正确，如何在 php 的 AllXsd 中格式化我的值？
Here is my code:
这是我的代码：
<?php
$start = isset($_GET['start']) ? $_GET['start'] : date("Y-m-d");
$end = isset($_GET['end']) ? $_GET['end'] : date("Y-m-d");
//define parameter array
$param = array('userName'=>'user', 'password'=>'pass', 'startDate' => $start, 'endDate' => $end, 'promotionId' => '');
//Get wsdl path
$serverPath = "https://webservices.filinet.com/affiliate/reports.asmx?WSDL";
//Declare Soap client
$client = new SoapClient($serverPath);
try {
//make the call
$result = $client->SubIdDetailsByOfferId($param);
//If error found display error
if(isset($fault))
{
echo "Error: ". $fault;
}
//If no error display response
else
{
//Used to display raw XML in the Web Browser
header("Content-Type: text/xml;");
//SubIdDetailsResult = XML results
echo $result->SubIdDetailsByOfferIdResult;
}
}
catch(SoapFault $ex) {
echo "<b>Exception:</b> ". $ex;
}
unset($client);
?>
回答by ólafur Waage
AllXsd values look something like this IIRC
AllXsd 值看起来像这个 IIRC
2010-05-24T18:13:00
2010-05-24T18:13:00
回答by dabaR
You need to use the ISO 8601 date format date('c', strtotime($my_date));
您需要使用 ISO 8601 日期格式 date('c', strtotime($my_date));
回答by Harry Slaughter
Cut to the chase and use
切入正题并使用
date('c');
回答by Dejan
// set the default timezone to use. Available since PHP 5.1
date_default_timezone_set('UTC');
// get the date
$startDate = date("Y-m-d") . 'T' . date("H:i:s");
回答by Jarrod Nettles
The problem is with the date format of either $start or $end. Instead of just grabbing the data from the query string with $_GET and sending it over, you need to do some integrity checking to make sure that the date matches the required format
问题在于 $start 或 $end 的日期格式。与其只是使用 $_GET 从查询字符串中获取数据并将其发送过来，您还需要进行一些完整性检查以确保日期与所需的格式匹配
2010-05-24T13:46:00
Instead of using date("Y-m-d") try using:
而不是使用 date("Ymd") 尝试使用：
$startDate = date("Y-m-d") . 'T' . date("H:i:s");
