[英]"date(): It is not safe to rely on the system's timezone settings..."
[英]PHP DateTime Error| date_create() [function.date-create]: It is not safe to rely on the system's timezone settings
我在重新加载页面时将日期发布到数据库中以获取日期时遇到问题。 我希望它自动为最后一个条目加上时间戳。
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<title>Location Tracker</title>
</head>
<body>
<hl> ACW Location Tracker </hl>
<?php
$server = 'SQL2008.net.dcs.hull.ac.uk';
$connectionInfo = array("Database"=>"rde_531545");
$conn = sqlsrv_connect($server,$connectionInfo);
$query='create table Location ';
$query .= '(Username int NOT NULL IDENTITY(500, 23), First_Name varchar(50) NOT NULL, Surname varchar(50) NOT NULL, Current_Location varchar(50) NOT NULL, Date datetime NOT NULL, PRIMARY KEY (Username))';
$result = sqlsrv_query($conn, $query);
if (!$result)
{
if( ($errors = sqlsrv_errors() ) != null)
{
foreach( $errors as $error ) {
echo "<p>Error: ".$error[ 'message']."</p>";
}
}
}
else {
echo "<p>DB successfully created</p>";
}
sqlsrv_close($conn);
$connectionInfo = array( "Database"=>"rde_531545");
$conn = sqlsrv_connect($server,$connectionInfo);
$insert_query = "INSERT INTO Location (First_Name, Surname, Current_Location, Date) VALUES (?, ?, ?,? )";
$params = array("John","Doe","Hull", Date);
$result = sqlsrv_query($conn,$insert_query,$params);
$params = array("Jane","Doe","Hull", Date);
$result = sqlsrv_query($conn,$insert_query,$params);
$LocationQuery='SELECT Username, First_Name, Surname, Current_Location, Date FROM Location';
$results = sqlsrv_query($conn, $LocationQuery);
if ($results) while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
{
echo '<p>'.$row['Username'].' '.$row['First_Name'].' '.$row['Surname'].' '.$row['Current_Location'].' '.$row['Date'].'</p>';
}
?>
</body>
</html>
我得到的错误是:
注意:使用未定义的常量日期-在第44行的C:\\ RDEUsers \\ NET \\ 531545 \\ Location.php中假定为'Date'
警告:date_create()[function.date-create]:依靠系统的时区设置并不安全。 您需要使用date.timezone设置或date_default_timezone_set()函数。 如果您
enter code here
这些方法的任何enter code here
但仍然收到此警告,则很可能拼写了时区标识符。 我们在第49行的C:\\ RDEUsers \\ NET \\ 531545 \\ Location.php中将“欧洲/伦敦”选择为“ 0.0 /无DSTenter code here
”,而不是可捕获的致命错误:第51行的C:\\ RDEUsers \\ NET \\ 531545 \\ Location.php中无法将类DateTime的对象转换为字符串
您正在发送Date
类,而不是发送字符串'2017-02-03'。
该错误位于$params = array(....., Date);
在这段代码的第二行中:
$insert_query = "INSERT INTO Location (First_Name, Surname, Current_Location, Date) VALUES (?, ?, ?,? )";
$params = array("John","Doe","Hull", Date);
您需要创建一个Date对象$date = new DateTime();
并进入查询,您需要使用format
函数从其中提取字符串:
$dateStr = $date->format(''Y-m-d H:i:s'');
然后,在您的参数中使用$ dateStr: $params = array("John","Doe","Hull", $dateStr);
编辑1:
$date = new DateTime();
$dateStr = $date->format('Y-m-d H:i:s');
$params = array("John","Doe","Hull", $dateStr);
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