PHP sql UPDATE not working
PHP sql UPDATE not working
我试图更新我的数据库中的表,把我的更新查询不工作。我已经检查了很多次代码,但我看不出有什么问题。我调用一个php文件和发送数据到它与javascript。什么好主意吗?
这是我的php文件
include 'db-connect.php';
$db->autocommit(false);
$postdata = file_get_contents("php://input");
$request = json_decode($postdata);
$id = $db->real_escape_string($request->ID);
$from = $db->real_escape_string($request->From);
$to = $db->real_escape_string($request->To);
$from_ampm = $db->real_escape_string($request->FromAMPM);
$to_ampm = $db->real_escape_string($request->ToAMPM);
$sql = "update OpeningHours set From = '$from', To = '$to', FromAMPM = '$from_ampm', ToAMPM = '$to_ampm' ";
$sql .= "where ID = '$id'";
$res = $db->query($sql);
$message = array();
if($db->affected_rows <= 0)
{
$db->rollback();
$message['error'] = "Could not edit. Contact IT manager.";
echo json_encode($message);
die();
}
$db->commit();
$db->close();
$message['error'] = "";
$message['success'] = "Edit Success!";
echo json_encode($message);
这是javascript部分:
var request = $http({
method: "post",
url: "updateOpeningHoursItem.php",
data: {
ID: $scope.openingHoursID,
From: $scope.from,
To: $scope.to,
FromAMPM: $scope.fromAMPM,
ToAMPM: $scope.toAMPM
},
headers: { 'Content-Type': 'application/x-www-form-urlencoded' }
});
request.success(function (data) {
if(data.error === "")
{
$scope.emptyOpeningHoursFields();
$scope.resetOpeningHoursEditForm();
swal("Success", data.success, "success");
}
else
{
swal("Ops!", data.error, "error");
}
});
你好,
$host = 'localhost';
$user = 'root';
$password = '';
$database = 'youdatabasename';
$con = mysql_connect($host,$user,$password);
$res = mysql_select_db($database,$con);
$postdata = file_get_contents("php://input");
$request = json_decode($postdata);
$id = $db->real_escape_string($request->ID);
$from = $db->real_escape_string($request->From);
$to = $db->real_escape_string($request->To);
$from_ampm = $db->real_escape_string($request->FromAMPM);
$to_ampm = $db->real_escape_string($request->ToAMPM);
$sql = "UPDATE OpeningHours set From = '".$from."', To = '".$to."', FromAMPM = '".$from_ampm."', ToAMPM = '".$to_ampm."' where ID = '".$id."'";
$result = mysql_query($sql) or die(mysql_error());
if($result==0)
{
$message['error'] = "Could not edit. Contact IT manager.";
echo json_encode($message);
die();
}
请尝试与此代码可能是工作,如果有任何遗漏的东西。
您在UPDATE
语句中使用了保留字,因此您必须使用反引号正确地转义它们。
修改SQL语句为
$sql = "UPDATE OpeningHours
SET `From` = '".$from."',
`To` = '".$to."',
`FromAMPM` = '".$from_ampm."',
`ToAMPM` = '".$to_ampm."'
WHERE `ID` = '".$id."'";
您还在查询中使用了保留词。把它们放在反引号里。您可以通过回显查询来调试它。查询应该是这样的。
$sql = "UPDATE OpeningHours set From = '".$from."', To = '".$to."', FromAMPM = '".$from_ampm."', ToAMPM = '".$to_ampm."' where ID = '".$id."'";
$result = mysql_query($sql) or die(mysql_error());
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