Question

尝试使用jquery datetimepicker，我可以让插件工作，但是，当我发布数据时，它不会在数据库中更新。其他所有内容都会更新，但日期/时间会更新。

注意：我意识到其中一些仍然是mysql，我正在努力更新整个网站。

edit_release_form.php，我有

<script src="https://code.jquery.com/jquery-1.12.4.js"></script>
<script src="https://code.jquery.com/ui/1.12.0/jquery-ui.js"></script>
<script type="text/javascript" src="http://vmcnucmed.cvm.umn.edu/includes/js/jquery-ui-timepicker-addon.js"></script>
<link rel="stylesheet" href="http://vmcnucmed.cvm.umn.edu/includes/jquery-ui-timepicker-addon.css">
<link rel="stylesheet" href="//code.jquery.com/ui/1.12.0/themes/base/jquery-ui.css">
<script>
  $( function() {
    $('#datetime_released').datetimepicker({
    controlType: 'select',
    oneLine: true,
    timeFormat: 'hh:mm:ss tt'
});
  } );
  </script>

我的意见：

 <input tabindex="1" id="datetime_released" type="text" name="datetime_released" >

post.php中

<?php
// Database Conection Parameters
define('DB_SERVER', "hostname");
define('DB_USER', "username");
define('DB_PASSWORD', "password");
define('DB_TABLE', "database");

// Connection to the Database
$mysqli = mysqli_connect(DB_SERVER, DB_USER, DB_PASSWORD, DB_TABLE);
$mysqli->set_charset("utf8");
$mysqli->query("SET NAMES 'utf8'");

if (mysqli_connect_errno($mysqli)) {
    trigger_error('Database connection failed: '  . mysqli_connect_error(), E_USER_ERROR);
}

$datetime_released = $_POST['datetime_released']; 
$release_probeSN = $_POST['release_probeSN']; 
$release_reading = $_POST['release_reading'];
$release_reading_loc = $_POST['release_reading_loc'];
$releasedby = $_POST['releasedby'];
$id = $_POST['id'];


// Query
$sql = "
UPDATE xxx
SET datetime_released=?, 
release_probeSN=?, 
release_reading=?, 
release_reading_loc=?,
releasedby=?

WHERE patientdoseID=?";
if(!($stmt = $mysqli->prepare($sql)))
{
  die("Unable to prepare statement");
}
else
{
  $stmt->bind_param("sissii", $datetime_released, $release_probeSN, $release_reading, $release_reading_loc, $releasedby, $id);    
  if($stmt->execute())
  {
    echo 'Patient Updated Successfully. <br /><br />Database ID: <br>';
          header('Location: http://vmcnucmed.cvm.umn.edu', 
            TRUE, // rewrite existing Location header
            302 // set status code
);
  }
  else
  {
    die("Update failed");
  }
}

mysqli_close($mysqli);

?>

Answer 1

这是你如何将发布的js日期转换为mysql格式。 PHP DateTime对象可以接受各种格式，通常可以正确转换它们。为了便于阅读，您可以轻松地将其打包成一个函数。

<?php
// set your timezone somewhere prior to creating a DateTime object
// to prevent local vs UTC issues
date_default_timezone_set('America/Chicago');

// mysql date format
$format = "Y-m-d H:i:s";

// an example input
$input = '08/25/2016 05:19:20 am';

// convert input date into a PHP DateTime object
$my_date = new DateTime($input);

// create a string in the desired mysql format
$output = $my_date->format($format);

echo $output;

// outputs  2016-08-25 05:19:20  - just the way mysql likes it

发布时数据库中的日期时间选取器值不会更新

1 个答案: