打开带有动态内容的jQuery UI对话框

Gla*_*den 5 javascript php mysql jquery jquery-ui

我有一个关于jQuery UI对话框的问题,并显示数据库中的动态内容.

所以我得到了一个web应用程序,我还需要创建一个管理模块来管理所有用户和其他信息.我创建了一个页面,显示列表中的所有用户,在每一行中我也创建了一个编辑按钮.我想这样做,当你按下用户的编辑按钮时,会打开一个对话框,在对话框中显示所有用户信息和内容.

所以我的问题是,最好的方法是什么?我正在考虑制作一个PHP页面,我执行MySQL查询并在对话框中显示,但我相信有更好的方法..

编辑:这是现在页面的代码.我添加了一个用于测试目的的小占位符对话框.

使用Javascript:

script type="text/javascript"> 
    jQuery(document).ready( function(){       
        jQuery(".edit-button").click( showDialog );

            //variable to reference window
            $myWindow = jQuery('#myDiv');

            //instantiate the dialog
            $myWindow.dialog({ height: 600,
                    width: 800,
                    modal: true,
                    position: 'center',
                    autoOpen:false,
                    title:'Bewerk therapeut',
                    overlay: { opacity: 0.5, background: 'black'}
                    });
            }

    );
//function to show dialog   
var showDialog = function() {
    $myWindow.show(); 
    //open the dialog
    $myWindow.dialog("open");
    }

var closeDialog = function() {
    $myWindow.dialog("close");
}
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PHP:

<?php
//LEFT OUTER JOIN Vragen ON Vragen.bsn_nummer = Gebruikers.bsn_nummer
include_once 'classes/class.mysql.php';
$db = new Mysql();
$dbUsers = new Mysql();

$db->Query("SELECT * FROM USERS_users ORDER BY username ASC");
$db->MoveFirst();

echo "<table>";
echo "<tr><th> </th><th> </th><th>BSN Nummer</th><th>Gebruikersnaam</th>       <th>Voornaam</th><th>Achternaam</th></tr>";
while(! $db->EndOfSeek()) {
$row = $db->Row();
$dbUsers->Query("SELECT * FROM Gebruikers WHERE user_idnr = '{$row->user_idnr}'");
$rowUser = $dbUsers->Row();
echo "<tr><td><a class='del-button' href='#'><img src='afbeeldingen/edit-delete.png' /></a></td>
    <td><a class='edit-button' href='#'><img src='afbeeldingen/edit.png' /></a>  </td>
    <td>".@$rowUser->bsn_nummer."</td>      
    <td>".@$row->username."</td>
    <td>".@$rowUser->voornaam."</td>
    <td>".@$rowUser->achternaam."</td></tr>";
    }
    echo "</table>";
?>
<div id="myDiv" style="display: none">
<p>Gebruiker bewerken</p>
</div>
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Ohg*_*why 6

不.听起来你说得对.

弹出窗口的占位符 - >

<div id="popup"></div>
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jQuery ui对话框 - >

$('#popup').dialog({
  autoOpen: 'false',
  modal: 'true',
  minHeight: '300px',
  minWidth: '300px',
  buttons: {
    'Save Changes': function(){
      $.ajax({
        url: 'path/to/my/page.ext',
        type: 'POST',
        data: $(this).find('form').serialize(),
        success: function(data){
          //some logic to show that the data was updated
          //then close the window
          $(this).dialog('close');
        }
      });
    },
    'Discard & Exit' : function(){
      $(this).dialog('close');
    }
  }
});
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现在已经创建了默认设置,从php文件发送数据的ajax请求,并更新'popup'div中的内容.

$('.edit').click(function(e){
  e.preventDefault();
  $.ajax({
    url: 'path/to/my/page.ext',
    type: 'GET',
    data: //send some unique piece of data like the ID to retrieve the corresponding user information
    success: function(data){
      //construct the data however, update the HTML of the popup div 
      $('#popup').html(data);
      $('#popup').dialog('open');
    }
  });
});
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在PHP页面中,构造一个要发送回的表单 - >

<?php
  if(isset($_GET['id'])){
    //build the query, do your mysql stuff
    $query = mysql_query(sprintf("SELECT * FROM sometable WHERE id = %d", $_GET['id']));
    //construct constant objects outside of the array
?>
  <form>
  <?php
    while($row = mysql_fetch_array($query)){
  ?>
    <tr>
      <td>
        <input type="text" name="<?php echo $row['id']?>" value="<?php echo $row['name'] ?>" />
      </td>
    </tr>   
  <?php 
    }
  ?>
  </form>
<?php
  }    
?>
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