Question

我现在已经搜索了几个小时以了解我如何才能做到这一点，但遗憾的是一切都无济于事。

我正在尝试发送输入的信息，以便它可以通过MySQL运行并获取信息，然后在屏幕上的表格中回显。

问题（据我所知）必须使用我的JQuery代码：

    $("#btnCheckNoteIDs").click(function(){
    var noteUser = $("#noteUser").val();
    var noteDate = $("#noteDate").val();
  $("#LoadNoteIDs").load('check_noteIDs.php?noteDate='+noteDate + "&noteUser="+noteUser);
});

我的php代码如下：

$result = mysqli_query($con,"
SELECT ID, ClientID, Note, ToDoDate, CaseID
FROM ToDoNotes
WHERE ToDoStatus='0' and Deleted='0' and `ToDoDate`='".$_GET['noteIDsDate']."' and User='".$_GET['noteIDsDate']."'");
while($row = mysqli_fetch_array($result))
{
$ID = $row['ID'];
$ClientID = $row['ClientID'];
$Note = $row['Note'];
$ToDoDate = $row['ToDoDate'];
$CaseID = $row['CaseID'];
}

?>

<table class="table table-striped">
<thead>
<tr>
<th>Note ID</th>
<th>Client ID</th>
<th>Case ID</th>
<th>Note</th>
<th>To Do Date</th>
</tr>
</thead>
<tbody>
<tr>
<td><? echo $noteID; ?></td>
<td><? echo $ClientID; ?></td>
<td><? echo $CaseID; ?></td>
<td><? echo $Note; ?></td>
<td><? echo $ToDoDate; ?></td>
</tr>
</tbody>
</table>

这里有人可以提供任何帮助吗？任何帮助将不胜感激！

谢谢！

Answer 1

假设您有如下的form.php：

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Document</title>
</head>
<body>
    <form method="post" id="btnCheckNoteIDs">
        <label> User</label>
            <input type="text" name="user" id="noteUser"><br>
        <label>Date</label>
            <input type="date" name="date" id="noteDate"><br>
        <input type="submit" value="submit">
    </form>
    <br>
    <div id="LoadNoteIDs"></div>    
</body>
</html>

<script  src="https://code.jquery.com/jquery-1.12.4.min.js"
              integrity="sha256-ZosEbRLbNQzLpnKIkEdrPv7lOy9C27hHQ+Xp8a4MxAQ="
              crossorigin="anonymous"></script>
<script>

        $(document ).ready(function() { //make sure document is ready     
          $('#btnCheckNoteIDs').submit(function(e){
          e.preventDefault();//to prevent default behaviour

          var noteUser = $("#noteUser").val();
          var noteDate = $("#noteDate").val();
          if(noteUser==undefined){
            alert('No noteuser');
            return false;
          }else if(noteDate==undefined){
            alert('no note date');
             return false;//stop sending
          }
        //to convert %2F, use decodeURIComponent
          var param= 'noteDate='+decodeURIComponent(noteDate) + '&noteUser='+noteUser;
            $.ajax({
              url: "check_noteIDs.php",
              data: param,
              type: "post",
              success: function(data){
                    $('#LoadNoteIDs').html(data);
              }
            });
           });
         });

</script>

在check_noteIDs.php中的

，它必须位于放置form.php的同一文件夹中

require_once "dbconfig.php";

 $user= isset($_REQUEST['noteUser'])? $_REQUEST['noteUser']:'';
 $date = isset($_REQUEST['noteDate'])? date('Y-m-d', strototime($_REQUEST['noteDate'])):'';
 //You can validate here.

 $result = mysqli_query($con," SELECT ID, ClientID, Note, 
                                ToDoDate, CaseID
                                FROM `ToDoNotes`
                                WHERE `ToDoStatus`='0' AND `Deleted`='0' 
                                AND `ToDoDate`='$date' 
                                AND `User`='$user'");

$row = mysqli_fetch_assoc($result);

<table class="table table-striped">
  <thead>
    <tr>
       <th>Note ID</th>
       <th>Client ID</th>
       <th>Case ID</th>
       <th>Note</th>
       <th>To Do Date</th>
   </tr>
  </thead>
  <tbody>
    <tr>
       <td><?php echo $row['ID']; ?></td>
       <td><?php echo $row['ClientID']; ?></td>
       <td><?php echo $row['CaseID']; ?></td>
       <td><?php echo $row['Note']; ?></td>
      <td><?php echo $row['ToDoDate']; ?></td>
   </tr>
  </tbody>
 </table>
<?php
mysqli_free_result($result);    
mysqli_close($con);
?>

Answer 2

我认为您的问题是表单中的提交按钮，在脚本执行之前提交表单。如果是这种情况，您需要更新脚本，如下所示：

$("#btnCheckNoteIDs").click(function(event){
    event.preventDefault();
    var noteUser = $("#noteUser").val();
    var noteDate = $("#noteDate").val();
    $("#LoadNoteIDs").load('check_noteIDs.php?noteDate='+noteDate + "&noteUser="+noteUser);
});

JQuery在url中发送多个变量

2 个答案: