Question

我有这两个表，分别是table1.php和table2.php

我可以使用此脚本将数据传递给模式，该脚本位于table1.php中，并且可以通过模式将数据传递至table2.php，我现在关心的是如何将数据直接传递至table2.php的SQL查询？由于它的方式，我只能使用传递到HTML元素模式中的数据。有没有办法传递模式值并将其转换为PHP的SQL查询？

$(function(){
  $("body").on('click', '.edit', function (e){
    e.preventDefault();
    $('#edit').modal('show');
    var id = $(this).data('id');
    getRow(id);
  });

我正在尝试将选定的数据行传递到另一个PHP页面，以便它将成为过滤第二个PHP表的参数。

这是我的 table1.php

<form type="POST" action="table2.php">
              <table id="example1" class="table table-bordered">
                <thead>
                  <th>Series No.</th>
                  <th>Account Type</th>
                  <th>Tools</th>
                </thead>
                <tbody>
                  <?php
                    $sql = "SELECT * FROM accounttype";
                    $query = sqlsrv_query($conn, $sql, array(), array("Scrollable" => SQLSRV_CURSOR_KEYSET));
                    while($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)){
                      echo "
                        <tr>
                        <td>".$row['seriesno']."</td>
                          <td>".$row['accounttype']."</td>
                          <td>
                            <button class='btn btn-success btn-sm edit btn-flat' data-id='".$row['seriesno']."'><i class='fa fa-edit'></i> Edit</button>
                            <button class='btn btn-danger btn-sm delete btn-flat' data-id='".$row['seriesno']."'><i class='fa fa-trash'></i> Delete</button>
                          </td>
                        </tr>
                      ";
                    }
                  ?>
                </tbody>
              </table>
              </form>

这将是我的 table2.php

<table id="example2" class="table table-bordered">
                <thead>
                  <th>Series No.</th>
                  <th>Account Type</th>
                </thead>
                <tbody>
                  <?php
                    $id=$_POST['seriesno'];
                    $sql = "SELECT * FROM accounttype where seriesno='$id'";
                    $query = sqlsrv_query($conn, $sql, array(), array("Scrollable" => SQLSRV_CURSOR_KEYSET));
                    while($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)){
                      echo "
                        <tr>
                        <td>".$row['seriesno']."</td>
                          <td>".$row['accounttype']."</td>
                        </tr>
                      ";
                    }
                  ?>
                </tbody>
              </table>

问题在于 table2.php 无法收到 $ _ POST ，这似乎是问题所在？

编辑：这是我的模式（ table2.php ），我需要的是将数据传递到表中以便$id=$_POST['seriesno'];可以正常工作。

<div class="modal fade" id="edit">
    <div class="modal-dialog" style="width:100%;">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-label="Close">
                    <span aria-hidden="true">&times;</span></button>
            </div>
            <div class="modal-body">
                    <input type="hidden" class="decid" id="id" name="id">

  <table id="example2" class="table table-bordered">
                <thead>
                  <th>Series No.</th>
                  <th>Account Type</th>
                </thead>
                <tbody>
                  <?php
                    $id=$_POST['seriesno'];
                    $sql = "SELECT * FROM accounttype where seriesno='$id'";
                    $query = sqlsrv_query($conn, $sql, array(), array("Scrollable" => SQLSRV_CURSOR_KEYSET));
                    while($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)){
                      echo "
                        <tr>
                        <td>".$row['seriesno']."</td>
                          <td>".$row['accounttype']."</td>
                        </tr>
                      ";
                    }
                  ?>
                </tbody>
              </table>
            </div>
        </div>
    </div>
</div>

Answer 1

所以..问题是您必须在同一页面上提交：并在同一页面的模式窗口中解析$_POST数组，因为它包含在此处。试试这个：

table1.php：

<form method="POST" action="table1.php">
          <table id="example1" class="table table-bordered">
            <thead>
              <th>Series No.</th>
              <th>Account Type</th>
              <th>Tools</th>
            </thead>
            <tbody>
              <?php
                $sql = "SELECT * FROM accounttype";
                $query = sqlsrv_query($conn, $sql, array(), array("Scrollable" => SQLSRV_CURSOR_KEYSET));
                while($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)){
                  echo "
                    <tr>
                    <td>".$row['seriesno']."</td>
                      <td>".$row['accounttype']."</td>
                      <td>
                        <button class='btn btn-success btn-sm edit btn-flat' data-id='".$row['seriesno']."'><i class='fa fa-edit'></i> Edit</button>
                        <button class='btn btn-danger btn-sm delete btn-flat' data-id='".$row['seriesno']."'><i class='fa fa-trash'></i> Delete</button>
                      </td>
                    </tr>
                  ";
                }
              ?>
            </tbody>
          </table>
          </form>

table2.php

<div class="modal fade" id="edit">
    <div class="modal-dialog" style="width:100%;">
        <div class="modal-content">
            <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-label="Close">
                    <span aria-hidden="true">&times;</span></button>
            </div>
            <div class="modal-body">
                    <input type="hidden" class="decid" id="id" name="id">

  <table id="example2" class="table table-bordered">
                <thead>
                  <th>Series No.</th>
                  <th>Account Type</th>
                </thead>
                <tbody>
                  <?php
                    if(isset($_POST['seriesno'])){
                    $id=$_POST['seriesno'];
                    $sql = "SELECT * FROM accounttype where seriesno='$id'";
                    $query = sqlsrv_query($conn, $sql, array(), array("Scrollable" => SQLSRV_CURSOR_KEYSET));
                    while($row = sqlsrv_fetch_array($query, SQLSRV_FETCH_ASSOC)){
                      echo "
                        <tr>
                        <td>".$row['seriesno']."</td>
                          <td>".$row['accounttype']."</td>
                        </tr>
                      ";
                    }
                    }

                  ?>
                </tbody>
              </table>
            </div>
        </div>
    </div>
</div>

希望它能发挥作用

将所选的dataTable行传递到另一个php页中的另一个dataTable行

1 个答案: