我在我的数据库中有两个表,我在两个不同的php页面中显示它们。连接两个表 - 逐行关联
如何从显示第二个表的页面中逐行显示第一个表的内容。
我想要的是:第二个表的每一行都有一个按钮,onclick会显示(抛出模式弹出)第一个表中的信息。
所以ROW1-表2的按键会告诉我只ROW1-表1的信息等等....
我能够实现每一行的按钮,并在弹出的,但我只能显示整个第一个表的信息,而不显示关联的单个行的信息。
<div class="container">
<div class="row text-center">
<div class="col-md-12 col-sm- hero-feature">
<div class="thumbnail">
<?php
include("../includes/connection.php");
if ($link->connect_errno > 0) {
die('Unable to connect to database [' . $link->connect_error . ']');
}
if (isset($_POST['update'])) {
$results = $link>query("UPDATE job SET status='$_POST[status]', priority='$_POST[priority]' WHERE id='$_POST[hidden]'");
$results = $link>query("UPDATE **table2** SET status='$_POST[status]' WHERE id='$_POST[hidden]'");}
$sql = "SELECT * from job";
if (!$result = $link->query($sql)) {
die('There was an error running the query [' . $link->error . ']');
}
echo "…………./* Get field information for all columns */………… "
while ($row = $result->fetch_assoc()) {
echo "<form action='' method=post>";
echo "<tr class='info'>
<input type=hidden name=hidden value=" . $row['id'] . ">
<td>" . $row['id'] . "</td>
<td>" . $row['device'] . "</td>
<td>" . $row['model'] . "</td>
<td>" . $row['problem'] . "</td>
<td><select class='form-control col-sm-10' id='status' name='status'>
<option value=" . $row['status'] . " >" . $row['status'] . "</option>
<option value='new'>New</option>
<option value='progress'>Progress</option>
<option value='wait'>Wait</option>
<option value='done'>Done</option>
<option value='close'>Close</option>
</select></td>
<td><select class='form-control col-sm-10' id='priority' name='priority'>
<option value=" . $row['priority'] . " >" . $row['priority'] . "</option>
<option value='high'>High</option>
<option value='medium'>Medium</option>
<option value='low'>Low</option>
</select></td>
<td>" . $row['status'] . "</td>
<td>" . $row['priority'] . "</td>
**<td> <button type='submit' class='btn btn-primary btn-sm' name='update'>Update</button></td>**
**<td> <a class='btn btn-primary btn-sm' data-toggle='modal' datatarget='#myModal'>Info</a></td>**
</tr>"; echo "</form>";}echo " </tbody>
</table>";
?>
<div class="container">
!-- Trigger the modal with a button -->
<!-- Modal -->
<div class="modal fade" id="myModal" role="dialog">
<div class="modal-dialog modal-lg">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Customer Information</h4>
</div> <div class="modal-body">
<?php
include("../includes/connection.php");
if ($link->connect_errno > 0) {
die('Unable to connect to database [' . $link->connect_error . ']');
}
$sql = "SELECT * from **table2**";
if (!$result = $link->query($sql)) {
die('There was an error running the query [' . $link->error . ']');
}
echo "
<table class='table'>
<thead><tr>";
/* Get field information for all columns */
while ($finfo = $result->fetch_field()) {
echo "<th>" . $finfo->name . "</th>";}echo "
</tr></thead><tbody>";
while ($row = $result->fetch_assoc()) {
echo "<tr class='info'>
<td>" . $row['id'] . "</td>
<td>" . $row['name'] . "</td>
<td>" . $row['mail'] . "</td>
<td>" . $row['number'] . "</td>
<td>" . $row['price'] . "</td>
<td>" . $row['paymenttype'] . "</td>
<td>" . $row['faktura'] . "</td>
<td>" . $row['date'] . "</td>
</tr>";}echo "
</tbody>
</table>";
?> </div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div></div></div></div></div></div></div>
<!-- Button trigger modal -->
2个问题。 1)table2的代码看起来像你所指的table1,因为它具有faktura字段,你可以发布其他表的代码,特别是你如何做按钮? 2)你已经发布的代码是单独的文件到第一个表的代码还是它们在同一个文件中? –
我更新了我的代码。在同一页面中,我收集了table2和一个div(模式弹出窗口)的php代码,其中php代码收集table2的按钮“Info”的table1 onclick。 – gigi