从表单创建数组并更新mysql [英] Create array from form and update mysql

问题描述

这是我第一次尝试为单个表列发布多条记录，需要一些帮助.

This is my first attempt to post multiple records for a single table column and would need some help.

首先，如何将这些记录发布到数组中?

First, how can I post these records in an array?

<?php
  <form action="set_order.php" method="POST">
  <table>
    <tr>
  $query = mysql_query("SELECT * FROM table ORDER BY order");
  while ($result = mysql_fetch_assoc($query)) {
?>
      <td><?php echo $result['order']; ?></td> //show current order
      <td><input type="text" name="order[]" value="<?php echo $result['order']; ?>" /></td>  //input new order
      <td><input type="hidden" name="id[]" value="<?php echo $result['id']; ?>" /></td> //send related id
    </tr>
    <tr>
      <td colspan ="2"><input type="submit" value="save" />
    </tr>
  </table>
  </form>

第二个问题是如何将数组插入到表中.

Second question is how to insert array to table.

table

id | order
1  |  3
2  |  4
3  |  2
4  |  1

我发现了这个:从表单发布数组以更新 mysql 表

<?php
foreach ($_POST['id'] as $id) {
$order = $_POST['order']; // here is the problem
echo $id . ',' . $order . '<br />';
}
?>

但我无法使用 ECHO 获得结果.

But I can't get the result using ECHO.

我明白了:

1, Array
2, Array
3, Array
4, Array

如果我能做到这一点，我想我可以使用 FOREACH 相应地更新表格.

If I can make that works, I think I can manage to update the table accordingly using FOREACH.

推荐答案

$_POST['id'] 和 $_POST['order'] 都是数组，迭代 id 是个好主意，但随后你需要检索每次迭代时 $_POST['order'] 的对应元素.

Both $_POST['id'] and $_POST['order'] are arrays, it's a good idea to iterate over id's but then you need to retrieve the corresponding element of $_POST['order'] at each iteration.

当 HTTP 请求到达服务器时，两个数组同时填充，匹配的元素应该共享相同的键，这样应该可以工作:

As both arrays are filled in simultaneously when the HTTP request comes to the server, the matching elements should share the same keys, so that should work:

<?php
foreach ($_POST['id'] as $key=>$id) {
    $order = $_POST['order'][$key];
    echo $id . ',' . $order . '<br />';
}
?>

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