用按钮更新数据库表

Mower 发表于 Dev

割草机

我试图用一个按钮更新我的数据库，将可见性从 0 设置为 1。没有任何语法错误，但由于某种原因它不会更改数据库值。我的数据库示例：（可见性是 tinyint，默认值为 0）

 id-   name-   visibility
---------------------------
1   -   John    -    1
---------------------------
2   -   Ben     -    1
---------------------------
3   -   Terry   -    0
---------------------------

和我的 php 代码：

<?php
$sql = "SELECT id, name, image, description, address, phone, phone2, email, job, visibility FROM cards";
$resultset = mysqli_query($conn, $sql) or die("database error:". mysqli_error($conn));          
while( $record = mysqli_fetch_assoc($resultset) ) {
?>
<div class="col-md-4" <?php if ($record['visibility'] == 1) echo " style='display: none';"; ?>>I want this to be hidden here</div>

<button type="button" class="btn btn-success" name="update">Accept</button>
<?php

if(isset($_POST['update'])){
$allowed = mysqli_query($conn," UPDATE cards SET visibility = '1' WHERE id = '$id' ");
}
?>
//html stuff here
 <?php }
      ?>

罗宾·吉利策

您必须将按钮包装成 a<form>并设置其type to submit. 此外，您需要通过records id它。试试这个例子：

<?php

$sql = "SELECT id, name, image, description, address, phone, phone2, email, job, visibility FROM cards";
$resultset = mysqli_query($conn, $sql) or die("database error:". mysqli_error($conn));          
while( $record = mysqli_fetch_assoc($resultset) ) { ?>

  <div class="col-md-4" <?php if ($record['visibility'] == 1) echo " style='display: none';"; ?>>

    Record name: <?php echo $record['name']; ?>
    <form action="" method="POST">
      <input value="<?php echo $record['id']; ?>" name="id">
      <button type="submit" class="btn btn-success" name="update">Set visibile</button>
    </form>

  </div>

<?php } ?> 

<?php
if(isset($_POST['update'])){
  $id = $_POST['id'];
  $allowed = mysqli_query($conn," UPDATE cards SET visibility = '1' WHERE id = '$id' ");
}
?>

如果你让它工作，你应该搜索关于prepared statements.

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