Question

我正在尝试使用一个按钮将可见性从0设置为1来更新我的数据库。没有任何语法错误，但是由于某种原因它不会更改数据库值。我的数据库示例：（可见性为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 }
      ?>

Answer 1

您必须将按钮包装到<form>中并设置其type to submit。此外，您需要将records id与它一起传递。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' ");
}
?>

如果您可以使用此工具，则应该搜索prepared statements。

使用按钮更新数据库表

1 个答案: