Question

我被要求创建一个基于PHP和mySQL的时间表。

我创建了一个用于插入数据的页面和另一个用于显示数据和编辑/删除所显示数据的页面。

以下是显示数据并编辑/删除数据的页面：

<?php
    while($info = mysql_fetch_array( $data )) {
?>
<tr>
    <td> <input type="checkbox" name="job" id="<?php echo $info['job_code']?>" value="<?php echo $info['job_code']?>" /></td>
    <td><?php echo $info['job_code'] ?></td>
    <td><?php echo $info['job_desc'] ?> </td>
    <td><?php echo $info['job_client'] ?> </td>
    <td><?php echo $info['job_year'] ?> </td>
    <td><?php echo $info['job_month']?> </td>
    <td><?php echo $info['job_date']?> </td>
    <td><?php echo $info['job_category']?> </td>
    <td>EDIT</td>
    <td>DELETE</td>
</tr>
<?php } ?>

这是删除过程

 $this->db_connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

$admin_data_query = mysql_query("SELECT * FROM jobs")or die(mysql_error());
$job_code_query = $_POST['job'];
$job_code_query_items = 'IN ( ';
foreach ($job_code_query as $key => $value) {
    $job_code_query_items .= ($key > 0 ? ", '" . $value . "'" : "'" . $value . "'");
}
$job_code_query_items .= ' );';
while ($admin_data = mysql_fetch_array($admin_data_query)) {
    if (isset($_POST[$admin_data['job_code']])) {
        $admin_sql = "DELETE FROM jobs WHERE job_code = $job_code_query_items";
        $admin_query = mysql_query($admin_sql) or die(mysql_error());
        $this->messages[] = "Job deleted.";
    } else {
        $this->messages[] = "Failed to delete job.";
    }
}

根据我的理解，第2页中的job []变量没有捕获第1页中的用户提交复选框，或者第一页没有发送数据，因为复选框值（也许我写错了）

编辑：我找到了答案，这里是正确的代码：

第1页

<td>
    <input type="checkbox" name="<?php echo $info['job_code']?>"
           id="<?php echo $info['job_code']?>"
           value="<?php echo $info['job_code']?>"/>
</td>
<td><?php echo $info['job_code'] ?></td>
<td><?php echo $info['job_desc'] ?> </td>
<td><?php echo $info['job_client'] ?> </td>
<td><?php echo $info['job_year'] ?> </td>
<td><?php echo $info['job_month']?> </td>
<td><?php echo $info['job_date']?> </td>
<td><?php echo $info['job_category']?> </td>

处理页面（2）

 while ($admin_data = mysqli_fetch_array($admin_data_query)) {
    $job_code_query = $admin_data["job_code"];

    if (isset($_POST[$admin_data["job_code"]])) {

        $admin_sql = "DELETE FROM jobs WHERE job_code = '$job_code_query'";
        $admin_query = mysqli_query($this->db_connection, $admin_sql) or die(mysqli_error($this->db_connection));
        $this->messages[] = "Job deleted.";
    } else {
        $this->messages[] = "Failed to delete job.";
    }
}

感谢所有试图帮助或阅读它的人！

Answer 1

将$job_code_query = $_POST['job[]'];更改为$job_code_query = $_POST['job'];

但这会导致另一个问题，因为$job_code_query是一个数组，因此您无法将数组发送到查询$admin_sql ="DELETE FROM jobs WHERE job_code = '$job_code_query'";

你必须修复上面的定义，并在将数组添加到查询中之前将其转换为字符串

你的代码是：

$this->db_connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

    $admin_data_query = mysql_query("SELECT * FROM jobs")or die(mysql_error());
    $job_code_query = $_POST['job'];
    $job_code_query_items = 'IN ( ';
    foreach ($job_code_query as $key => $value) {
        $job_code_query_items .= ($key>0?", '".$value."'":"'".$value."'");
    }
    $job_code_query_items .= ' );';
    while ($admin_data = mysql_fetch_array($admin_data_query)) {
        if (isset($_POST['job_code'])) {
            $admin_sql ="DELETE FROM jobs WHERE job_code $job_code_query_items";
            $admin_query = mysql_query($admin_sql) or die(mysql_error());
            $this->messages[] = "Job deleted.";
        } else {
            $this->messages[] = "Failed to delete job.";
        }
    }

Answer 2

将第29行替换为

$job_code_query = $_POST['job'];

您还应该使用PDO或MySQLi而不是mysql_ *函数，这些函数现已弃用。使用MySQLi连接，您必须使用$dbconnection->query($sqlquery)而不是mysql_query

PHP获取要删除的自动sql数据

2 个答案: