1

我被要求创建一个基于 PHP 和 mySQL 的时间表。

我创建了一个页面来插入数据和另一个页面来显示数据并编辑/删除显示的数据。

这是显示数据和编辑/删除它们的页面：

<?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.";
    }
}

据我了解，第 2 页中的 job[] 变量没有捕捉到第 1 页中用户提交的复选框，或者第一页没有发送数据，因为复选框值（可能我写错了）

编辑：我找到了答案，这里是正确的代码：

在第 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>

在处理页面 (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.";
    }
}

感谢任何试图帮助或阅读它的人！

4

2 回答 2

0

更改$job_code_query = $_POST['job[]'];为$job_code_query = $_POST['job'];

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

您必须修复上述定义并将数组转换为字符串，然后再将其添加到查询中

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.";
        }
    }
于 2013-08-22T08:34:39.610 回答
0

用那个替换你的第 29 行

$job_code_query = $_POST['job'];

您还应该使用 PDO 或 MySQLi 代替现在已弃用的 mysql_* 函数。使用 MySQLi 连接，您必须使用$dbconnection->query($sqlquery)并且没有mysql_query

于 2013-08-22T08:35:34.063 回答