Question

我使用BlueImp jQuery File Upload进行数据库集成，调整script found here。

我改变了handle_file_upload()这样的方法：

$sql = 'INSERT INTO '.$this->options['db_table'].' (`path`, `type`, `tab`, `item`) VALUES (?, ?, ?, ?)';
$query = $this->db->prepare($sql);
$query->bind_param(
    'ssii',
    $path = $this->options['upload_url'] . $file->name,
    $file->type,
    $file->tab,
    $file->item
);

但是，在$path = $this->options['upload_url'] . $file->name行上，我有一个Strict Standards: Only variables should be passed by reference错误

好吧，我不知道如何改变这个......

我试着看看其他答案，但它们不适合我的情况：请帮忙吗？感谢

Answer 1

由于它是通过引用传递的，因此变量需要在绑定之前存在

 $path = $this->options['upload_url'] . $file->name;
    $sql = 'INSERT INTO '.$this->options['db_table'].' (`path`, `type`, `tab`, `item`) VALUES (?, ?, ?, ?)';
    $query = $this->db->prepare($sql);
    $query->bind_param(
        'ssii',
        $path,
        $file->type,
        $file->tab,
        $file->item
    );

Answer 2

这应该有用......

$sql = 'INSERT INTO '.$this->options['db_table'].' (`path`, `type`, `tab`, `item`) VALUES (?, ?, ?, ?)';
$query = $this->db->prepare($sql);
$p = $this->options['upload_url'] . $file->name;
$query->bind_param(
    'ssii',
    $p,
    $file->type,
    $file->tab,
    $file->item
);

您在参数绑定中设置了$ path的值。所以它不会返回任何东西。

Answer 3

在此处阅读更多内容：Strict Standards: Only variables should be passed by reference

$sql = 'INSERT INTO '.$this->options['db_table'].' (`path`, `type`, `tab`, `item`) VALUES (?, ?, ?, ?)';
$query = $this->db->prepare($sql);
$upload_url = $this->options['upload_url']  . $file->name;
$query->bind_param(
    'ssii',
    $path = $upload_url,
    $file->type,
    $file->tab,
    $file->item
);

PHP错误：只应通过引用传递变量

3 个答案: