最近用jquery-file-upload?來改善網(wǎng)站上傳的體驗

https://github.com/blueimp/jQuery-File-Upload/wiki/PHP-MySQL-database-integration
上傳時按照他的參考文檔，立馬就完成了，一開始也按照他的sql?架構(gòu)先試試

結(jié)果上傳后，也能成功插入，json傳回頁面一切正常！

但問題來了，他的sql?架構(gòu)...有個叫url

但作者好像在PHP的SQL中沒有處理

那我就改改吧，....

先新增了一些基本配置

$dir = $_COOKIE["uid"].'/'.date("Y").'/'.date("m").'/'.date("d").'/';$dirUP =  "../../../att/".$dir;$dirLink =  $dir;$options=array(    'upload_dir' => $dirUP,    'upload_url' => $dirLink,    'delete_type' => 'POST',    'db_host' => 'localhost',    'db_user' => 'root',    'db_pass' => '*****',    'db_name' => '*****',    'db_table' => 'files');

應(yīng)該就是這段了....

我嘗試多次，加入url字段都不成功 [原本的文檔代碼]

    protected function handle_file_upload($uploaded_file, $name, $size, $type, $error,            $index = null, $content_range = null) {        $file = parent::handle_file_upload(            $uploaded_file, $name, $size, $type, $error, $index, $content_range        );        if (empty($file->error)) {            $sql = 'INSERT INTO `'.$this->options['db_table']                .'` (`name`, `size`, `type`, `title`, `description`)'                .' VALUES (?, ?, ?, ? , ?)';            $query = $this->db->prepare($sql);            $query->bind_param(                'sisss',                $file->name,                $file->size,                $file->type,                $file->title,                $file->description            );            $query->execute();            $file->id = $this->db->insert_id;        }        return $file;    }

都給我顯示：
Warning: mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables in

這是什么意思，說我的數(shù)量有問題? 是指我加少了嗎?
我已經(jīng)改成...這樣，5處的type字段也都加了url也說是數(shù)量問題?

    protected function handle_file_upload($uploaded_file, $name, $size, $type,$url, $error,            $index = null, $content_range = null) {        $file = parent::handle_file_upload(            $uploaded_file, $name, $size, $type,$url, $error, $index, $content_range        );        if (empty($file->error)) {            $sql = 'INSERT INTO `'.$this->options['db_table']                .'` (`name`, `size`, `type`, `url`, `title`, `description`)'                .' VALUES (?, ?, ?, ?,? , ?)';            $query = $this->db->prepare($sql);            $query->bind_param(                'sisss',                $file->name,                $file->size,                $file->type,                $file->url,                $file->title,                $file->description            );            $query->execute();            $file->id = $this->db->insert_id;        }        return $file;    }

我需要保存成
$url?=?$_COOKIE["uid"].'/'.date("Y").'/'.date("m").'/'.date("d").'/'.?filename

要怎么改??直接用$file->url就可以嗎?




另外...因為這玩意，還弄到一個append取值問題，熟jq的朋友也可以去這幫幫小弟吧
http://bbs.csdn.net/topics/390862894

回復(fù)討論(解決方案)

警告：mysqli_stmt：：bind_param()：在類型定義字符串不匹配的綁定變量的元素個數(shù)
這還不清楚嗎？
$query->bind_param(
????????????????' sisss',?//怎么只有?5?個類型聲明？
????????????????$file->name,?//1
????????????????$file->size,?//2
????????????????$file->type,?//3
????????????????$file->url,?//4
????????????????$file->title,//5
????????????????$file->description?//6?共6個
????????????);

警告：mysqli_stmt：：bind_param()：在類型定義字符串不匹配的綁定變量的元素個數(shù)
這還不清楚嗎？
$query->bind_param(
????????????????' sisss',?//怎么只有?5?個類型聲明？
????????????????$file->name,?//1
????????????????$file->size,?//2
????????????????$file->type,?//3
????????????????$file->url,?//4
????????????????$file->title,//5
????????????????$file->description?//6?共6個
????????????);

喔...原來這個問題

已成功解決了！

以前用完mysql_query后就轉(zhuǎn)用pdo了，沒用過mysqli，PDO好像就沒怎么提過這寫法

謝謝，學(xué)習(xí)了