jquery-file-upload 的php mysql插入問題
Jun 23, 2016 pm 01:51 PM 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

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