本文介绍了为什么我不能通过PHP从数据库中检索图像?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 29岁程序员,3月因学历无情被辞! 这是我的代码。你能纠正吗?在此先感谢。 <? php echo <<< _END < DOCTYPE html > < html > < head > < meta charset = utf-8 > < / head > _END; require_once'login.php'; $ connection = new mysqli($ db_hostname,$ db_username,$ db_password,$ db_database); if($ connection-> connect_error)die($ connection-> connect_error); if( isset($ _ POST ['title'])&& isset($ _ POST ['category'])&& isset($ _ POST ['name'])&& isset($ _ POST ['telno'])&& isset($ _ POST ['price'])&& isset($ _ POST ['info'])) { $ title = get_post($ connection,'title'); $ category = get_post($ connection,'category'); $ name = get_post($ connection,'name'); $ telno = get_post($ connection,'telno'); $ price = get_post($ connection,'price'); $ info = get_post($ connection,'info'); if(is_uploaded_file($ _ FILES ['img1'] ['tmp_name'])) { $ imgData = addslashes(file_get_contents($ _ FILES [' IMG1 '] [' tmp_name的值'])); $ query =INSERT INTO avto(title,category,img1,name,telno,price,info)VALUES。 ('$ title','$ category','$ imgData','$ name','$ telno','$ price','$ info'); $ result = $ connection-> query($ query); if(!$ result) echoКушиббулмаяпти:$ query < br > 。 $ connect_error。 < br > < br > ; } } echo < << _END < form action = index.php 方法 = post enctype = multipart / form-data > < pre > 标题< input type = text 名称 = title > 类别< 输入 type = text 名称 = 类别 > 主要img < 输入 type =' file' 名称 =' img1' > 名称< input type = text name = name > 电话号码< 输入 type = text 名称 = telno > 价格< 输入 type = text 名称 = 价格 > ; 信息< textarea 行 = 6 cols = 30 名称 = 信息 > < / textarea > < 输入 类型 = 提交 value = 提交 > < / pre > < / form > _END; $ query =SELECT * FROM avto; $ result = $ connection-> query($ query); 如果(!$ result)死($ connection->错误); $ rows = $ result-> num_rows; for($ j = 0; $ j < $ rows; ++ $ j ) { $ result - > data_seek($ j); $ row = $ result-> fetch_array(MYSQLI_NUM); echo < << _END < pre > $ row [0] $ row [1] $ row [ 2] $ row [3] $ row [9] $ row [10] $ row [11] $ row [12] $ row [13] < / pre > _END; } $ result-> close(); $ connection-> close(); 函数get_post($ connection,$ var) {返回$ connection-> real_escape_string($ _ POST [$ var]); } echo < << _END < / html > _END; ?> 解决方案 connection = new mysqli( db_hostname, db_username, This is my code. Could you correct? Thanks in advance.<?phpecho <<<_END<DOCTYPE html><html><head><meta charset="utf-8"></head>_END;require_once'login.php';$connection=new mysqli($db_hostname, $db_username, $db_password, $db_database);if($connection->connect_error) die($connection->connect_error);if ( isset($_POST['title'])&& isset($_POST['category'])&& isset($_POST['name'])&& isset($_POST['telno'])&& isset($_POST['price'])&& isset($_POST['info'])){ $title=get_post($connection, 'title'); $category=get_post($connection, 'category'); $name=get_post($connection,'name'); $telno=get_post($connection,'telno'); $price=get_post($connection,'price'); $info=get_post($connection,'info'); if (is_uploaded_file($_FILES['img1']['tmp_name'])) { $imgData=addslashes(file_get_contents($_FILES['img1']['tmp_name'])); $query="INSERT INTO avto(title, category, img1, name, telno, price, info) VALUES" . "('$title', '$category', '$imgData', '$name', '$telno', '$price', '$info')"; $result = $connection->query($query); if(!$result) echo "Кушиб булмаяпти: $query<br>". $connect_error. "<br><br>"; }}echo <<<_END <form action="index.php" method="post" enctype="multipart/form-data"><pre> Title <input type="text" name="title"> Category <input type="text" name="category"> Main img <input type='file' name='img1' > Name <input type="text" name="name"> Tel NO <input type="text" name="telno"> Price <input type="text" name="price"> Info <textarea rows="6" cols="30" name="info"></textarea> <input type="submit" value="Submit"> </pre></form>_END;$query="SELECT * FROM avto";$result=$connection->query($query);If (!$result) die ($connection->error);$rows=$result->num_rows;for ($j=0;$j<$rows;++$j){$result->data_seek($j);$row=$result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> $row[0] $row[1] $row[2] $row[3] $row[9] $row[10] $row[11] $row[12] $row[13] </pre>_END;}$result->close();$connection->close();function get_post($connection, $var){ return $connection->real_escape_string($_POST[$var]);}echo <<<_END</html>_END;?> 解决方案 connection=new mysqli(db_hostname,db_username, 这篇关于为什么我不能通过PHP从数据库中检索图像?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云!
09-05 20:12
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