为什么我不能通过PHP从数据库中检索图像? [英] Why I cant retreive image from database via php?
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问题描述
这是我的代码。你能纠正吗?在此先感谢。
<? php
echo <<< _END
< DOCTYPE < span class =code-summarycomment> 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 < span class =code-attribute> 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 < span class =code-keyword> = name >
电话号码< 输入 type = text 名称 = < span class =code-keyword> 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 < span class =code-keyword>>
_END;
?>
解决方案
connection = new mysqli(
db_hostname,
db_username,
This is my code. Could you correct? Thanks in advance.
<?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']));
$query="INSERT INTO avto(title, category, img1, name, telno, price, info) VALUES" .
"('$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,
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