html:
<form method="post" action="upimage.php" enctype="multipart/form-data">
描述:
<input type="text" name="form_description" size="40">
<input type="hidden" name="MAX_FILE_SIZE" value="1000000"> <br>
上传文件到数据库:
<input type="file" name="form_data" size="40"><br>
<input type="submit" name="submit" value="提交">
</form>
upimage.php:
<?php
if (isset($_POST['submit'])) {
$form_description = $_POST['form_description'];
$form_data_name = $_FILES['form_data']['name'];
$form_data_size = $_FILES['form_data']['size'];
$form_data_type = $_FILES['form_data']['type'];
$form_data = $_FILES['form_data']['tmp_name'];
$conn = mysqli_connect('localhost', 'root', 'root', 'upload', '8080');
$data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));
//echo "mysqlPicture=".$data;
$result = $conn->query("INSERT INTO ccs_image (description,bin_data,filename,filesize,filetype)
VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");
if ($result) {
echo "图片已存储到数据库";
} else {
echo "请求失败,请重试";
}
}
getimage:
<?php
$id =2;// $_GET['id']; 为简洁,直接将id写上了,正常应该是通过用户填入的id获取的
$conn = mysqli_connect('localhost', 'root', 'root', 'upload', '8080');
$query = "select bin_data,filetype from ccs_image where id=2";
$result = $conn->query($query);
$result = $result->fetchAll(2);
//var_dump($result);
$data = $result[0]['bin_data'];
$type = $result[0]['filetype'];
Header( "Content-type: $type");
echo $data;
?>
在网上找的相关教程,然后根据自己的修改,可以上传图片保存到数据库中,但是读取数据库里的图片出现了问题,报
Fatal error: Call to undefined method mysqli_result::fetchAll() in E:BoboCodeuploadUploadImagegetimage.php on line 6
错误,是getimage里这个$result = $result->fetchAll(2);
,原代码用的是PDO连接数据库,我用了连接不上,就换成了mysqli,然后获取图片的php就报错了,网上没搜到相关,这个该怎么修改?
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