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HarmonyOS
建表语句：
CREATE TABLE `image` (
`id` int(3) NOT NULL AUTO_INCREMENT,
`name` varchar(100) CHARACTER SET utf8 NOT NULL,
`pic` blob NOT NULL,
`type` varchar(50) CHARACTER SET utf8 NOT NULL,
`date` datetime NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1conn.php
<?php
$conn = mysqli_connect('localhost', 'root', 'root', 'duoxuankuangceshi', '8080');
?>upimg.php
<?php
include('./config.php');
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en_US" xml:lang="en_US">
<head>
<meta http-equive="Content-Type" content=text/html charset=utf-8>
<title> </title>
</head>
<body>
<form method='post' action='./upload.php' enctype="multipart/form-data">
<input type="file" name="image" />
<input type="submit" name="submit" value="上传" />
</form>
<!-----------显示图片--------------------->
<table>
<?php
$ret = mysqli_query($conn,'select * from image order by id desc');
if ($ret) {
while ($row = mysqli_fetch_array($ret)) {
?>
<tr>
<td style='width:170px;'>
<img src="image.php?id=<?php echo $row['id'];?>" width="170" height="150" border="0">
<div style='text-align:center;'><?php echo $row['name'];?></div>
<?php echo $row['date'];?>
</td>
</tr>
<?php
}
}
?>
</table>
<!-----------/显示图片--------------------->
</body>
</html>upload.php
<?php
include('./config.php');
if ($_POST['submit']) {
if ($_FILES['image']['size']) {
$names = $_FILES['image']['name'];
$arr = explode('.', $names);
$name = $arr[0]; //图片名称
$date = date('Y-m-d H:i:s'); //上传日期
$fp = fopen($_FILES['image']['tmp_name'], 'rb');
$type = $_FILES['image']['type'];
if (!$fp) {
showInfo('读取图片失败！');
} else {
$image = addslashes(fread($fp, filesize($_FILES['image']['tmp_name'])));
if ($image) {
$q = "insert into image (name, pic, type, date) values ('$name','$image','$type','$date')";
$result = mysqli_query($conn,$q);
if ($result) {
showInfo('上传成功！');
} else {
showInfo('上传失败！');
}
} else {
showInfo('请选择要上传的文件！');
}
}
} else {
showInfo('请选择要上传的文件！');
}
}
function showInfo($info)
{
echo "<html><head><meta http-equiv=\"Content-Type\" content=\"text/html; charset=utf-8\" />";
echo "<meta http-equiv='refresh' content='1;url=upimg.php'>";
echo "</head>";
echo "<body>" . $info . "……</body>";
echo "</html>";
}
?>image.php
<?php
include('./config.php');
$id = $_GET['id'];
$sql = "select * from image where id='$id'";
$result = mysqli_query($conn,$sql);
if (!$result)
die("读取图片失败！");
$num = mysqli_num_rows($result);
if ($num < 1)
die("暂无图片");
while ($row = mysqli_fetch_assoc($result)) {
$data = $row['pic'];
$type = $row['type'];
}
/*$data = mysqli_result($result, 0, 'pic');
$type = mysqli_result($result, 0, 'type');*/
mysqli_close($id);
Header("Content-type: $type");
echo $data;
?>数据库：
效果图：
在做图片上传回显测试，现在图片已经上传成功了，但是图片的回显却没有显示出来，只显示框框，能获取图片对应的id，也没有报错，在地址栏中打开页面/image.php?id=1或2，两个打开页面都是只有页面正中间有个白色边框的小框，图片都不能显示，数据库里面的pic打开却有图片，请问该怎么让图片显示出来？
mysql
你单独把图片请求的地址在浏览器里打开,你看输出的是啥,, 你的截图里,两个图片的请求,虽然状态码是200,但是 content-type 和 size 都不对,应该已经出错了.