功能:php实现将图片存入mysql并取出预览:
数据库结构:
代码
<?php
// 连接数据库
//echo realpath('..');//获得上级目录 两个点 获得当前目录 一个点
include(realpath('..').'\controller\sqlConnect.php');
// 判断action
$action = isset($_REQUEST['action'])? $_REQUEST['action'] : '';
$tableName="photo";
// 上传图片
if($action=='add'){
//防止出现‘ “” 等字符,对其进行/ 转义
//$image = mysqli_real_escape_string($link,file_get_contents($_FILES['photo']['tmp_name']));
$image=base64_encode(file_get_contents($_FILES['photo']['tmp_name']));
// $_FILES['photo']['tmp_name'] 文件存储路径名 file_get_contents: 由路径获得文件内容
//$_FILES['photo']['type'] 获取文件类别
$type = $_FILES['photo']['type'];
//ob_clean();
//echo $image;
//echo $type."\n";
$sql = "INSERT INTO photo (userId,photoType,photoBinaryData) VALUES(1,'$type','$image')";
//谨记插入式变量需加引号!!!
if(mysqli_query($link,$sql)or die(mysqli_error($link)))
{
echo "save successfully!";
header('url=addPhoto.php');
}
else
{
echo "insert failed";
echo mysql_error();
exit();
}
}
elseif($action=='show'){
$photoId = isset($_GET['photoId'])? intval($_GET['photoId']) : 0;//intval:转化为整数
//echo $photoId;
//echo $photoId;
//console.log($photoId);
$sqlstr = "select * from photo where photoId=$photoId";
$query = mysqli_query($link,$sqlstr) or die(mysqli_error($link));
$thread = mysqli_fetch_assoc($query);
if($thread){
$photoShow=base64_decode($thread['photoBinaryData']);
header('content-type:'.$thread['photoType']);
echo $photoShow;
exit();
}
}else{
// 显示图片列表及上传表单
?>
<!DOCTYPE HTML >
<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8">
<title> upload image to db demo </title>
</head>
<body>
<form name="form1" method="post" action="addPhoto.php" enctype="multipart/form-data">
<p>图片:<input type="file" name="photo"></p>
<p><input type="hidden" name="action" value="add"><input type="submit" name="b1" value="提交"></p>
</form>
<?php
$sqlstr = "select * from photo order by photoId desc";
$query = mysqli_query($link,$sqlstr) or die(mysqli_error($link));
$result = array();
while($thread=mysqli_fetch_assoc($query)){
$result[] = $thread;
}
foreach($result as $val){
echo '<p><img src="addPhoto.php?action=show&photoId='.$val['photoId'].'&t='.time().'" width="150" height="150"></p>';
echo "<br/>".$val['photoType'];
}
?>
</body>
</html>
<?php
}
?>
结果:
问题:为什么mysql数据库的图片不显示? add和调用函数已写好,我感觉是if action=show 后面代码的问题 但不清楚请指出!