php使用blob存取图片的信息（含源码）
BLOB是一种MySQL数据类型，称为二进制大对象。正如它的名字它是用来存储类似MYSQL二进制和VARBINARY类型的、大量的字符串数据。
MySQL BLOB分类
MySQL BLOB类型 最大存储长度(字节)
TINYBLOB (1)(2 ^ 8)
blob ((2 ^ 16)1)
MEDIUMBLOB ((2 ^ 24)1)
LONGBLOB ((2 ^ 32)1)
在这篇教程中，我们学习如何使用PHP插入和读取MySQL BLOB字段。
（PS：T不错的PHP Q扣峮：276167802，验证：csl）
首先,我们需要创建一个MySQL表与一个BLOB字段。
CREATE TABLE IF NOT EXISTS `output_images` ( `imageId` tinyint(3) NOT NULL AUTO_INCREMENT, `imageType` varchar(25) NOT NULL DEFAULT '', `imageData` mediumblob NOT NULL, PRIMARY KEY (`imageId`))
插入数据
将图片信息插入MySQL BLOB字段中。
1、上传图像文件.
2、获取图像属性(图像数据、图像类型等等。)
3、图像文件插入BLOB。
PHP实现脚本：
imageUpload.php
<?phpif(count($_FILES) > 0) {if(is_uploaded_file($_FILES['userImage']['tmp_name'])) {mysql_connect("localhost", "root", "");mysql_select_db ("phppot_examples");$imgData =addslashes(file_get_contents($_FILES['userImage']['tmp_name']));$imageProperties = getimageSize($_FILES['userImage']['tmp_name']);$sql = "INSERT INTO output_images(imageType ,imageData)VALUES('{$imageProperties['mime']}', '{$imgData}')";$current_id = mysql_query($sql) or die("<b>Error:</b> Problem on Image Insert<br/>" .mysql_error());if(isset($current_id)) {header("Location: listImages.php");}}}?><HTML><HEAD><TITLE>Upload Image to MySQL BLOB</TITLE><link href="imageStyles.css" rel="stylesheet" type="text/css" /></HEAD><BODY><form name="frmImage" enctype="multipart/form-data" action="" method="post"class="frmImageUpload"><label>Upload Image File:</label><br/><input name="userImage" type="file" class="inputFile" /><input type="submit" value="Submit" class="btnSubmit" /></form></div></BODY></HTML>
执行这个脚本后上传表单将显示如下：
提交表单,PHP获取内容图像的文件并将其作为二进制数据存储到MySQL BLOB列。
显示图片
在浏览器上显示BLOB图像,我们必须：
1、从MySQL BLOB获得图像数据和类型
2、将类型设置为图像(image/jpg, image/gif, …)使用PHP header()函数。
3、输出图像内容。
imageView.php<?php$conn = mysql_connect("localhost", "root", "");mysql_select_db("phppot_examples") or die(mysql_error());if(isset($_GET['image_id'])) {$sql = "SELECT imageType,imageData FROM output_images WHERE imageId=" . $_GET['image_id'];$result = mysql_query("$sql") or die("<b>Error:</b> Problem on Retrieving Image BLOB<br/>". mysql_error());$row = mysql_fetch_array($result);header("Content-type: " . $row["imageType"]);echo $row["imageData"];}mysql_close($conn);?>
上面的PHP代码将显示MySQL BLOB存储的图片。从HTML图像标签我们可以参考这个PHP文件与相应image_id作为参数。例如：
<img src="imageView.php?image_id=<?php echo $row["imageId"]; ?>" />
完成代码如下：
listImages.php
<?php$conn = mysql_connect("localhost", "root", "");mysql_select_db("phppot_examples");$sql = "SELECT imageId FROM output_images ORDER BY imageId DESC";$result = mysql_query($sql);?><HTML><HEAD><TITLE>List BLOB Images</TITLE><link href="imageStyles.css" rel="stylesheet" type="text/css" /></HEAD><BODY><?phpwhile($row = mysql_fetch_array($result)) {?><img src="imageView.php?image_id=<?php echo $row["imageId"]; ?>" /><br/><?php}mysql_close($conn);?></BODY></HTML>
感谢大家的阅读，希望大家收益多多。
原文链接：https://blog.csdn.net/u012275531/article/details/1791499
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