我如何显示图像存储为MySQL中的blob数据类型与PHP？ [英] How do i display images stored as blob data type in mysql with php?

问题描述

如何将存储在mySQL数据库中的图像显示为BLOB？
到目前为止所尝试的是：
1.创建一个新的php函数/文件来获取图片（getpicture.php）。
2.在html中，我有以下代码：

 < img src =getpicture.php？ id = 2border =0height =250width =250/> 
 
 
 
 
 / *下面是getpicture.php * / 
<？php 
 
 @ $ db =新的MySQLi（'localhost'，'root'，''，'myDatabase'）; 
 
 if（mysqli_connect_errno（））{
 echo'连接数据库失败：'。mysqli_connect_error（）; 
 exit（）; 
 
 
 if（isset（$ _ GET ['People_Id']））{
 
 $ id = mysqli_real_escape_string（$ _ GET ['People_Id']）; 
 $ query = mysqli_query（SELECT * FROM'people'WHERE'People_Id'='$ id'）; 
 
 while（$ row = mysqli_fetch_assoc（$ query））{
 $ imageData = $ row ['image']; 
} 
 
 header（content-type：image / jpeg）; 
 echo $ imageData; 
 echo $ id; 
} 
 
 else {
 echoError！; 
 echo $ id; 
} 
 
？> 
  

代码有什么问题？请帮忙！

解决方案

b

以下是getpicture.php：

 <？php 
 
 $ db = new MySQLi（'localhost'，''，''，'mydatabase'）; 
 
 if（$ db-> connect_errno）{
 echo'连接数据库失败：'。 $ DB-> connect_error; 
 exit（）; 
 
 
 if（isset（$ _ GET ['id']））{
 
 $ id = $ db-> real_escape_string（$ _ GET ['id ]）; 
 
 $ query =SELECT`Picture` FROM member WHERE`Id` ='$ id'; 
 
 $ result = $ db-> query（$ query）; 
 
 while（$ row = mysqli_fetch_array（$ result））{
 $ imageData = $ row ['Picture']; 
 header（Content-type：image / jpeg）; 
 echo $ imageData; 
} 
  

}
？>

检索上面的getpicture.php的php脚本如下所示：

  echo'< img src =getpicture.php？id ='。htmlspecialchars（$ _ GET [id]）。'border =0height =250width =250/>'; 
  

为您提供帮助

How do i display an image stored in mySQL database as BLOB ? What it tried so far: 1. Created a new php function/file to get picture (getpicture.php). 2. In the html, I have the following code:

<img src="getpicture.php?id=2" border ="0" height="250" width="250" /> 




/*below is the getpicture.php*/
<?php

@ $db = new MySQLi('localhost','root','','myDatabase');

if(mysqli_connect_errno()) {
    echo 'Connection to database failed:'.mysqli_connect_error();
    exit();
}

if(isset($_GET['People_Id'])) {

    $id = mysqli_real_escape_string($_GET['People_Id']);
    $query = mysqli_query("SELECT * FROM 'people' WHERE 'People_Id' = '$id'");

    while($row = mysqli_fetch_assoc($query)) {
        $imageData =$row['image'];
    }

    header("content-type: image/jpeg");
    echo $imageData;
    echo $id;
}

else {
    echo "Error!";
    echo $id;
}

?>

What's wrong with the codes ? Please help!

解决方案

I answered my own question, it's working now..

Below is the getpicture.php:

<?php

$db = new MySQLi('localhost', '', '', 'mydatabase');

if ($db->connect_errno) {
    echo 'Connection to database failed: '. $db->connect_error;
    exit();
}

if (isset($_GET['id'])) {

$id = $db->real_escape_string($_GET['id']);

$query = "SELECT `Picture` FROM member WHERE `Id` = '$id'";

$result = $db->query($query);

while($row = mysqli_fetch_array($result)) {
    $imageData = $row['Picture'];
    header("Content-type:image/jpeg");
    echo $imageData;
}

} ?>

The php script which retrieve the getpicture.php above looks like this:

echo '<img src="getpicture.php?id=' . htmlspecialchars($_GET["id"]) . '"border ="0" height="250" width="250" />';

Thaank you all for the help

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