我正在尝试在表'用户'列'image'下的数据库'cemembers'中显示jpg图像。有了这个逻辑,我创建了自定义代码,但是当我在php中运行我的页面时,只回显存储在数据库中但没有图形视图的文件名。例如'cdda_dd.jpg'我稍后检查了浏览器中的图像,我可以看到问题可能是编码错误的图像src ...请查看我的代码并帮助我。
<?php
//setting connection variables
$hostname = "localhost";
$username= "root";
$password ="";
$db= "cemembers";
//connect to database
$mysqli_db = new mysqli( $hostname, $username, $password, $db);
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = mysqli_query($mysqli_db, "SELECT * FROM users WHERE image ");
$row= mysqli_fetch_array($sql);
$photo= $row['image'];
echo $photo;
?>
答案 0 :(得分:0)
使用此代码..修改在第18行&amp; 21.从列名称中选择图片&#34;图像&#34; id = 1或任何内容而不是select * from users ..
<?php
//setting connection variables
$hostname = "localhost";
$username= "root";
$password ="";
$db= "cemembers";
//connect to database
$mysqli_db = new mysqli( $hostname, $username, $password, $db);
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$sql = mysqli_query($mysqli_db, "SELECT image FROM users WHERE id=1 ");
$row= mysqli_fetch_array($sql);
$photo= $row['image'];
header("Content-type:image/jpeg"); // add this
echo $photo;
?>