循环未完成

发布于 2025-01-04 15:52:13 字数 591 浏览 1 评论 0原文

在这里我写了一个小代码...在此循环中未完成..仅显示第一张图像..实际上有 7 张图像...请帮助我

 <?php

$username = "root";
$password = "root";
$host = "localhost";
$database = "test";
error_reporting(E_ERROR | E_PARSE);
mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

mysql_select_db($database) or die("Can not select the database: ".mysql_error());


header('Content-type: image/jpg');

$query = mysql_query("SELECT * FROM tbl_images");
while($row = mysql_fetch_array($query))
{
echo $content = $row['image']."</br>";
echo $content;
}
?>

原文

Here i have written a small code...in this looping is not done..only first image is displaying..Actually there are 7 images...Please help me on this

 <?php

$username = "root";
$password = "root";
$host = "localhost";
$database = "test";
error_reporting(E_ERROR | E_PARSE);
mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

mysql_select_db($database) or die("Can not select the database: ".mysql_error());


header('Content-type: image/jpg');

$query = mysql_query("SELECT * FROM tbl_images");
while($row = mysql_fetch_array($query))
{
echo $content = $row['image']."</br>";
echo $content;
}
?>

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

随梦而飞# 2025-01-11 15:52:13

你的数据库里到底有什么？如果存在图像路径，则应像

echo '';

一样输出它，并且如果存在带有以下内容的 blob 数据图像 - 您只能使用 header('Content-type: image/jpg'); 输出其中一个图像。这样您就可以创建一个打印所需图像的页面。

回复收藏 0 原文

烧了回忆取暖 2025-01-11 15:52:13

试试这个：

<?php

error_reporting(E_ERROR | E_PARSE);

$username = "root";
$password = "root";
$host = "localhost";
$database = "test";

mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

mysql_select_db($database) or die("Can not select the database: ".mysql_error());

//header('Content-type: image/jpg');

$query = mysql_query("SELECT * FROM tbl_images");

while($row = mysql_fetch_array($query)){
    echo '<img src="'.$row['image'].'" width=40 height=40>';// remove width and height later
}
?>

try this:

<?php

error_reporting(E_ERROR | E_PARSE);

$username = "root";
$password = "root";
$host = "localhost";
$database = "test";

mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

mysql_select_db($database) or die("Can not select the database: ".mysql_error());

//header('Content-type: image/jpg');

$query = mysql_query("SELECT * FROM tbl_images");

while($row = mysql_fetch_array($query)){
    echo '<img src="'.$row['image'].'" width=40 height=40>';// remove width and height later
}
?>

回复收藏 0 原文

~没有更多了~