Question

这里我写了一个小代码......在这个循环中没有完成..只有第一张图片正在显示..实际上有7张图片...请帮我这个

 <?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;
}
?>

Answer 1

您的数据库中的实际内容是什么？如果有图像路径，则应输出

echo '<img src="'.$row['image'].'">';

如果有带图像的blob数据 - 您只能使用header('Content-type: image/jpg');输出其中一个。这样你就可以创建一个可以打印所需图像的页面。

Answer 2

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
}
?>

循环没有完成

2 个答案: