Question

我创建了一个名为movie的数据库，并且有一个名为test_image的表（id int autoincrement，name varchar（30），image blob）。借助php代码插入数据。现在我已经借助以下代码显示了图像：

<?php
$host="yourhostname";
$user="username";
$pass="password";
$db="movie";

// just so we know it is broken
error_reporting(E_ALL);

//connect to the db
$link = mysql_connect("$host", "$user", "$pass") or die("Could not connect: " . mysql_error());

// select our database
mysql_select_db("$db") or die(mysql_error());

// get the image from the db
$sql = "SELECT image FROM test_image;";

// the result of the query
$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());

// set the header for the image
header("Content-type: image/jpeg");

echo mysql_result($result, 0);

// close the db link
mysql_close($link);

＆GT;

它只显示一张图片现在如果我想在数据库中幻灯片显示我必须做出哪些改变？

Answer 1

你在代码中遗漏了一点

的mysqli

<?
//no more mysql_ means start using mysqli
$mysqli = new mysqli('host', 'username', 'password', 'database'); //connect
$query = "SELECT image FROM test_image"; // your query
$result = $mysqli->query($query); // Fetch Result
//now do something with that result
foreach ($result as $row){
   echo "<img src=\"$row['image']\">";
}
?>

Answer 2

现在我已经为幻灯片显示了js。我编写了这样的代码：

<?php
include "file_constants.php";//in this file connection is established with db
header('Content-Type: text/html; charset=utf-8');
$sql="SELECT * FROM test_image;";
$result=mysqli_query($con,$sql) or die("invalid query ".mysqli_error());
$array=array();
while($row=mysqli_fetch_array($result))
{
$array[]=$row['image'];
}
 ?>


           <html>
           <body>

    <title>Slideshow</title>
    <h1>Slideshow of images stored in mysql</h1>
    <script language="javascript">
        var delay=1000; 
        var curindex=0;


        var randomimages=["$array[0]","$array[1]","$array[2]","$array[3]","$array[4]","$array[5]"];

        var preload=[],img,curindex,tempindex;

        for (var n=0;n<randomimages.length;n++)
        {
            img=new Image();
            img.src=randomimages[n];
            preload.push(img);
        }

        document.write('<img name="defaultimage" src="'+randomimages[Math.floor(Math.random()*(randomimages.length))]+'">');

        function rotateimage()
        {
            if (curindex==(tempindex=Math.floor(Math.random()*(randomimages.length))))
            {
                curindex=curindex==0 ? 1 : curindex-1;
            } else
            {
                curindex=tempindex;
            }

            document.images.defaultimage.src=randomimages[curindex];
        }

        window.setInterval(rotateimage,delay);
    </script>
</body>
     </html>

但是这段代码没有显示图片......

Answer 3

简单步行抛出结果集：

while($row = mysql_fetch_assoc($result)){
    echo $row[0];
}

请注意你使用了obsolte功能。请使用mysqli（http://php.net/manual/de/book.mysqli.php）

php + mysql图片幻灯片

3 个答案: