Question

我有一个名为images的表，字段为id, name, photo，photo字段为varchar，因为图片使用php表单存储在服务器上。在我的代码只显示表中的1张图片时，请有人帮助我，以便它使用查询对所有图像进行幻灯片放映：

SELECT * FROM images WHERE name='$pagetitle'

$pagetitle在脚本的其他地方定义。

问题出在index.php文件中

的index.php

＆＃13;

<!-- Image Slide Show Start -->
						
						<div style="display: flex; justify-content: center;">
						<img align="middle" src="" name="slide" border=0 width=300 height=375>
				

<?php 
require('mysqli.php');
$data = $conn->query("SELECT * FROM images WHERE name= '$pagetitle'");
$image = mysqli_fetch_array( $data );

$directory = ("/images/".$image['photo'] . "");

$conn->close();

?>				
<script>


//configure the paths of the images, plus corresponding target links
slideshowimages("<?php echo $directory ?>")

//configure the speed of the slideshow, in miliseconds
var slideshowspeed=2000

var whichlink=0
var whichimage=0
function slideit(){
if (!document.images)
return
document.images.slide.src=slideimages[whichimage].src
whichlink=whichimage
if (whichimage<slideimages.length-1)
whichimage++
else
whichimage=0
setTimeout("slideit()",slideshowspeed)
}
slideit()


</script> </div><br><br>
						<!-- Image Slide Show End -->

＆＃13;

Answer 1

解决的代码如下：

<?php

// Connect to the database
       require('mysqli.php');
        
// Query for a list of all existing files

$sql = "SELECT * FROM images WHERE name= '$pagetitle'";
$result = $conn->query($sql);

$directory = '';
while( $image = $result->fetch_assoc() )
    $directory .= ($directory != '' ? "," : '') . ('"/images/'.$image["photo"] . '"');



// Check if it was successfull
	if($directory != '') {
		
    // if there are images for this page, run the javascript
	?><script>


	//configure the paths of the images, plus corresponding target links			
	slideshowimages(<?php print $directory ?>)

	//configure the speed of the slideshow, in miliseconds
	var slideshowspeed=2000

	var whichlink=0
	var whichimage=0
	function slideit(){
	if (!document.images)
	return
	document.images.slide.src=slideimages[whichimage].src
	whichlink=whichimage
	if (whichimage<slideimages.length-1)
	whichimage++
	else
	whichimage=0
	setTimeout("slideit()",slideshowspeed)
	}
	slideit()


	</script>
	<?
   }  else {
        // If there are not any images for this page, leave the space blank
        echo "";
		} 
 
// Close the mysql connection
$conn->close();
?>

Answer 2

<?php 
$servername = "";
$username = "";
$password = "";
$dbname = "";
$conn = new mysqli($servername, $username, $password, $dbname);

//Make connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

//SQL query
$sql = "SELECT * FROM TABLE_NAME";
$result = $conn->query($sql);

if ($result->num_rows > 0) {

    while($row = $result->fetch_assoc()) {
        $dir= "" . $row["field"]. "";
       echo" <div class='content section' style='max-width:500px'>
             <img class='slides' src='$dir' style=width:100%></div>";
    }
} else {
    echo "Empty Gallery";
}
?>

<html>
<body>
<script>
var index = 0;
slideshow();
function slideshow() {
  var i;
  var x = document.getElementsByClassName("slides");
  for (i = 0; i < x.length; i++) {
    x[i].style.display = "none";  
  }
  index++;
  if (index > x.length) {index = 1}    
  x[index-1].style.display = "block";  
  setTimeout(slideshow, 3000); // Change slides every 3 seconds
}
</script>
</body>
</html>

如何显示mysql数据库中的图像幻灯片

的index.php

2 个答案: