Question

我有一个显示图像的目录：

<img src="(image path)" style="width:304px;height:228px">

如何使用PHP获取图像路径？

我还想检查目录是否存在，我得到图像路径，如果没有显示错误。

如果我得到图像路径，我想使用HTML显示图像：

class MonsterNode: GameObjectNode {

var monsterType: MonsterType!
let monsterSound = SKAction.playSoundFileNamed("MONSTER.mp3", waitForCompletion: false)

override func collisionWithPlayer(player: SKNode) -> Bool {



    if player.physicsBody?.velocity.dy < 0 {

        runAction(monsterSound, completion: {
        })

        player.physicsBody?.velocity = CGVector(dx: player.physicsBody!.velocity.dx, dy: 450.0)

       self.removeFromParent()


    } else if player.physicsBody?.velocity.dy > 0 {


        player.physicsBody?.velocity = CGVector(dx: -player.physicsBody!.velocity.dx, dy: -450.0)
        player.hidden = true


    }


    return false
}

Answer 1

假设您将图像路径转换为变量＆gt; $ IMAGE_PATH

只需这样做......

<img src="<?php echo $image_path; ?>" style="width:304px;height:228px">

Answer 2

http://php.net/manual/en/function.file-exists.php

<?php 
$file = "C:\myfile.jpg";
if(file_exists($file)){
        if(is_readable($file)){
        echo '<img src="'.$file.'">';
        }
    }
    else
    {
    // UNABLE TO FIND FILE;
    }
?>

Answer 3

亲爱的雷克斯这样做，

<?php
$path = "www/item";
$folder = opendir($path);
$pic = false;
while(false !== ($file = readdir($folder))){
    $fileName = $path.$file;
    $info = pathinfo($path.$file);
    $ext = strtolower($info['extension']);
    $allowed =array("jpg","bmp","png","gif");
    if(filetype($fileName) == "file"){
        if(in_array($ext,$allowed)){
            echo"
            <div style='display:inline-block'>
            <img src='$fileName' style='width:96px;height:70px;' title='Filename: $file'>
            <a target = '_blank' href='$fileName'>Download</a>
            </div>
            ";
            $pic = true;
            }
        }
    }
if($pic == false){
    echo "There are no images in $path";
    }
?>

php ---使用php获取图像路径，并使用HTML显示图像

3 个答案: