Question

如何在函数/类中将图像作为参数传递，或者这是不可能的，如果可以的话我该如何解决？例如：

session_start();
require_once("twitteroauth/twitteroauth.php"); //Path to twitteroauth library

$twitteruser = "xxx";
$notweets = 30;
$consumerkey = "xxxx";
$consumersecret = "xxx";
$accesstoken = "xxx";
$accesstokensecret = "xxx";



//use the php similar_text function to remove similar tweets

function removeSimilar ($array, $minSimilarity = 60) {
    $result = [];

    foreach ($array as $outerValue) {
        $append = true;
        foreach ($result as $key => $innerValue) {
            $similarity = null;
            similar_text($innerValue, $outerValue, $similarity);
            if ($similarity >= $minSimilarity) {
                if (strlen($outerValue) > strlen($innerValue)) {
                    // always keep the longer one
                    $result[$key] = $outerValue;
                }
                $append = false;
                break;
            }
        }

        if ($append) {
            $result[] = $outerValue;
        }
    }

    return $result;
}


function getConnectionWithAccessToken($cons_key, $cons_secret, $oauth_token, $oauth_token_secret) {
  $connection = new TwitterOAuth($cons_key, $cons_secret, $oauth_token, $oauth_token_secret);
  return $connection;
}

$connection = getConnectionWithAccessToken($consumerkey, $consumersecret, $accesstoken, $accesstokensecret);

$tweets = $connection->get("https://api.twitter.com/1.1/search/tweets.json?q=london+exclude:retweets+exclude:replies&since=2019-03-01&count=100");

$encode = json_encode($tweets);
$decode = json_decode($encode,true);

$decode = removeSimilar($decode);

foreach($decode['statuses'] as $tweets) {
    echo '<p>' . $tweets['text'] . '</p>';
    }

我已经尝试过dimage.src和其他方法，但我使用的任何方法似乎都不起作用：/

Answer 1

像其他任何方式一样传递它，但是要显示它，请使用appendChild而不是innerHTML：

var tree = new Image();
tree.src = "https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Ash_Tree_-_geograph.org.uk_-_590710.jpg/220px-Ash_Tree_-_geograph.org.uk_-_590710.jpg";
function showTree(dimage) {
  document.getElementById("div").appendChild(dimage);
  console.log(dimage);
}

showTree(tree);

<div id="div"></div>

如果您确实想使用innerHTML，请制作一个<img>为src的{{1}}：

dimage.src

var tree = new Image();
tree.src = "https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Ash_Tree_-_geograph.org.uk_-_590710.jpg/220px-Ash_Tree_-_geograph.org.uk_-_590710.jpg";
function showTree(dimage) {
  document.getElementById("div").innerHTML = "<img src='" + dimage.src + "'>";
  console.log(dimage);
}

showTree(tree);

Answer 2

您必须命名函数并通过将图像作为参数传递来调用它。如果您使用的是innerHTML，则必须创建图像标签，并将接收到的图像源作为参数添加到函数中

var tree = new Image();
tree.src = "img/statobj/tree.png"
function a(dimage){
document.getElementById("myitems").innerHTML ='<div id="invetoryitem" ><img src="'+dimage.src+ '"></div>'
console.log(dimage) //gets undefined
}
a(tree);

<body id="myitems"></body>

将图像对象作为函数和类参数传递

2 个答案: