這是我的uploadImage.php。我想在他們上傳之後立即顯示圖片。 問題是,我不是什麼似乎是問題。我試圖改變它的一些,並來到這個代碼。但仍然沒有得到我預期的結果。任何人都在意幫助我?上傳後顯示圖片
<?php
$tempFile = $_FILES["image"]["tmp_name"];
$name = $_FILES["image"]["name"];
$uploadDirectory = "/home/trainee/Desktop/tmp/";
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["image"]["type"] == "image/gif")
|| ($_FILES["image"]["type"] == "image/jpeg")
|| ($_FILES["image"]["type"] == "image/jpg")
|| ($_FILES["image"]["type"] == "image/pjpeg")
|| ($_FILES["image"]["type"] == "image/x-png")
|| ($_FILES["image"]["type"] == "image/png"))
&& ($_FILES["image"]["size"] < 20000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["image"]["error"] > 0)
{
echo "Return Code: " . $_FILES["image"]["error"] . "<br>";
}
else
{
echo "Upload: " . $name. "<br>";
echo "Type: " . $_FILES["image"]["type"] . "<br>";
echo "Size: " . ($_FILES["image"]["size"]/1024) . " kB<br>";
echo "Temp file: " . $tempFile . "<br>";
if (file_exists("$uploadDirectory" . $name))
{
echo $name . " already exists. ";
}
else
{
move_uploaded_file($tempFile, "$uploadDirectory/$name"),
$name
echo "Stored in: " ."$uploadDirectory". $name;
}
}
}
else {
echo "invalid File!";
}
?>
<img src="upload/<?php=$name?>" height=200 width=100">
旁註:這條線路是壞了'move_uploaded_file($臨時文件, 「$ uploadDirectory/$名稱」),$名稱回聲 「中存儲的:」 「$ uploadDirectory」 $名;' - [閱讀()函數中的手冊](http://php.net/move_uploaded_file)。一旦你建立了正確的變量,你只需要回顯它 –
你在那裏迷失了我的好友,將會像回聲「存儲在:」.move_uploaded_file($ tempFile,「$ uploadDirectory/$ name」),這是否正確? – user3335903
請參閱下面給出的答案。 –