Question

所以我几乎看了每个问题并尝试使用下面的PHP脚本来显示我的图像。但它不起作用。我还试图看看是否使用PHPInfo（）启用了GD库，它也正常工作。我是PHP的新手，但似乎无法实现这一点。谢谢你的帮助！

编辑：点击上传文档按钮后，我收到了一张损坏的图片图标。

<?php
 if($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_FILES['photo'])
        && is_uploaded_file($_FILES['photo']['tmp_name'])
        && $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
            foreach ($_FILES['photo'] as $key => $value) {
                echo "$key : $value<br />";
            }
            if($_FILES['photo']['type'] == 'image/jpeg') {
                $tmp_img = $_FILES['photo']['tmp_name'];
                $image = imagecreatefromjpeg($tmp_img);
                header('Content-Type: image/jpeg');
                imagejpeg($image,NULL);
                imagedestroy($image);
            } else {
                echo "Uploaded file ewas not a jpg image.";
            }
                echo "no photo uploaded.";
              }
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>

Answer 1

您可以尝试在die();之后添加imagedestroy($image);以防止除图像数据之外的任何其他输出。同时将imagejpeg($image,NULL);更改为imagejpeg($image);。

Answer 2

您无法使用tmp_name直接访问图片。首先，您应该将图像保存到服务器，之后可以使用imagecreatefromjpeg

<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
if(isset($_FILES['photo'])
        && is_uploaded_file($_FILES['photo']['tmp_name'])
        && $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
            if($_FILES['photo']['type'] == 'image/jpeg') {
                $tmp_img = $_FILES['photo']['tmp_name'];
                $uniq_name = uniqid().".".explode(".",$tmp_img)[1];
                move_uploaded_file($_FILES['photo']['tmp_name'],$uniq_name);
                $image = imagecreatefromjpeg($uniq_name);
                header('Content-Type: image/jpeg');
                imagejpeg($image);
                imagedestroy($image);
            } else {
                echo "Uploaded file ewas not a jpg image.";
            }
                echo "no photo uploaded.";
              }
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>

Answer 3

我认为问题在于echo和PHP部分之后的输出字符串（HTML代码）。删除foreach循环并仅在HTML是新页面或出现错误时输出HTML。

<?php
  if($_SERVER['REQUEST_METHOD'] == 'POST') {
    if(isset($_FILES['photo'])
        && is_uploaded_file($_FILES['photo']['tmp_name'])
        && $_FILES['photo']['error'] == UPLOAD_ERR_OK) {
            if($_FILES['photo']['type'] == 'image/jpeg') {
                $tmp_img = $_FILES['photo']['tmp_name'];
                $image = imagecreatefromjpeg($tmp_img);
                header('Content-Type: image/jpeg');
                imagejpeg($image,NULL);
                imagedestroy($image);
                exit; // Add this to stop the program here.
            } else {
                echo "Uploaded file was not a jpg image.";
            }
    } else {
      echo 'No file was sent.';
    }
}
?>
<form enctype="multipart/form-data" action="book.php" method="post">
<input type="file" name="photo">
<input type="submit" value="upload a doc">
</form>

另外一件事，如果您使用记事本进行编码，请确保文件开头没有隐藏的字符。

PHP imagejpeg不起作用

3 个答案: