PHP imagejpeg无法正常工作

Question

所以我看了几乎每个问题，并尝试使用下面的PHP脚本来显示我的图像。但它不起作用。我也试着看看GD库是否使用PHPInfo（）来启用，并且它也可以工作。我对PHP很陌生，但似乎无法得到这个工作。谢谢你的帮助！

编辑︰我点击上传文档按钮后，我得到一个破碎的图像图标。

<?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imagedestroy($image);之后添加die();防止除了图像数据的任何其他输出。还将imagejpeg($image,NULL);更改为imagejpeg($image);。

Answer 2

您无法直接访问图像tmp_name。首先，你应该将图像保存到服务器后，您可以用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

使用我认为问题出在PHP部分（HTML代码）后echo和输出字符串。删除foreach循环，并仅当它是新页面或出现错误时才输出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> 

还有一件事，如果您使用记事本进行编码，请确保文件开头处没有隐藏字符。