Question

我制作了一个名为createCaptcha.php的php文件，代码如下：

create_image(); 
exit(); 

function create_image() { 
    //Let's generate a totally random string using md5 
    $md5_hash = md5(rand(0,999)); 
    //We don't need a 32 character long string so we trim it down to 5 
    $security_code = substr($md5_hash, 15, 5); 

    //Set the session to store the security code
    $_SESSION["security_code"] = $security_code;

    //Set the image width and height 
    $width = 100; 
    $height = 20;  

    //Create the image resource 
    $image = ImageCreate($width, $height);  

    //We are making three colors, white, black and gray 
    $white = ImageColorAllocate($image, 255, 255, 255); 
    $black = ImageColorAllocate($image, 0, 0, 0); 
    $grey = ImageColorAllocate($image, 204, 204, 204); 

    //Make the background black 
    ImageFill($image, 0, 0, $black); 

    //Add randomly generated string in white to the image
    ImageString($image, 3, 30, 3, $security_code, $white); 

    //Throw in some lines to make it a little bit harder for any bots to break 
    ImageRectangle($image,0,0,$width-1,$height-1,$grey); 
    imageline($image, 0, $height/2, $width, $height/2, $grey); 
    imageline($image, $width/2, 0, $width/2, $height, $grey); 

    //Tell the browser what kind of file is come in 
    header("Content-Type: image/jpeg"); 

    //Output the newly created image in jpeg format 
    ImageJpeg($image); 

    //Free up resources
    ImageDestroy($image); 
}

现在，当我尝试include('createCaptcha.php');时，我收到了这个警告：

Warning: Cannot modify header information - headers already sent by (output started at C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\gtw\index.php:14) in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\gtw\registration\createCaptcha.php on line 43

输出是一系列难以理解的位（我认为是那些出现在图像中的位）。实际上我之前设置了标题。我该如何解决这个问题？

Answer 1

不要包含'createCaptcha.php'来调用它

<img src="path/to/your/script/createCaptcha.php"/>

你的index.php包含'createCaptcha.php'首先发送标题并且它与jpg的标题冲突 - Sjoerd更好地解释它

并且还不需要exit()

Answer 2

在调用create_image（）之前不要输出任何内容。显然，index.php正在第14行输出内容。

如果已经有内容发送到客户端，则无法再修改标头，因此服务器无法告诉浏览器该内容实际上是JPEG文件而不是HTML文件。

Answer 3

已发送的标头意味着您已经在页面上输出了一些内容。

如果在发送header（）之前回显某些内容，浏览器将自动发送标题。

PHP - 显示GD图像的麻烦

3 个答案: