我如何用打印调试PHP图像？ [英] how do i debug PHP Image with print?

问题描述

我想用print调试php生成的图像算法。
问题是它不会输出我打印的文本。
如何输出变量，以便我可以调试？
谢谢，
Furtano

i want to debug an php generated image algorithm with print. The problem is that it doesn't output the text i put in print. How can i output variables, so that i can debug? Thanks, Furtano

 public function drawPicture()
    {
        $im = imagecolorallocate ($this->picture, 255, 0, 255);
        imagettftext($this->picture, $this->fontSize , 0, 100, 100,$im , "cooperm.TTF", $this->name);



        # int ImageCopy ( resource $dst_im , resource $src_im , int $dst_x , int $dst_y , int $src_x , int $src_y , int $src_w , int $src_h )
        //imagecopy($this->picture, $this->pika, $this->wappen['pika']['dst_x'], $this->wappen['pika']['dst_y'], 0, 0, $this->pika_size[0], $this->pika_size[1]);

        $zufall = rand(1,99999999);

        #header("Content-Type: image/jpeg");
        imagepng($this->picture);

        $this->checkFontSize();

        imagedestroy($this->picture);

        print "WHY_DOESNT_PRINT?";


    }

推荐答案

您正在向浏览器发送.png图像，因此浏览器会尝试显示图像。附加文本仅被视为无效图像数据，因此无法显示。

You are sending a .png image to the browser, so the browser tries to display an image. The additional text is just seen as invalid image data and thus not displayed.

您的问题的解决方案是使用 header（）调用将调试消息发送到您的浏览器，或使用Firefox + Firebug + PHP firebug适配器。 Firebug使用标头来传输信息，因此它在图像生成功能中是安全的。

A solution to your problem would be using header() calls to send debug messages to your browser, or use Firefox+Firebug+ a PHP firebug adapter. Firebug works with headers to transmit information, so it's safe in image generation functions.

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