想要渲染图像而不使用PHP GD lib将其保存到磁盘 [英] Want to render an image without saving it to disk using PHP GD libs

问题描述

您好，具有以下功能

function renderBusinessCard($details){
        //Getting the template for the business card
        $filename = Templates::model()->getTemplateFileName($details['BusinessCards']['dp_id']);
        header("Content-type: image/jpeg");

        $image = $_SERVER['DOCUMENT_ROOT'].'resources/templates/'.$filename;


//        header("Content-Type: image/jpeg");
        //Getting the width and height of the image
        list($width,$height) = getimagesize($image);


//echo $width;die;
        //Creating a copy of a loaded image
        $create = imagecreatefromjpeg($image);


        //Creating a blank template to work from
        $template = imagecreatetruecolor($width,$height);
        imagecopyresized($template, $create, 0, 0, 0, 0, $width, $height, $width, $height);

        imagejpeg($template, null, 100);

    }

我想要实现的是在图像标签中显示图像，如下所示:

What I want to achieve is to display the image within an image tag like the following

<img src="<?php renderBusinessCard($details); ?>"

，但是总会以某种方式在屏幕上呈现乱码，而不是我想要的结果.这是可能吗?以及如何在不使用单独文件的情况下，将其保留在函数或方法中.

but somehow it's always rendering garbled code to the screen and not the result I want. Is this at all possible? And how without using a seperate file, I want this to remain within a function or method.

推荐答案

您需要在单独的资源中渲染图像.没有在HTML文档中添加图像数据的好方法. (有人想到推荐data: URI:这是一个糟糕的主意.)

You need to render the image in a separate resource. There is no good way of adding the image data inside the HTML document. (Anybody thinking of recommending data: URIs: It's a terrible idea.)

<img src="renderBusinessCard.php?param1=1&param2=2">

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