调整图像大小并保持纵横比以适合 iPhone 的算法 [英] Algorithm to resize image and maintain aspect ratio to fit iPhone

问题描述

我正在为 iPhone 应用创建一个网络服务来与之交互.

I'm creating a web service for an iPhone app to interact with.

当我的客户端在服务器端上传图像时，我希望我的 php 脚本调整图像大小，同时保持纵横比，使其适合 iPhone 屏幕.(即最长边 <= 960，最短边 <= 640

When my client uploads images server-side, I want my php script to resize the image, whilst maintaining the aspect ratio, so that it will fit onto the iPhone screen. (i.e. the longest side is <= 960 and the shortest <= 640

我用 JS 创建了一个模型，只是因为我发现它更容易快速完成.

I've created a mock-up in JS, simply because I find it easier to do quickly.

我很确定，虽然我可能错了，但这不是最有效的方法.有人可以用更好的逻辑(尤其是开始时的那一点)或更数学的方法来纠正我吗?

I'm pretty sure, though I may be wrong, that this isn't the most efficient way of doing it. Could someone correct me with either better logic (especially the bit at the start), or a more mathematical way of approaching this?

var w = 960, h = 960, new_w, new_h;
if (w >= h && w > 960 || h >= w && h > 960 || w >= h && h > 640 || h >= w && w > 640) {
    if (w > h) {
        if (w>960) {
            new_w = 960;
            new_h = h*(new_w/w);
        }
        if (h>640) {
            new_h = 640;
            new_w = w*(new_h/h);
        }
    }
    else {
        if (h>960) {
            new_h = 960;
            new_w = w*(new_h/h);
        }
        if (w>640) {
            new_w = 640;
            new_h = h*(new_w/w);
        }
    }
}

推荐答案

我认为以下内容应该能让您有所了解.它不是使用任何特定语言，而是类似于 C 的伪代码.

I think the following should give you the idea. It's not in any particular language, but rather a C-like pseudo code.

shortSideMax = 640;
longSideMax = 960;
function Resize(image)
{
    if (image.width >= image.height)
    {
        if (image.width <= longSideMax && image.height <= shortSideMax)
            return image;  // no resizing required
        wRatio = longSideMax / image.width;
        hRatio = shortSideMax / image.height;
    }
    else
    {
        if (image.height <= longSideMax && image.width <= shortSideMax)
            return image; // no resizing required
        wRatio = shortSideMax / image.width;
        hRatio = longSideMax / image.height;
    }

    // hRatio and wRatio now have the scaling factors for height and width.
    // You want the smallest of the two to ensure that the resulting image
    // fits in the desired frame and maintains the aspect ratio.
    resizeRatio = Min(wRatio, hRatio);

    newHeight = image.Height * resizeRatio;
    newWidth = image.Width * resizeRatio;

    // Now call function to resize original image to [newWidth, newHeight]
    // and return the result.
}

这段代码的效率，或者你所拥有的，都不是问题.实际调整图像大小所需的时间比进行几次比较、两次除法和两次乘法所需的时间相形见绌.

The efficiency of this code, or what you have, won't be an issue. The time it takes to actually resize the image will dwarf the time it takes to do a couple of comparisons, two divides, and two multiplies.

这是一种更数学化"的方法吗?我想，因为它将您的四个案例折叠为两个.但方法本质上是一样的.

Is this a "more mathematical" way to do it? I suppose, in that it collapses your four cases into two. But the approach is essentially the same.

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