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Amazon Product Advertising API を使用する Web アプリケーションを構築しています。問題は、API が商品ごとに異なるサイズと縦横比の画像を返すことです。すべての画像のサイズを静的な幅/高さに変更すると、比率が変わるために一部の画像が奇妙に見えます。画像はそれぞれ 4 つの行に配置されます。APIが奇妙なサイズの画像を返す場合に備えて、縦横比を同じに保ち、高さの最大しきい値を設定しながら、画像の幅のみを同じにすることを考えていました。

しかし、ここにいる人々が以前にこれに出くわしたことがあるかどうか、またこの設計上の問題についてどう考えているかを知りたかったのです。

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2 に答える 2

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なんて偶然。私も同様の問題に直面していましたが、これが私が前進することを決めた方法です。それは最善の解決策ではないかもしれませんが、私にとってはうまくいきます。

まず、画像の元の高さと幅を取得します

list($width, $height) = getimagesize("path to image");

次に、2つの最大公約数を見つけて、幅と高さの比率を次のような変数に格納します$wr。$hr

$wr > $hr次に、水平方向と$hr > $wr垂直方向を比較して、画像の方向（水平または垂直）を確認します。

水平の場合、親指のサイズが特定の値、たとえば120pxを超えないようにし、アスペクト比に基づいて高さを120pxに対応させます。向きが垂直の場合も同じです。

于 2012-07-31T08:23:13.837 に答える
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I came across the same problem trying to standardize logos. At first, I thought that I could choose a standard area and resize all the images to that area, but there are two problems with that strategy. First, you have to set limits for height and width, so sometimes you end up images with a smaller area because they are very wide or tall. Secondly, that might make your layout look ugly. In my case I have a header on top, an image and then text below. I realized that a very wide image would leave a big gap between the header and the text. Because of that, I needed a solution that would give preference to width. What I mean by that is that the image can change a lot in width without changing the height a lot. Different applications may have different requirements, but I think my solution allows for any situation such as a preference for height as well:

function width_to_height($width, $slope, $icpt){
    if ($width == 0){
        return FALSE;
    }
    return floor(($icpt + sqrt(pow($icpt, 2) + 4*$slope*pow($width, 2)))/(2*$width));
} 

function ratio_to_height($ratio, $slope, $icpt){
    if ($ratio == 0){
        return FALSE;
    }
    $area = $ratio*$slope + $icpt;
    return floor(sqrt($area/$ratio));
}   

function calc_dims($width, $height, $max_w=168, $max_h=100){
    $slope = 2500;
    $icpt = 6000;
    $ratio = $width/$height;
    $max_ratio = $max_w/$this->width_to_height($max_w, $slope, $icpt);
    if ($ratio > $max_ratio){
        $ht = floor($max_w/$ratio);
        return array('width' => $max_w, 'height' => $ht);
    }
    $ht = $this->ratio_to_height($ratio, $slope, $icpt);
    if ($ht > $max_h){
        $wd = floor($max_h*$ratio);
        return array('width' => $wd, 'height' => $max_h);
    }
    $wd = floor($ht*$ratio);
    return array('width' => $wd, 'height' => $ht);
}

The main function is calc_dims() which calls the other two functions. Since I almost always work with classes, these functions are called with the "this->" operator. If you don't use classes, you can simply delete the operator.

You can see that I hard-coded some variables. (In my real life application, these are called from a configuration file.) $max_w and $max_h are pretty self explanatory as the maximum height and width of the images. However $slope and $icpt may be a little more difficult to understand. As you can see in the ratio_to_height() function, $slope and $icpt (intercept) are components in a linear relationship between area and the aspect ratio of the image.

You can use the values that I provide ($slope = 2500; $icpt = 6000), or you can calculate your own. I thought of automating the process a little more, but since the parameters are highly subjective, it didn't seem very practical. In order to calculate them, it's necessary to define areas for two different instances of the aspect ratios, where the ratio is $width/$height. For example, when the ratio is 1 ($width = $height), you might want the area to be 8000 pixels^2, and when the ratio is 2 ($width = 2*$height) the area could be 12000 pixels^2. With these values we have:

$A1 = 8000;
$A2 = 12000;
$r1 = 1;
$r2 = 2;

You can calculate slope and intercept as follows:

$slope = ($A2 - $A1)/($r2 - $r1);
$icpt = $A1 - $slope*$r1;
于 2013-01-10T10:15:27.373 に答える