2

我在 indesign cs6 中编写 js 脚本以导出我的格式化图像时遇到问题。下面的代码（在本网站上找到并稍作修改）仅打开文档。

理想情况下，脚本将遍历我文档中所有格式化/裁剪的图像，并将它们导出到桌面上的新文件夹中，但使用原始文件名。

任何帮助将非常感激：

test();
function test(){

var myDoc = app.open('/Users/StudioA/Desktop/file.indd'); 
var myGroups = myDoc.groups;

//for each group...
for (var i = 0;i < myGroups.length; i++){
    // for each rectangle in the group...
    for(var r = 0; r< myGroups[i].rectangles.length; r++){

         var myRect = myGroups[i].rectangles[r];
           app.jpegExportPreferences.exportResolution = 300;
           app.jpegExportPreferences.jpegQuality = JPEGOptionsQuality.MAXIMUM;

           //give it a unique name
           var myFile = new File('/Users/StudioA/Desktop/Export/' + myRect.name + '.jpg');

           myRect.exportFile(ExportFormat.JPG, myFile);

           }
       }

 }
4

2 回答 2

5

文件名不在矩形上，而是在与放置的图形相关的链接上。给定一个打开的文档，这应该可以满足您的要求：

test();



function test() {

    var myDoc = app.activeDocument, apis = myDoc.allPageItems, rect, fileName;


    while ( rect = apis.pop() )
    {
        if ( !(rect instanceof Rectangle) || !rect.graphics[0].isValid ){ continue; }

        fileName = File ( rect.graphics[0].itemLink.filePath ).name;
        fileName = fileName.replace( /\.[a-z]{2,4}$/i, '.jpg' );

        app.jpegExportPreferences.exportResolution = 300;
        app.jpegExportPreferences.jpegQuality = JPEGOptionsQuality.MAXIMUM;

        //give it a unique name
        var myFile = new File (Folder.desktop+"/"+ fileName);

        rect.exportFile(ExportFormat.JPG, myFile);
    }
}
于 2012-06-11T16:57:27.427 回答
1

只需添加我的详细版本，它适用于 InDesign 中的当前选择并提供控制台反馈。它使用前缀“crop_”重命名图像并将它们保存到 ~/temp

exportSelectedImages();

function exportSelectedImages() {
    // configure export settings
    app.jpegExportPreferences.exportResolution = 72;
    app.jpegExportPreferences.jpegQuality = JPEGOptionsQuality.HIGH;

    // collect selected objects
    var selected = app.activeDocument.selection;
    $.writeln("Got " + selected.length + " selected objects...");

    // process selected objects
    for (var i = 0; i < selected.length; i++) {
        var cursor = selected[i];
        var img = cursor.images;

        $.writeln("Processing #" + (i+1) + "/" + selected.length);
        $.writeln("\t Type: " + cursor.constructor.name);

        // verify if object contains an image or not
        if (cursor.images.length > 0) {     
            var img = cursor.images[0];
            $.writeln("\t Contains image of type " + img.imageTypeName);
            var imageFileName = cursor.images[0].itemLink.name;
            $.writeln("\t File Name: " + imageFileName);
        } else {
            $.writeln("\t Not an image");
        }

        // save the object to a jpeg in path specified below
        var myFile = new File('~/temp/' + "crop_" + imageFileName + '.jpg');
        cursor.exportFile(ExportFormat.JPG, myFile);

     }

    $.writeln("Done.");
}
于 2014-08-27T04:07:57.507 回答