Question

当要搜索的文本大于255个字符时，

context.document.body.search()会在Word 2016中为桌面引发InvalidArgument个异常。这似乎与Word Online一起使用。

var searchResults = context.document.body.search("TextMoreThan255Chars");
context.load(searchResults);
return context.sync().then(function() {

});

Answer 1

您遇到的是Word（对于台式机）的限制。例如，某些more information在Greg Maxey的网站上可用。 VBA的建议可以适应Office JS方法。

关键是将搜索词分解成少于255个字符的限制。搜索术语的第一部分（如果找到），搜索下一部分。如果找到了它，请检查它是否紧跟在第一个找到的范围之后。重复直到搜索完所有部分。

下面的示例代码在我的测试中很适合我-但是我不是JS人员，因此它的结构可能不是最佳的。它至少应提供有关如何执行任务的基础知识。最重要的概念是使用range对象：找到的所有range的副本都需要扩展到文档正文范围的末尾。这提供了搜索术语下一部分的基础范围。

如果找到下一个零件，请确定它是否紧邻原始找到的范围（compareLocationWith）。如果是这样，请继续循环播放，直到处理完整个搜索词为止。

最后，突出显示找到的术语。

async function basicSearch() {
    await Word.run(async (context) => {
        let maxNrChars = 254;
        let searchterm = "";
        let shortSearch = true; //search string < 255 chars
        let fullSearchterm = "Video provides a powerful way to help you prove your point. When you click Online Video, you can paste in the embed code for the video you want to add. You can also type a keyword to search online for the video that best fits your document. Aösdlkvaösd faoweifu aösdlkcj aösdofi "
        let searchTermNrChars = fullSearchterm.length;
        let nrSearchCycles = Number((searchTermNrChars / maxNrChars).toFixed(0));
        let nrRemainingChars = searchTermNrChars - (nrSearchCycles * maxNrChars);
        //console.log("Number of characters in search term: " + searchTermNrChars
        //    + "\nnumber of search cycles required: " + nrSearchCycles
        //    + "\nremaining number of characters: " + nrRemainingChars);

//numerous ranges are required to extend original found range
        let bodyRange = context.document.body.getRange();
        bodyRange.load('End');
        let completeRange = null;
        let resultRange = null;
        let extendedRange = null;
        let followupRange = null;

        let cycleCounter = 0;
        let resultText = "";
        if (searchTermNrChars > maxNrChars) {
            searchterm = fullSearchterm.substring(0, maxNrChars);
            cycleCounter++;
            shortSearch = false;
        }
        else { searchterm = fullSearchterm; }

        let results = context.document.body.search(searchterm);
        results.load({ select: 'font/highlightColor, text' });

        await context.sync();

        // short search term, highlight...
        if (shortSearch) {
            for (let i = 0; i < results.items.length; i++) {
                results.items[i].font.highlightColor = "yellow";
            }
        }
        else {
            //console.log("Long search");
            for (let i = 0; i < results.items.length; i++) {
                resultRange = results.items[i];
                resultRange.load('End');
                extendedRange = resultRange.getRange('End').expandTo(bodyRange.getRange('End'));

                await context.sync();

                //search for the remainder of the long search term
                for (let cycle = 1; cycle < nrSearchCycles; cycle++) {
                    searchterm = fullSearchterm.substring((cycle * maxNrChars), maxNrChars);
                    //console.log(searchterm + " in cycle " + cycle);
                    let CycleResults = extendedRange.search(searchterm);
                    CycleResults.load({ select: 'text, Start, End' });
                    await context.sync();
                    followupRange = CycleResults.items[0];

                    //directly adjacent?
                    let isAfter = followupRange.compareLocationWith(resultRange);
                    if (isAfter.value == Word.LocationRelation.adjacentAfter) {
                        resultRange.expandTo(followupRange);
                        extendedRange = resultRange.getRange('End').expandTo(bodyRange.getRange('End'));
                    }
                    await context.sync();
                }

                if (nrRemainingChars > 0) {
                    console.log("In remaining chars");
                    searchterm = fullSearchterm.substring(searchTermNrChars - nrRemainingChars);
                    console.log(searchterm);
                    let xresults = extendedRange.search(searchterm);
                    xresults.load('end, text');
                    await context.sync();

                    let xresult = xresults.items[0];

                    let isAfter = xresult.compareLocationWith(resultRange);

                    await context.sync();
                    console.log(isAfter.value);
                    if (isAfter.value == Word.LocationRelation.adjacentAfter) {
                        completeRange = resultRange.expandTo(xresult);
                        completeRange.load('text');
                        //completeRange.select();
                        await context.sync();
                        resultText = completeRange.text.substring(0, fullSearchterm.length);
                        console.log("Result" + cycleCounter + ": " + resultText);
                    }  
                }
                else {
                    //No remeaining chars
                    resultRange.load('text');
                    //resultRange.select();
                    await context.sync();
                    resultText = resultRange.text.substring(0, fullSearchterm.length);
                    completeRange = resultRange; 
                }

                //long search successful?
                if (resultText == fullSearchterm) {
                    completeRange.font.highlightColor = "yellow";
                }
                else {
                    console.log("Else! " + resultText + "  /  " + fullSearchterm);
                }
                completeRange = null;
            }
        }  
    });

body.search word桌面问题

1 个答案: