2

我有一个名为categoriescolumns的 SQLite FTS3 虚拟表category, id, header, content。我想搜索它，并让结果同时包含列数据和snippet()结果。

最终期望的行为是显示搜索词的每个匹配项的类别和 ID，然后如果在内容列中找到搜索词，则还显示每个片段。像这样：

Search term "virus":

Category: Coughing
ID: Coughing Symptoms
Snippet 1: "the common cold, which is spread by the cold <b>virus</b> blah blah etc"
Snippet 2: "lorem ipsum some other things <b>virus</b> and then some other stuff"

Category: Coughing
ID: Coughing treatment
Snippet 1: "...coughing can be treated by managing the symptoms. If a <b>virus</b> blah etc"

Category: Headache
ID: Headache Symptoms
Snippet: "I think you get the idea now <b>virus</b> more things"

我现在可以将搜索行结果和片段作为单独的函数获取。这将搜索表格并正确打印出 row.elements：

 function SearchValueInDB() {

            db.transaction(function(transaction) {
               var search = $('#txSearch').val(),
                   query = "SELECT * FROM guidelines WHERE guidelines MATCH '" + search + "*';";

               transaction.executeSql(query,[], function(transaction, result) {
                   if (result != null && result.rows != null) {
                       if (result.rows.length == 0) {
                          //no results message
                       } else {
                           for (var i = 0; i < result.rows.length; i++) {
                               var row = result.rows.item(i);
                               $('#lbResult').append('<br/> Search result:' + row.category + ' ' + row.id + ' ' + row.header + '<br/>');

                           }
                       }
                   }

               },nullHandler,errorHandler);
            });

        }

这将搜索content列中的片段，并打印出它们的列表：

function SearchForSnippet() {

            db.transaction(function(transaction) {
                var search = $('#txSearch').val(),
                        query = "SELECT snippet(guidelines) FROM guidelines WHERE content MATCH '" + search + "*';";

                transaction.executeSql(query,[], function(transaction, result) {

                    if (result != null && result.rows != null) {
                        $('#lbResult').html('');

                        for (var i = 0; i < result.rows.length; i++) {
                            var row = result.rows.item(i);

                            for(var key in row) {
                                var value = row[key];
                                $('#lbResult').append('<br/> ' + i + value );
                            }
                        }
                    }

                },nullHandler,errorHandler);
            });

        }

到目前为止，我可以想象两种可能的方法：要么我可以以某种方式组合 SELECT 查询——尽管我找不到任何使用片段（）函数的 JOINing 查询示例。或者，我可以创建一个新函数findSnippet()，并在每次遍历result.rows数组后调用它。这些方法中的任何一种都可能奏效，还是有更好的方法来处理这个问题？

4

1 回答 1

1

只需列出要为每个匹配记录获取的所有结果列：

SELECT category,
       id,
       header,
       snippet(guidelines) AS snip
FROM guidelines
WHERE content MATCH 'snail*'
于 2013-12-01T08:57:10.743 回答