Question

编辑 - 解决方案：将(function() { // input here contains c.data from client script if (input && input.cases) { for (var i = 0; i < input.cases.length; i++) { // write to system log gs.info(input.cases[i]); } } });移至第一次之前
我制作了这个代码来展示比赛的记分牌第一次：获取所有类别。
第二个时间：从类别中获取所有子类别第三个：获取子类别的所有记录问题是：
如果DESC的第一个循环顺序是仅打印第一个结果并对此类别执行如果我改为ASC，它什么都不打印注意：并非每个子类别都有记录。

$sub = '';

Answer 1

也许它可以帮助重新格式化查询以反映蝙蝠的变量值（而不是以后处理它们 - 在软件中可能是也可能不是这样）：取代

$ querysub = $ db-＆gt; query（＆＃34; SELECT id，id_category，name FROM category_sub WHERE id_category = $ categoryid＆＃34;）;

与

$ querysub = $ db-＆gt; query（＆＃34; SELECT id，id_category，name FROM category_sub WHERE id_category =＆＃34;。$ categoryid）;

并替换

$ record = $ db-＆gt; query（＆＃34; SELECT records.id as id，competitor.name as competitionitor，category.name as category，category_sub.name as subcategory，records.amplifier as amplifier，records。扬声器作为扬声器，records.battery作为电池，records.decibels作为分贝，records.car作为汽车FROM记录LEFT JOIN竞争者ON竞争者.id = records.id_competitor LEFT JOIN类别ON records.id_category = category.id LEFT JOIN category_sub ON记录.id_subcategory = category_sub.id WHERE category_sub.name =＆＃39;。$ subcategoryname。＆＃39; ORDER BY points ASC＆＃34;）;

与

$ record = $ db-＆gt; query（＆＃34; SELECT records.id as id，competitor.name as competitionitor，category.name as category，category_sub.name as subcategory，records.amplifier as amplifier，records。扬声器作为扬声器，records.battery作为电池，records.decibels作为分贝，records.car作为汽车FROM记录LEFT JOIN竞争者ON竞争者.id = records.id_competitor LEFT JOIN类别ON records.id_category = category.id LEFT JOIN category_sub ON记录.id_subcategory = category_sub.id WHERE category_sub.name =＆＃39;＆＃34;。$ subcategoryname。＆＃34;＆＃39; ORDER BY points ASC＆＃34;）;

此外，您要停止在此行执行整个代码：

if(!is_numeric($subcategory['id'])){
    continue;
}

将continue替换为break以仅停止该次迭代并继续下一次迭代：

if(!is_numeric($subcategory['id'])){
    break;
}

虽然在里面而在里面

1 个答案: