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 = '';答案 0 :(得分：0)
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;
}