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我正在使用 Msqli 进行多重查询，查询的工作方式与我在 phpmyadmin 中想要的完全一样。

它不再适用于 mysqli。查询在服务器之间没有改变。

下面的查询在以前的 LAMP 安装中有效，但在当前无效。

$SQLquery ='set @type = \'\';
set @num  = 1;
SELECT 
RA.`DATE` as DATES, 
RA.`ADDR` as ADDR,
RA.`QID` as QID, 
RT.`TAGS` as TAG,
Q.`id` AS QUID,
Q.`ADDR` AS QADDR, 
Q.`ORIGINALTEXT` AS QTEXTS, 
Q.`DATE` AS QDATES, 
cs.`id` AS CUID,
cs.`ADDR` AS CADDR, 
cs.`ORIGINALTEXT` AS CTEXTS, 
cs.`DATE` AS CDATES,
sol.`id` AS SUID, 
sol.`ORIGINALTEXT` AS STEXTS,
sol.`ADDR` AS SADDR, 
sol.`DATE` AS SDATES,
prj.`id` AS PUID,  
prj.`ORIGINALTEXT` AS PTEXTS, 
prj.`ADDR` AS PADDR,
prj.`DATE` AS PDATES, 
Max(Q.`DATE` ) AS Q, 
Max(cs.`DATE` ) AS C, 
Max(sol.`DATE` ) AS S, 
Max(prj.`DATE` ) AS P,
#@num as row_number 
@num:= if(@type = RA.`ADDR`, 1+@num, 1)  as v_number,
@type := RA.`ADDR` as dummy
FROM        (SELECT `id`,`TAGS`, `QID`    from `REL_TAG` ) AS RT 
 inner Join (SELECT `DATE`, `ADDR`, `QID` from `REL_ADDR` order by DATE) AS RA ON ( RT.`QID` = RA.`QID`) 
 Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `QUESTION`) AS Q ON ( RT.`QID` = Q.`QID`) and Q.`ADDR` = RA.`ADDR`
 Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `CASES` order by `CASES`.`DATE`) AS cs ON ( RT.`QID` = cs.`QID`) and cs.`ADDR` = RA.`ADDR` 
 Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `SOLUTION` order by `SOLUTION`.`DATE`) AS sol ON ( RT.`QID` = sol.`QID`) and sol.`ADDR` = RA.`ADDR`
 Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `PROJECT`  order by `PROJECT`.`DATE`) AS prj ON ( RT.`QID` = prj.`QID`) and prj.`ADDR` = RA.`ADDR`
where RT.`QID` = \''.NbOnly($Fetchmodifier).'\' Group by `QID`, addr, v_number LIMIT '.$Anstart.' ,'.$Ansnb.';';

更新 查询不会在日志中返回任何错误，它只是不返回任何内容（null）。

这是执行 MySQLi 多查询的 PHP 代码

$mysqlin = new mysqli("localhost", "user", "pass", "db");
if (mysqli_error($mysqlin)) {
     outputdataXML(sprintf("Connect Problem : %s\n", mysqli_error($mysqlin)));
     exit();
}
if ($mysqlin->multi_query($SQLquery)) {
    do {
          if ($result = $mysqlin->store_result()) {
               while ($row = $result->fetch_array(MYSQLI_BOTH)) {
                   if( $Foundrows = $row[0]){ //Maybe the problem is here ?
                     $Outputvalue[] = FormatFile($row);
                   }
               }
               $result->free();
          }
          if ($mysqlin->more_results()) {
          }
       } while ($mysqlin->more_results() && $mysqlin->next_result());
}
$mysqlin->close();
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1 回答 1

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我发现了问题，一些但不是所有的表都是EMPTY，如没有任何记录，

我为每个表添加了一个虚拟记录，有些表有记录，但有些没有，现在即使查询不匹配连接现在仍然正确制作。

我希望这对很多人有帮助。

当您使用 MySQL "Join"时，您需要在每个和所有连接表中都有1 条记录，无论它是否会被使用，例如：0、00-00-0000、Null、0、空

只有这样，查询中包含的所有连接表才会起作用。

于 2012-09-21T08:10:44.173 回答