我正在使用Msqli进行多查询,该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更改。
下面的查询在以前的LAMP安装中工作,但在当前安装中不工作。
$SQLquery ='set @type = '''';
set @num = 1;
SELECT
RA.`DATE` as DATES,
RA.`ADDR` as ADDR,
RA.`QID` as QID,
RT.`TAGS` as TAG,
Q.`id` AS QUID,
Q.`ADDR` AS QADDR,
Q.`ORIGINALTEXT` AS QTEXTS,
Q.`DATE` AS QDATES,
cs.`id` AS CUID,
cs.`ADDR` AS CADDR,
cs.`ORIGINALTEXT` AS CTEXTS,
cs.`DATE` AS CDATES,
sol.`id` AS SUID,
sol.`ORIGINALTEXT` AS STEXTS,
sol.`ADDR` AS SADDR,
sol.`DATE` AS SDATES,
prj.`id` AS PUID,
prj.`ORIGINALTEXT` AS PTEXTS,
prj.`ADDR` AS PADDR,
prj.`DATE` AS PDATES,
Max(Q.`DATE` ) AS Q,
Max(cs.`DATE` ) AS C,
Max(sol.`DATE` ) AS S,
Max(prj.`DATE` ) AS P,
#@num as row_number
@num:= if(@type = RA.`ADDR`, 1+@num, 1) as v_number,
@type := RA.`ADDR` as dummy
FROM (SELECT `id`,`TAGS`, `QID` from `REL_TAG` ) AS RT
inner Join (SELECT `DATE`, `ADDR`, `QID` from `REL_ADDR` order by DATE) AS RA ON ( RT.`QID` = RA.`QID`)
Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `QUESTION`) AS Q ON ( RT.`QID` = Q.`QID`) and Q.`ADDR` = RA.`ADDR`
Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `CASES` order by `CASES`.`DATE`) AS cs ON ( RT.`QID` = cs.`QID`) and cs.`ADDR` = RA.`ADDR`
Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `SOLUTION` order by `SOLUTION`.`DATE`) AS sol ON ( RT.`QID` = sol.`QID`) and sol.`ADDR` = RA.`ADDR`
Left outer Join (SELECT `id`,`DATE`, `ADDR`, `QID`, `ORIGINALTEXT` FROM `PROJECT` order by `PROJECT`.`DATE`) AS prj ON ( RT.`QID` = prj.`QID`) and prj.`ADDR` = RA.`ADDR`
where RT.`QID` = '''.NbOnly($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,只是不返回任何内容(null)。
以下是执行MySQLi Multiquery 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();
我发现了问题,一些但不是所有的表都是空的,在没有任何记录的情况下,
我给每个表添加了一个伪记录,有些表有记录,但有些没有,现在,即使查询不匹配,连接现在仍在正确生成。
我希望这能帮助很多人。
当您使用MySQL"Join"时,您需要在中的每个和所有中有一条记录,无论它是否会被使用,例如:0,00-00-0000,Null,0,空
只有到那时,查询中包含的所有联接表才能工作。