Question

最后，我需要结果来包含来自所有3个表的所有数据。如何加入3个表，所以最终都存在。以下是我到目前为止的情况：

if ($results = $db->query("SELECT * FROM HWData
                        LEFT JOIN ClassData
                        ON HWData.class = ClassData.class
                        LEFT JOIN Judges
                        ON Judges.groupnum = ClassData.groupnum
                        ORDER BY HWData.entrynum",
                        MYSQLI_USE_RESULT)) {
$result_set = mysqli_fetch_all($results, MYSQLI_ASSOC);
echo json_encode($result_set);

结果与预期不符。奇怪的是console.log[i]['groupnum'];显示正确的groupnum但$("#output").append("<p>"+data[i]['groupnum']);只显示1的

3个表格如下：

ClassData

groupnum    abbr
0   DQB
0   DQC
1   SUM
1   PP
1   FOS
1   VANF
1   VANS
2   AFCF
2   AFCS
2   ICCS
2   CCS
3   LSUF
3   LTMS
4   HL
4   MC

士师记

judge   group
Coleen  4
Daniel  3
Aaron   2
Jeff    1

HWDATA

entrynum    fname   class
1   Esteban FOS
2   Dalia   PP
3   Sheri
4   Sheri   HL
5   Sheri   MC
10  Danita  ICCS
11  Chris   AFCS

Answer 1

好的，所以我发现你可以使用多个连接。

if ($results = $db->query("SELECT * FROM HWData
                           LEFT JOIN ClassData
                           ON HWData.Abbr = ClassData.Abbr
                           LEFT JOIN Judges
                           ON Judges.groupnum = ClassData.groupnum",
                            MYSQLI_USE_RESULT)) {
$result_set = mysqli_fetch_all($results, MYSQLI_ASSOC);
echo json_encode($result_set);
$results->close();
}

如何使用3个表进行mysqli内连接

1 个答案: