Question

我有这个问题：

SELECT `ca`.*, `a`.* FROM `container_asset` AS `ca` LEFT JOIN `asset` AS `a` ON ca.asset_id = a.id WHERE `container_id` = '4' AND `deleted` = '0' AND `reason` IN ('main', 'previous_version') ORDER BY `reason` ASC, `asset_id` DESC

我希望返回两行，但是当我用这个命令执行查询时：

$conn =new  \mysqli('localhost', 'root', '12345', 'assetdb') or die("Error " . mysqli_error($link));
$res3 = $conn->query("SELECT `ca`.*, `a`.* FROM `container_asset` AS `ca` LEFT JOIN `asset` AS `a` ON ca.asset_id = a.id WHERE `container_id` = '4' AND `deleted` = '0' AND `reason` IN ('main', 'previous_version') ORDER BY `reason` ASC, `asset_id` DESC");
while ($r = mysqli_fetch_array($res3))
        {
            echo $r['id']."\n";
        }

在获取结果时返回1行但是当我直接执行查询时（在heidisql或phpmyadmin中），它会返回预期的结果，即2行。

我无法解释这个问题？

修改

我做了一些测试，如果我期待5行它返回4，所以它返回1行。

EDIT2

数据： container_asset表，asset_id foreing key in asset-＆gt; id：

"container_id"  "asset_id"  "reason"
"1"          "2"    "main"
"2"          "1"    "main"
"3"          "3"    "main"
"4"          "3"    "previous_version"
"4"          "4"    "main"

资产表

"id"    "asset_class"   "deleted"   "file_name"
"1" "video"          "0"             "ÄÖ.jpg"
"2" "image"          "0"             "èungrandeguitar.jpg"
"3" "image"          "0"              "3.jpg"
"4" "image"          "0"              "çàù.jpg"

编辑3

即使我删除了联接，结果仍然是错误的：

SELECT `ca`.* FROM `container_asset` AS `ca` WHERE ca.`container_id` = '4'  AND `reason` IN ('main', 'previous_version') ORDER BY `reason` ASC, `asset_id` DESC

mysqli返回1行而不是2行

0 个答案: