在php中从mysqli连接查询访问行
我有以下代码:在php中从mysqli连接查询访问行,php,mysql,mysqli,left-join,Php,Mysql,Mysqli,Left Join,我有以下代码: // db connection info set up earlier $sql= "SELECT `TABLE_1.ID`, `TABLE_2.ID`, `POTATO` FROM `TABLE_1.ID` LEFT JOIN `TABLE_2` ON `TABLE_1`.`ID` = `TABLE_2`.`ID_OF_OTHER_TABLE`;"; $rows = mysqli_query($connection, $sql); foreach ($rows as $row
// db connection info set up earlier
$sql= "SELECT `TABLE_1.ID`, `TABLE_2.ID`, `POTATO` FROM `TABLE_1.ID` LEFT JOIN `TABLE_2` ON `TABLE_1`.`ID` = `TABLE_2`.`ID_OF_OTHER_TABLE`;";
$rows = mysqli_query($connection, $sql);
foreach ($rows as $row){
$potato = $row["POTATO"];
$id = $row["TABLE_2.ID"];
}
我找不到表2.ID。我尝试过打印以获得正确的格式,但它说这是一个mysqli对象,我没有得到更多的信息。然而,我可以得到土豆。所以我猜这是一个调用语法问题,但我搜索了多个站点(包括stack和google),没有找到明确的答案。那么,我需要做什么来代替
$id = $row["TABLE_2.ID"];
?为具有相同名称的列指定别名
$sql= "SELECT `TABLE_1`.`ID` AS t1_id, `TABLE_2`.`ID` AS t2_id, `POTATO`
FROM `TABLE_1.ID`
LEFT JOIN `TABLE_2` ON `TABLE_2`.`ID_OF_OTHER_TABLE` = `TABLE_1`.`ID`;";
$rows = mysqli_query($connection, $sql);
foreach ($rows as $row){
$potato = $row["POTATO"];
$id = $row["t2_id"];
}
为具有相同名称的列指定别名
$sql= "SELECT `TABLE_1`.`ID` AS t1_id, `TABLE_2`.`ID` AS t2_id, `POTATO`
FROM `TABLE_1.ID`
LEFT JOIN `TABLE_2` ON `TABLE_2`.`ID_OF_OTHER_TABLE` = `TABLE_1`.`ID`;";
$rows = mysqli_query($connection, $sql);
foreach ($rows as $row){
$potato = $row["POTATO"];
$id = $row["t2_id"];
}
不能将一个表与它自己连接,也不能有一个未连接的表=来自另一个表的不能与它自己连接的东西 这将有助于:
// db connection info set up earlier
$sql= "SELECT TABLE_1.ID, TABLE_2.ID, POTATO
FROM
TABLE_1.ID
LEFT JOIN TABLE_2 ON TABLE_1.ID = TABLE_2.ID";
$rows = mysqli_query($connection, $sql);
while ($row = mysqli_fetch_assoc($rows)) {
echo ($row['ID']);
}
mysql_free_result($rows);
不能将一个表与它自己连接,也不能有一个未连接的表=来自另一个表的不能与它自己连接的东西 这将有助于:
// db connection info set up earlier
$sql= "SELECT TABLE_1.ID, TABLE_2.ID, POTATO
FROM
TABLE_1.ID
LEFT JOIN TABLE_2 ON TABLE_1.ID = TABLE_2.ID";
$rows = mysqli_query($connection, $sql);
while ($row = mysqli_fetch_assoc($rows)) {
echo ($row['ID']);
}
mysql_free_result($rows);
我怀疑将表与自身连接只是一个输入错误。您没有使用正确的列进行连接。应该是
表2。其他表的id\u
@Barmar正确,这是一个打字错误,抱歉。我修复了它。我怀疑将表本身连接只是一个输入错误。您没有使用正确的列进行连接。应该是表2。其他表的id\u
@Barmar正确,这是一个打字错误,抱歉。我修好了。