Php 如何在mysqli中将一列添加到多个表中?
首先,我选择一个表从id列读取值Php 如何在mysqli中将一列添加到多个表中?,php,mysqli,Php,Mysqli,首先,我选择一个表从id列读取值 $db_hand1 = mysqli_connect($server, "root", "password",$database1); $result = mysqli_query($db_hand1, "SELECT id FROM table"); 现在,我想将一列添加到另一个数据库的表中: $db_hand2 = mysqli_connect($server, "root", "password",$database2); while($row = m
$db_hand1 = mysqli_connect($server, "root", "password",$database1);
$result = mysqli_query($db_hand1, "SELECT id FROM table");
现在,我想将一列添加到另一个数据库的表中:
$db_hand2 = mysqli_connect($server, "root", "password",$database2);
while($row = mysqli_fetch_array($result)){
$user_id= $row['id'];
$result = mysqli_query($db_hand2,"ALTER TABLE $user_id ADD us_id INT( 1 ) NULL DEFAULT '1'");
}
然而,这段时间不起作用。它总是在第一个表中添加一列。当我手动操作时:
$db_hand2 = mysqli_connect($server, "root", "password",$database2);
$user_id= "table_name";
$result = mysqli_query($db_hand2,"ALTER TABLE $user_id ADD us_id INT( 1 ) NULL DEFAULT '1'");
并手动为其工作的表名设置$user\u id。您覆盖
$result
,更改如下
$db_hand2 = mysqli_connect($server, "root", "password",$database2);
while($row = mysqli_fetch_array($result)){
$user_id= $row['id'];
$result_alter = mysqli_query($db_hand2,"ALTER TABLE $user_id ADD us_id INT( 1 ) NULL DEFAULT '1'");
}
您遇到了什么错误?您正在对这两个查询使用
$result
,请将其更改为另一个变量,可能$result2=mysqli\u query($db\u hand2,“ALTER TABLE$user\u id ADD us\u id INT(1)NULL默认值“1”)代码>