Question

$status = 1;
$user_name = aaa;
$mysql_qry = "INSERT INTO l_status (user_id , status) values ('select id from user where username in (''$user_name'')','$status')";
if($conn->query($mysql_qry) === TRUE ) {
echo "inserted.";
}
else {
echo "Error" . $mysql_qry . "<br>" . $conn->error;
}

我想在l_status表中插入id和status，我需要从用户表中选择id。

值已插入，但每次都是id = 0。在用户表中id对于不同的用户是不同的。

请帮助我，这是php的新手。

Answer 1

拆分：

$mysql_qry = 'SELECT id FROM user WHERE username = "'.$user_name.'"';
while($res = $conn->query($mysql_qry){
$id = $res['id'];
$mysql_qry2 = "INSERT INTO l_status (user_id , status) values ('$id','$status')";
if($conn->query($mysql_qry2) === TRUE ) {
echo "inserted.";
}else {
echo "Error" . $mysql_qry2 . "<br>" . $conn->error;
    }
}

或在一个查询中：

$mysql_qry = "INSERT INTO l_status (user_id , status) values ((select id from user where username = '$user_name') ,'$status')";

请等一下你的$ var ='test'; - ＆GT;你必须引用它们。看看你的

$user_name = 'aaa';

Answer 2

$user_name = 'aaa';
$mysql_qry = "INSERT INTO chaosgam_ghs.login_status (user_id , status) values ((select user_id from chaosgam_ghs.user_table where username = '$user_name'),'$status')";

此代码中显示了2个错误。

Android-PHP，与其他查询数据一起插入表中

2 个答案: