我查询了数据库表'users'中的'user_id'。并获得一个id数组。
$sel = "SELECT user_id FROM users WHERE status='Approved'";
$result = @mysqli_query ($dbcon, $sel);
然后在另一个表income中插入所有这些用户id的值。
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$ins = "INSERT INTO income (user_id, income_amount) VALUES ('$row', '100')";
$giv = @mysqli_query ($dbcon, $ins);
}
注意:数组到字符串的转换在E:'xampp'htdocs'project't.p p第109行
有谁能帮我解决这个问题吗?$sel = "SELECT user_id FROM users WHERE status='Approved'";
$result = @mysqli_query ($dbcon, $sel);
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$ins = "INSERT INTO income (user_id, income_amount) VALUES ('" . $row['user_id'] . "', '100')";
$giv = @mysqli_query ($dbcon, $ins);
}
首先，检查$results是否在数组中。你可以把一些错误处理检查is_array($result)。
如果它是好的，然后传递给mysqli_fetch_array()。开发时不要添加suppress @ error
我想建议您对此进行单个查询，以便在此之后您无需使用while循环将数据插入收入表:
试试吧:
INSERT INTO income (user_id,income_amount) SELECT user_id,'100' AS income_amount FROM users WHERE status = 'Approved';
你可以这样使用:
$sel = "INSERT INTO income (user_id,income_amount) SELECT user_id,'100' AS income_amount FROM users WHERE status = 'Approved'";
$result = @mysqli_query ($dbcon, $sel);