传递变量值
问题描述:
我可以知道如何从$ query中传递值,如果其中有值的话。如果它是空的,我仍然需要传递变量。虽然变量确实存在于sql数据库中,但我仍然收到未定义变量的错误。传递变量值
<?php
include("dbconnect.php");
include("header.php");
if (isset($_POST['btn'])) {
$uname = $MySQLi_CON->real_escape_string(trim($_POST['user_name']));
$email = $MySQLi_CON->real_escape_string(trim($_POST['user_email']));
$upass = $MySQLi_CON->real_escape_string(trim($_POST['password']));
$enroller_id_n = $MySQLi_CON->real_escape_string(trim($_POST['enroller_id_n']));
$enrolled_id_n= $MySQLi_CON->real_escape_string(trim($_POST['enrolled_id_n']));
$direction = $MySQLi_CON->real_escape_string(trim($_POST['direction'])) ;
$new_password = password_hash($upass, PASSWORD_DEFAULT);
$query = $MySQLi_CON->query("select * from personal where enroller_id='".$enroller_id_n."'");
if($query){
while ($row = $query->fetch_array()) {
$enroller_id3 = $row['enroller_id'];
$left_mem = $row['left_mem'];
$right_mem = $row['right_mem'];
$test = "left_mem";
$test2 = "right_mem";
$direc = $direction;
}
}
}
?>
答
您需要在这里尝试一些调试。例如,根据该意见
var_dump($enroller_id_n); // Check if the variable is not empty
$query = $MySQLi_CON->query("select * from personal where enroller_id='".$enroller_id_n."'") or die($MySQLi_CON->error);
$row = $query->fetch_array(MYSQLI_ASSOC);
echo "<pre>";
print_r($row); // Check your result array
,你需要删除WHERE从查询条件,将其更改为简单:
$query = $MySQLi_CON->query("select * from personal");
从'$ query'值传递给*,其中*? –
您是否在$ query上尝试了print_r?结果是什么? –
传递$ query的变量,例如$ left mem和$ right mem。这里的问题是我得到未定义的变量,因为我的个人登记员ID是空的 – *