"Undefined Variable" 尝试读取行时
"Undefined Variable" while trying to read rows
我正在尝试读取一堆行并在没有通常 "ARRAY =>" 废话的情况下显示它们,但我得到的只是这个错误
Notice: Undefined variable: data in C:\xampp\htdocs\sql.php on line 0
这是我的代码:
<?php virtual('/Connections/TDBS_local.php');
$query = "SELECT sub_category FROM sub_category WHERE main_category_id = 2 ";
$result = mysqli_query($TDBS_local, $query);
if ($result = mysqli_query($TDBS_local, $query)) {
/* fetch associative array */
if($row = mysqli_fetch_assoc($result)) {
$data = unserialize($row['sub_category']);
}
/* free result set */
mysqli_free_result($result);
}
print "<pre>";
print_r($data);
print "</pre>";
?>
我哪里错了?
我认为你必须使用 require 或 include 而不是 virtual:
<?php
require('Connections/TDBS_local.php');
$data = '';
$query = "SELECT sub_category FROM sub_category WHERE main_category_id = 2 ";
$result = mysqli_query($TDBS_local, $query);
if ($result = mysqli_query($TDBS_local, $query)) {
/* fetch associative array */
if($row = mysqli_fetch_assoc($result)) {
$data = unserialize($row['sub_category']);
}
/* free result set */
mysqli_free_result($result);
}
print "<pre>";
print_r($data);
print "</pre>";
?>
以下是使用 pdo、mysqli 和 mysql
的方法
MYSQL我
$con=mysqli_connect("localhost","username","password","dbname");
$sql="SELECT sub_category FROM sub_category WHERE main_category_id = 2";
$result=mysqli_query($con,$sql);
$new_array = array();
while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){
$new_array[] = $row;
}
print_r($new_array);
mysqli_free_result($result);
mysqli_close($con);
MYSQL
mysql_connect("localhost","username","password");
mysql_select_db("dbname");
$result = mysql_query("SELECT sub_category FROM sub_category WHERE main_category_id = 2");
$new_array = array();
while ($row = mysql_fetch_array($result)) {
$new_array[] = $row;
}
print_r($new_array);
PDO
$pdo = new PDO('mysql:host=localhost;dbname=dbname', 'username', 'password');
$statement = $pdo->query("SELECT sub_category FROM sub_category WHERE main_category_id = 2");
$new_array = array();
while ($row = $statement->fetch(PDO::FETCH_ASSOC)) {
$new_array[] = $row;
}
print_r($new_array);
我正在尝试读取一堆行并在没有通常 "ARRAY =>" 废话的情况下显示它们,但我得到的只是这个错误
Notice: Undefined variable: data in C:\xampp\htdocs\sql.php on line 0
这是我的代码:
<?php virtual('/Connections/TDBS_local.php');
$query = "SELECT sub_category FROM sub_category WHERE main_category_id = 2 ";
$result = mysqli_query($TDBS_local, $query);
if ($result = mysqli_query($TDBS_local, $query)) {
/* fetch associative array */
if($row = mysqli_fetch_assoc($result)) {
$data = unserialize($row['sub_category']);
}
/* free result set */
mysqli_free_result($result);
}
print "<pre>";
print_r($data);
print "</pre>";
?>
我哪里错了?
我认为你必须使用 require 或 include 而不是 virtual:
<?php
require('Connections/TDBS_local.php');
$data = '';
$query = "SELECT sub_category FROM sub_category WHERE main_category_id = 2 ";
$result = mysqli_query($TDBS_local, $query);
if ($result = mysqli_query($TDBS_local, $query)) {
/* fetch associative array */
if($row = mysqli_fetch_assoc($result)) {
$data = unserialize($row['sub_category']);
}
/* free result set */
mysqli_free_result($result);
}
print "<pre>";
print_r($data);
print "</pre>";
?>
以下是使用 pdo、mysqli 和 mysql
的方法MYSQL我
$con=mysqli_connect("localhost","username","password","dbname");
$sql="SELECT sub_category FROM sub_category WHERE main_category_id = 2";
$result=mysqli_query($con,$sql);
$new_array = array();
while ($row = mysqli_fetch_array($result,MYSQLI_ASSOC)){
$new_array[] = $row;
}
print_r($new_array);
mysqli_free_result($result);
mysqli_close($con);
MYSQL
mysql_connect("localhost","username","password");
mysql_select_db("dbname");
$result = mysql_query("SELECT sub_category FROM sub_category WHERE main_category_id = 2");
$new_array = array();
while ($row = mysql_fetch_array($result)) {
$new_array[] = $row;
}
print_r($new_array);
PDO
$pdo = new PDO('mysql:host=localhost;dbname=dbname', 'username', 'password');
$statement = $pdo->query("SELECT sub_category FROM sub_category WHERE main_category_id = 2");
$new_array = array();
while ($row = $statement->fetch(PDO::FETCH_ASSOC)) {
$new_array[] = $row;
}
print_r($new_array);