这是 database.php
每次收到此错误时都编写代码:
Notice: Undefined variable: dbConn in C:\xampp\htdocs\couriermanagement\database.php on line 15
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\couriermanagement\database.php on line 15
Warning: mysqli_error() expects exactly 1 parameter, 0 given in C:\xampp\htdocs\couriermanagement\database.php on line 15
<?php
// database connection config
$dbHost = 'localhost';
$dbUser = 'root';
$dbPass = '';
$dbName = 'courier_db';
$dbConn = mysqli_connect($dbHost, $dbUser, $dbPass, $dbName) or die ('MySQL
connect failed. ' . mysqli_error($dbConn));
//mysqli_select_db($dbConn ,$dbName) or die('Cannot select database. ' .
mysqli_error($dbConn));//
function dbQuery($sql)
{
$result = mysqli_query($dbConn,$sql) or die(mysqli_error());
return $result;
}
function dbAffectedRows()
{
global $dbConn;
return mysqli_affected_rows($dbConn);
}
function dbFetchArray($result, $resultType = MYSQL_NUM) {
return mysqli_fetch_array($result, $resultType);
}
function dbFetchAssoc($result)
{
return mysqli_fetch_assoc($result);
}
function dbFetchRow($result)
{
return mysqli_fetch_row($result);
}
function dbFreeResult($result)
{
return mysqli_free_result($result);
}
function dbNumRows($result)
{
return mysqli_num_rows($result);
}
function dbSelect($dbName)
{
return mysqli_select_db($dbName);
}
function dbInsertId()
{
return mysqli_insert_id();
}
?>
最佳答案
“链接”丢失,我认为你从 mysql 升级到 mysqli
它应该看起来像这样:
$dbConn = mysqli_connect("localhost", "my_user", "my_password", "world");
mysqli_query($dbConn, "SELECT * FROM test");
阅读:
http://php.net/manual/en/mysqli.query.php
在程序样式下
关于php - 警告 : mysqli_query() expects at least 2 parameters, 1 在 C :\xampp\htdocs\couriermanagement\database. php 第 15 行给出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46741753/