php mysqli_query 结果没有[重复]

Question

我正在尝试使用 mysqli_query 从表中获取数据。 当我使用以下命令时，它可以正常工作：

$hostname = "********";
$username = "*******";
$password = "********";
$databaseName = "**************";
$dbConnected = mysqli_connect($hostname, $username, $password, $databaseName);

当我尝试使用上述代码包含文件时 include('../htconfig/dbConfig.php');然后我没有得到任何结果：

...“0 个结果”

<?php

include('../htconfig/dbConfig.php');
$dbConnected = mysqli_connect($db['hostname'], $db['username'], $db['password'], $db['databaseName']); 

if(!$dbConnected) {
    die('Connect Error (' . mysqli_connect_errno() . ') '
            . mysqli_connect_error());
}
echo 'Success... ' . mysqli_get_host_info($dbConnected) . "\n". "<br>";
mysqli_set_charset($dbConnected, "utf8");

$tPerson_SQLselect = "SELECT  ";
$tPerson_SQLselect .= "ID, Salutation, FirstName, LastName, CompanyID ";    
$tPerson_SQLselect .= "FROM ";
$tPerson_SQLselect .= "tPerson ";           

$result = mysqli_query($dbConnected, $tPerson_SQLselect);

if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
    echo "Salutation: ".$row["Salutation"]. "-FirstName: ".$row["FirstName"]." ".$row["LastName"]."  -CompanyID: ".$row["CompanyID"]. "<br>";
}
} else {
    echo "0 results";
}

mysqli_close($dbConnected);
?>

我找不到我的错误.. 请帮忙！

dbConfig.php 文件：

<?php
$db = array(
'hostname' => '*****',
'username' => '*****',
'password' => '*****',
'database' => '*****',
); 
?>

Answer 1

如果您仍在使用相同的变量名，您的 $dbConnected 应该是这样的：

include('../htconfig/dbConfig.php');
$dbConnected = mysqli_connect($hostname, $username, $password, $databaseName);

如果你希望它是 $db['field'] 那么你的 ../htconfig/dbConfig.php 应该是这样的：

$db = array('hostname' => 'xxxx',
            'username' => 'xxxx',
            'password' => 'xxxx',
            'databaseName' => 'xxxx');

编辑： 您在 dbConfig.php 中的数组显示“数据库”，但您在 $dbConnected mysqli_connect 函数中使用了“数据库名称”？