N.K*_*.K. 10 php mysql include
我试图使用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);
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当我尝试包含上述代码的文件包括('../ htconfig/dbConfig.php'); 然后我没有得到任何结果:
......"0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);
?>
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我找不到我的错误..请帮忙!
dbConfig.php文件:
<?php
$db = array(
'hostname' => '*****',
'username' => '*****',
'password' => '*****',
'database' => '*****',
);
?>
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如果您仍使用相同的变量名,则$ dbConnected应该是这样的:
include('../htconfig/dbConfig.php');
$dbConnected = mysqli_connect($hostname, $username, $password, $databaseName);
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如果您希望它是$ db ['field'],则您的../htconfig/dbConfig.php应该如下所示:
$db = array('hostname' => 'xxxx',
'username' => 'xxxx',
'password' => 'xxxx',
'databaseName' => 'xxxx');
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编辑: 您在dbConfig.php中的数组说'database',但是您在$ dbConnected mysqli_connect函数中使用了'databaseName'?
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