php mysqli_query results nothing
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我正在尝试使用mysqli_查询从表中获取数据。当我使用以下命令时,它工作正常:
1 2 3 4 5 | $hostname ="********"; $username ="*******"; $password ="********"; $databaseName ="**************"; $dbConnected = mysqli_connect($hostname, $username, $password, $databaseName); |
当我试图用上述代码包括文件时("../htconfig/dbconfig.php");那么我不会得到任何结果:
…"0个结果"
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | <?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($dbConnected) ." ".""; mysqli_set_charset($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"].""; } } else { echo"0 results"; } mysqli_close($dbConnected); ?> |
我找不到我的错误。请帮助!
dbconfig.php文件:
1 2 3 4 5 6 7 8 | <?php $db = array( 'hostname' => '*****', 'username' => '*****', 'password' => '*****', 'database' => '*****', ); ?> |
如果仍然使用相同的变量名,则$dbconnected应该如下所示:
1 2 | include('../htconfig/dbConfig.php'); $dbConnected = mysqli_connect($hostname, $username, $password, $databaseName); |
如果您希望它是$db['field'],那么../htconfig/dbconfig.php应该如下所示:
1 2 3 4 | $db = array('hostname' => 'xxxx', 'username' => 'xxxx', 'password' => 'xxxx', 'databaseName' => 'xxxx'); |
编辑:dbconfig.php中的数组显示"database",但在$dbconnected mysqli_connect函数中使用了"databasename"?