MySQLi - Debug
语法
bool mysqli_debug ( string $message )
定义和用法
用于执行调试操作。
示例
试试下面的例子 −
<?php $servername = "localhost:3306"; $username = "root"; $password = ""; $dbname = "TUTORIALS"; $conn = new mysqli($servername, $username, $password, $dbname); if ($conn->connect_error) { die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error()); } echo 'Success... ' . mysqli_get_host_info($conn) . " "; mysqli_debug("d:t:o,debug.txt"); mysqli_autocommit($conn,FALSE); mysqli_query($conn,"INSERT INTO tutorials_auto (id,name) VALUES (10,'sai')"); mysqli_commit($conn); mysqli_close($conn); ?>
上述代码的示例输出应该是这样的 −
Success... localhost:3306 via TCP/IP
在上面的示例中,它执行调试操作并将信息存储在 debug.txt 文件中