如何执行SQL并从PHP获得结果？

我有这个密码

  $tnid = mysql_query("SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1");

上面的查询将返回训练表中的最后一行。 如果我回显$tnid，它将显示“资源id 5”。 如果我加上

    $d = mysql_fetch_array($tnid);

然后我回显$d，它将显示一条错误消息

中的数组到字符串转换 第32行的U:\XAMPP\htdocs\pds\action\doInsertSchedule.php

如何显示查询的准确结果？ 有人请帮忙。

试试这个

$con = mysql_connect("host_name", "user_name", "password");
mysql_select_db("database_name", $con);
$query="SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1";
$res=mysql_query($query,$con);
$farow=mysql_fetch_array($res);
$answer=$farow['TrainingID'];

---

应该使用msqli而不是mysql。@dannmate这由他决定。你不应该推广不好的做法。mysql库已被删除，正如@dannmate所说，您应该升级PDO或MySQLiit，并在第30行的U:\XAMPP\htdocs\pds\action\doInsertSchedule.php中显示一条错误消息Undefined index:SubjectTrainingID。请参阅此处，@user3152496欢迎您：

<?php
    $con=mysqli_connect("example.com","peter","abc123","my_db");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $result = mysqli_query($con,"SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1");
    while($row = mysqli_fetch_array($result)) {
      echo  " Training ID - " . $row['TrainingID '] .;
      echo "<br>";
    }
    mysqli_close($con);
?>