Question

我有这段代码

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

上面的查询将返回训练表的最后一行。如果我回显$ tnid，它将显示'资源ID＃5'。如果我添加

    $d = mysql_fetch_array($tnid);

然后我回显$ d，它会显示错误信息

数组转换为字符串第32行的U：\ XAMPP \ htdocs \ pds \ action \ doInsertSchedule.php

如何显示查询的确切结果？有人请帮忙。

Answer 1

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'];

Answer 2

<?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);
?>

如何执行SQL并从PHP获取结果？

2 个答案: