Question

我有两张桌子

学生

S_ID, LastName, FirstName, MiddleName

takensubject

S_ID, SubjectCode, Time

我在下面有一个代码，使用session来显示从takensubject获取Subject但是它只显示ID我想要的是显示使用来自takensubject的ID的学生的全名我的代码下面显示了一些错误警告：mysqli_num_rows（）期望参数1是mysqli_result，boolean。任何人都可以帮我纠正我的疑问吗？

<?php
    include'database.php';
    $sescode = $_SESSION['sessioncode'];
    $sestime = $_SESSION['sessiontime'];
    $conn = mysqli_connect($server, $dbusername, $dbpassword, $database);
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }
    $sql = "SELECT * FROM takensubject tb2, student tb1 where tb2.S_ID=tb1.S_ID and SchoolYear ='$Sy' and Semester ='$Sem' and SubjectCode='$sescode' and Time='$sestime'";
    $no = 0;
    $result = mysqli_query($conn, $sql);
    if (mysqli_num_rows($result) > 0) {
        while ($row = mysqli_fetch_assoc($result)) {
            $no++;
            echo"<td>$no ).</td> ";
            echo"{$row['S_ID']}";
            echo"{$row['LastName']}<br>";
        }
    } else {
        echo "No Record Results";
    }
    mysqli_close($conn);
?>

它会继续显示此警告：

mysqli_num_rows（）期望参数1为mysqli_result，boolean

Answer 1

我没有在你的任何表中看到SchoolYear，Semester列。你还需要在列名旁边指定表，如tb2.Time。同样回应您的查询并直接在phpmyadmin中尝试检查错误。希望它有所帮助:)尝试像

这样的东西

SELECT takensubject.*,student.* FROM takensubject INNER JOIN student ON takensubject.S_ID=student.S_ID where takensubject.SchoolYear ='$Sy' and takensubject.Semester ='$Sem' and takensubject.SubjectCode='$sescode' and takensubject.Time='$sestime'

Answer 2

我猜您的查询正在返回dispatch_async。尝试这样的事情。

FALSE

PHP查询从table2获取记录到表1

2 个答案: