Question

我如何显示student.Student而不是student_id

 <?php
for ($i = 0; $i < $chkcount; $i++) {
    $id = $chk[$i];
    $res = $MySQLiconn->query("SELECT Student.ID, student.Student, student.School,student.PR, Jumps.dis_ft_1, Jumps.dis_in_1, Jumps.dis_ft_2, Jumps.dis_in_2, Jumps.dis_ft_3, Jumps.dis_in_3, Jumps.dis_ft_4, Jumps.dis_in_4, Jumps.dis_ft_5, Jumps.dis_in_5, Jumps.dis_ft_6, Jumps.dis_in_6
            FROM Student LEFT JOIN Jumps ON Student.ID = Jumps.student_id WHERE ID=" . $id);
    while ($row = $res->fetch_array()) {
        ?>
        <tr>
            <td>
                <input type="hidden" name="id[]" value="<?php echo $row['ID']; ?>"/>
                Student ID: <input type="text" name="st[]" value="<?php echo $row['student_id']; ?>"
                                   class="form-control"/>
                Attempt 1 <input type="text" name="df1[]" value="<?php echo $row['dis_ft_1']; ?>"
                                 class="form-control"/>
                <input type="text" name="di1[]" value="<?php echo $row['dis_in_1']; ?>"
                       class="form-control"/>
                Attempt 2 <input type="text" name="df2[]" value="<?php echo $row['dis_ft_2']; ?>"
                                 class="form-control"/>
                <input type="text" name="di2[]" value="<?php echo $row['dis_in_2']; ?>"
                       class="form-control"/>
                Attemp 3 <input type="text" name="df3[]" value="<?php echo $row['dis_ft_3']; ?>"
                                class="form-control"/>
                <input type="text" name="di3[]" value="<?php echo $row['dis_in_3']; ?>"
                       class="form-control"/>
                Attempt 4 <input type="text" name="df4[]" value="<?php echo $row['dis_ft_4']; ?>"
                                 class="form-control"/>
                <input type="text" name="di4[]" value="<?php echo $row['dis_in_4']; ?>"
                       class="form-control"/>
                Attempt 5 <input type="text" name="df5[]" value="<?php echo $row['dis_ft_5']; ?>"
                                 class="form-control"/>
                <input type="text" name="di5[]" value="<?php echo $row['dis_in_5']; ?>"
                       class="form-control"/>
                Attempt 6 <input type="text" name="df6[]" value="<?php echo $row['dis_ft_6']; ?>"
                                 class="form-control"/>
                <input type="text" name="di6[]" value="<?php echo $row['dis_in_6']; ?>"
                       class="form-control"/>
            </td>
        </tr>
        <?php
    }
}
?>

不确定是否可以完成。但学生姓名会优于学生ID。不会发布更多的文字，所以只需输入，它将发布。我正在学习PHP。

Answer 1

如果只有一个名为Student的字段，即Student.Student，则可以输出名称<?php echo $row['Student']; ?>。

但是，如果存在名称冲突，您始终可以为该字段指定要访问的PHP的新名称。

SELECT Student.ID, Student.Student AS some_new_name FROM ....

<?php echo $row['some_new_name']; ?>

以输入形式显示来自不同表的字段

1 个答案: