Question

我有两个表job，其中包含一个名为employer_id_job的属性，将其链接到employer表。我可以使用PHP使用以下代码在表中显示job的内容

<?php
    $query = "SELECT * FROM job";
    $result = mysql_query($query);
    $num = mysql_numrows($result);

    $count = 0;
    while ($count < $num)
    {
        $title = mysql_result ($result, $count, "Title");
        $date_posted = mysql_result ($result, $count, "Date_posted");
        $application_deadline = mysql_result ($result, $count, "Application_deadline");
        $description = mysql_result ($result, $count, "Description");
        $years_of_experience = mysql_result ($result, $count, "Years_of_experience");
        $education_level = mysql_result ($result, $count, "Education_level_required");
        $contract = mysql_result ($result, $count, "Contract_type");
        $company = mysql_result ($result, $count, "Company_name");
?>

        <tr>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $count + 1; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $title; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $company; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $description; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $date_posted; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $application_deadline; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $education_level; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $years_of_experience; ?></font></td>
        <td><font face = "Arial, Helvetica, sans-serif"><? echo $contract; ?></font></td>
<?
        $count ++;
    }
?>

我在尝试为$company变量赋值时将此方法分解，该变量保存在employer表中。我的假设只是使用与其余变量相同的表示法将导致脚本跟随employer表的外键并在那里获取属性，但它没有这样做。

我应该如何访问通过外键相关的表的内容？

Answer 1

select * from job as j, company as c where j.company == c.company

我不确定你的田地名称是什么，所以我做了最好的猜测。这允许您访问这两个字段。

使用PHP通过外键关系显示MySQL表属性

1 个答案: