php - 查询多个数据库表的数据到表单表中(MYSQL,PHP)

Question

由于我是PHP新手，我想知道如何将不同数据库表中的数据放入页面上的一个表格形式中。我的代码如下，

<?php


include('DBconnect.php');
mysql_query("USE onlinerecruitment");

$username =$_SESSION['user'];

$result = mysql_query("SELECT * FROM application_data_file");

$rows = mysql_fetch_array($result, MYSQL_ASSOC);

$pos_id = $rows['Position_ID'];

$resultt = mysql_query("SELECT * FROM position WHERE Position_ID = '".$pos_id."' ");

$resulttt = mysql_query("SELECT * FROM resume_data_file WHERE App_Email = '".$pos_id."' ");

?>

<TABLE border ='1'>
<table style="width:100%">
<tr>

<th>Application ID</th>
<th>Applicant E-mail</th>
<th>Position Selected</th>
<th></th>
<th></th>
<th></th>

</tr>

<?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC) &  $rowss = mysql_fetch_array($resultt, MYSQL_ASSOC)){

echo "<TR>";

echo "<TD>".$row['App_Data_ID']."</TD>";
echo "<TD>".$row['App_Email']."</TD>";
echo "<TD>".$rowss['Position_Name']."</TD>";
echo "<TD><a href='view-app-form.php?app_mail=".$row['App_Email']."'>View Application Data</a></TD>";
echo "<TD><a href='view-resume-form.php?app_mail=".$row['App_Email']."'>View Resume Data</a></TD>";
echo "<TD><a href='view-test-score.php?app_mail=".$row['App_Email']."'>View Testing Score Data</a></TD>";

echo "</TR>";
}

?>
</table>

我将重点关注这里。

<TABLE border ='1'>
<table style="width:100%">
<tr>

<th>Application ID</th>
<th>Applicant E-mail</th>
<th>Position Selected</th>
<th></th>
<th></th>
<th></th>

</tr>

<?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC) &  $rowss = mysql_fetch_array($resultt, MYSQL_ASSOC)){

echo "<TR>";

echo "<TD>".$row['App_Data_ID']."</TD>";
echo "<TD>".$row['App_Email']."</TD>";
echo "<TD>".$rowss['Position_Name']."</TD>";
echo "<TD><a href='view-app-form.php?app_mail=".$row['App_Email']."'>View Application Data</a></TD>";
echo "<TD><a href='view-resume-form.php?app_mail=".$row['App_Email']."'>View Resume Data</a></TD>";
echo "<TD><a href='view-test-score.php?app_mail=".$row['App_Email']."'>View Testing Score Data</a></TD>";

echo "</TR>";
}

?>
</table>

但是，如果我没有关注的部分有任何问题，我仍然感谢您的解决方案。

提前谢谢您。

Answer 1

为此，您需要在 sql 语句中使用 JOIN。

mysql_query("SELECT resume_data_file.App_Email, position.Position_ID FROM position INNER JOIN resume_data_file ON position.Position_ID = position.Position_ID  WHERE position.Position_ID = '".$pos_id."'   ");

http://www.w3schools.com/sql/sql_join.asp