Question

由于我是PHP的新手，我想知道如何将来自不同数据库表的数据放到页面上的一个表格中。我的代码到目前为止，

<?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>

我将把重点放在这里。

<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。

Answer 1

为此，您需要在sql语句中使用JOIN。

function totalTime($times=array()) {
     if(!is_array($times)) {
         $times = array($times);
     }
     $seconds = 0;
     foreach ($times as $time) {
       list($hour,$minute,$second) = explode(':', $time);
       $seconds += $hour*3600;
       $seconds += $minute*60;
       $seconds += $second;
     }
     $hours = floor($seconds / 3600);
     $seconds -= $hours * 3600;
     $minutes  = floor($seconds / 60);
     $seconds -= $minutes * 60;

     // return "{$hours}:{$minutes}:{$seconds}";
     return sprintf('%02d:%02d:%02d', $hours, $minutes, $seconds);
}


$yourArray = array(...);

//clean your array, so each array value will be a duration

echo totalTime($yourArray);

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

查询来自多个数据库表的数据到表单表（MYSQL，PHP）

1 个答案: