Question

如果与mysql_query($sqlStr3)关联的结果的值为零，则下面的HTML表mysql_query($sqlStr1)（其行中带有“已加入”一词的那个）不会回显。即使mysql_query($sqlStr3)返回结果，也会发生这种情况。

换句话说，如果给定的loginid在MySQL表“login”中有一个条目，但在MySQL表“submission”中没有，则与mysql_query($sqlStr3)相关联的表不会回显。

我不明白为什么MySQL“submission”表会对mysql_query($sqlStr3)产生任何影响，因为$ sqlStr3只处理另一个名为“login”的MySQL表，如下所示。

为什么会发生这种情况？

提前致谢，

John W。

<?php

echo '<div class="profilename">User Profile for </div>';
echo '<div class="profilename2">'.$profile.'</div>';

$tzFrom = new DateTimeZone('America/New_York'); 
$tzTo = new DateTimeZone('America/Phoenix'); 

  $profile = mysql_real_escape_string($_GET['profile']);

  $sqlStr = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl
               FROM submission AS s
         INNER JOIN login AS l
                 ON s.loginid = l.loginid
              WHERE l.username = '$profile'
           ORDER BY s.datesubmitted DESC";

  $result = mysql_query($sqlStr);

$arr = array(); 
echo "<table class=\"samplesrec1\">";
while ($row = mysql_fetch_array($result)) { 
    $dt = new DateTime($row["datesubmitted"], $tzFrom); 
    $dt->setTimezone($tzTo);
    echo '<tr>';
    echo '<td class="sitename3">'.$dt->format('F j, Y &\nb\sp &\nb\sp g:i a').'</a></td>'; 
    echo '<td class="sitename1"><a href="http://www.'.$row["url"].'">'.$row["title"].'</a></td>';
    echo '</tr>';
    }
echo "</table>";    

  $sqlStr1 = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl, l.created, count(s.submissionid) countSubmissions
               FROM submission AS s
         INNER JOIN login AS l
                 ON s.loginid = l.loginid
              WHERE l.username = '$profile'";


  $result1 = mysql_query($sqlStr1);

$arr1 = array(); 
echo "<table class=\"samplesrec2\">";
while ($row1 = mysql_fetch_array($result1)) { 
    echo '<tr>';
    echo '<td class="sitename5">Submissions: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row1["countSubmissions"].'</td>';
    echo '</tr>';
    }
echo "</table>";    


  $sqlStr2 = "SELECT l.username, l.loginid, c.loginid, c.commentid, c.submissionid, c.comment, c.datecommented, l.created, count(c.commentid) countComments
               FROM comment AS c
         INNER JOIN login AS l
                 ON c.loginid = l.loginid
              WHERE l.username = '$profile'";


  $result2 = mysql_query($sqlStr2);

$arr2 = array(); 
echo "<table class=\"samplesrec3\">";
while ($row2 = mysql_fetch_array($result2)) { 
    echo '<tr>';
    echo '<td class="sitename5">Comments: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row2["countComments"].'</td>';
    echo '</tr>';
    }
echo "</table>";    


$tzFrom3 = new DateTimeZone('America/New_York'); 
$tzTo3 = new DateTimeZone('America/Phoenix'); 


  $sqlStr3 = "SELECT created, username
               FROM login 
              WHERE username = '$profile'";


  $result3 = mysql_query($sqlStr3);

$arr3 = array(); 
echo "<table class=\"samplesrec4\">";
while ($row3 = mysql_fetch_array($result3)) { 
    $dt3 = new DateTime($row3["created"], $tzFrom3); 
    $dt3->setTimezone($tzTo3);
    echo '<tr>';
    echo '<td class="sitename5">Joined: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$dt->format('F j, Y').'</td>';
    echo '</tr>';
    }
echo "</table>";    

?>

</body>
</html>

Answer 1

如果我正确理解了问题，您需要在代码的最后一行中回显$dt3->format('F j, Y')而不是$dt->format('F j, Y')。

Answer 2

while ($row3 = mysql_fetch_array($result3)) { 
    $dt3 = new DateTime($row3["created"], $tzFrom3); 
    $dt3->setTimezone($tzTo3);
    echo '<tr>';
    echo '<td class="sitename5">Joined: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$dt->format('F j, Y').'</td>';
    echo '</tr>';
    }
echo "</table>";

您是否希望在此回音中调用$ dt3-＆gt;格式（'F j，Y'），而不是$ - ＆gt;格式（'F j，Y'）？ $ dt3我没有看到任何其他用途。

表不响应另一个表是否为零值

2 个答案: