php - 如果另一个表的值为零，则表不回显

如果与mysql_query($sqlStr3)关联的结果的值为零，则下面带有mysql_query($sqlStr1)的HTML表（行中带有单词“ Joined”的HTML表）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”中具有一个条目，但在MySQL表“ submission”中没有一个条目，则与mysql_query($sqlStr3)关联的表不会回显。

我不明白为什么MySQL“提交”表会对mysql_query($sqlStr3)产生任何影响，因为$ sqlStr3仅处理另一个称为“登录”的MySQL表，如下所示。

任何想法为什么会这样？

提前致谢，

约翰·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>

最佳答案

如果我正确理解了问题，则需要在代码的最后一行之一中回显$dt3->format('F j, Y')而不是$dt->format('F j, Y')。