PHP 只选择一项显示答案

【问题标题】：PHP select only one entry showsPHP 只选择一项显示
【发布时间】：2013-02-24 17:22:13
【问题描述】：

我怎么会用这段代码：

$emailc = 'thomas990428@me.com_classes';
$emaila = 'thomas990428@me.com_assignments';
$emailp = 'thomas990428@me.com_projects';

$resulty = mysqli_query($con,"SELECT * FROM `$emailc` ORDER BY period"); if (!$result) echo mysqli_error(); else // ok, do your thing.
$resulti = mysqli_query($con,"SELECT * FROM `$emaila` ORDER BY duehw"); if (!$result) echo mysqli_error(); else // ok, do your thing.
$resulto = mysqli_query($con,"SELECT * FROM `$emailp` ORDER BY dueproj"); if (!$result) echo mysqli_error(); else // ok, do your thing.

$classcount = 1;


while($row = mysqli_fetch_array($resulty))
  {

  $period = $row['period'];
  $teacher = $row['teacher'];
  $subject = $row['subject'];
  $subjecto = strtolower($subject);
  $subjecto = str_replace(' ', '', $subjecto);
  $grade = $rowy['grade'];

  echo "<section id='" . $subjecto . "'> \n";
  echo "<p class='title'>" . $subject . "</p> \n";
  echo "<a style='cursor:pointer;' onclick='homework" . $classcount . "()'>Homework&nbsp;</a>|<a style='cursor:pointer;' onclick='projects" . $classcount . "()'>&nbsp;Projects</a> \n";
  echo "<div id='homework" . $classcount . "'><br /><a onclick='addassignment()'>Add Assignment</a><br />";
  echo "<table class='homework'>";
  echo "<tr>";
  echo "<th class='title'>";
  echo "Title";
  echo "</th>";
  echo "<th class='duedate'>";
  echo "Due Date";
  echo "</th>";
  echo "</tr>";

  while($row = mysqli_fetch_array($resulti))
    {

      $subjecthw = $row['subjecthw'];
      $namehw = $row['namehw'];
      $duehw = $row['duehw'];

      echo "<tr>";
      echo "<td class='title'>";
      echo $namehw;
      echo "</td>";
      echo "<td class='duedate'>March ";
      echo $duehw;
      echo "</td>";
      echo "</tr>";

    }

  echo "</table>";
  echo "</div> \n";
  echo "<div id='projects" . $classcount . "'><br /><a onclick='addassignment()'>Add Assignment</a><br />";
  echo "<table class='homework'>";
  echo "<tr>";
  echo "<th class='title'>";
  echo "Title";
  echo "</th>";
  echo "<th class='duedate'>";
  echo "Due Date";
  echo "</th>";
  echo "</tr>";

  while($row = mysqli_fetch_array($resulto))
    {

      $subjectproj = $row['subjectproj'];
      $nameproj = $row['nameproj'];
      $dueproj = $row['dueproj'];

      echo "<tr>";
      echo "<td class='title'>";
      echo $nameproj;
      echo "</td>";
      echo "<td class='duedate'>March ";
      echo $dueproj;
      echo "</td>";
      echo "</tr>";

    }

  echo "</table>";
  echo "</div> \n";
  echo "</section> \n";

  $classcount += 1;

  }

我只在一个部分中获得了硬件和项目。我做错了吗？谢谢！我知道它可能永远不会起作用，但是我该如何解决呢？我有 3 个表，我需要从所有表中获取数据。

【问题讨论】：

你为什么要让你的服务器如此努力地让 PHP 回显静态元素？
它们是动态的，来自数据库
看来代码绝对没问题，但想知道DB中的肠炎是怎么产生的……
显示的部分，但硬件的东西只显示在一个部分...我有多个部分
行$grade = $rowy['grade']; 应该是$grade = $row['grade'];

标签： php mysql sql select mysqli

【解决方案1】：

不要在while 循环中使用$row。

例如

while($row = mysql_fetch_row($resulti)){
    while($row2 = mysql_fetch_row($resulto)){

    }
}

也没有时间浏览代码，但循环运行 SQL 绝不是一个好主意。

【讨论】：