PHP表可链接列

Question

您好，我有一个查询来列出所有用户，我希望能够单击用户名，然后转到用户信息页面，所以我的问题是如何将用户名传递给其他php。

    while($row = mysql_fetch_array($result)) {
  echo "<tr>";
  echo "<td><font color='white'>" . ++$i . "</font></td>";
  echo "<td><a href='page2.php'><font color='white'>" . $row['1'] . "</font></a></td>";
  echo "<td><font color='white'>" . $row['2'] . "</font></td>";
  echo "<td><font color='white'>" . $row['3'] . "</font></td>";
  echo "<td><font color='white'>" . $row['4'] . "</font></td>";
  echo "</tr>";
}

该表很好，我只需要一种方法即可将选择的用户名传递给page2.php。

提前致谢

编辑：

    while($row = mysql_fetch_array($result)) {
  echo "<tr>";
  echo "<td><font color='white'>" . ++$i . "</font></td>";
  echo "<td><a href='page2.php?nome='".$row['nome']."><font color='white'>" . $row['nome'] . "</font></a></td>";
  echo "<td><font color='white'>" . $row['sexo'] . "</font></td>";
  echo "<td><font color='white'>" . $row['idade'] . "</font></td>";
  echo "<td><font color='white'>" . $row['diabetes'] . "</font></td>";
  echo "</tr>";
}

page2.php：

    include_once 'ligacao.php';
$name = $_GET['nome'];
echo "<table border='1' align='center'>
<tr>
<th><font color='white'>Registo</font></th>
<th><font color='white'>Name</font></th>
</tr>";

echo "<tr>";
  echo "<td><font color='white'>" . ++$i . "</font></td>";
  echo "<td><font color='white'>" . $name . "</font></td>";

  echo "</tr>";

编辑2：

如何在页面中间并排对齐2张桌子？

    echo "<table border='1'>
<tr>
<th><font color='white'>Glicemia</font></th>
<th><font color='white'>Hidratos</font></th>
<th><font color='white'>Peso</font></th>
<th><font color='white'>Descri&ccedil&atildeo</font></th>
<th><font color='white'>Timestamp</font></th>
</tr>";

while($row = mysql_fetch_array($result)) {
echo "<tr>";
  echo "<td><font color='white'>" . $row['glicemia'] . "</font></td>";
  echo "<td><font color='white'>" . $row['hidratos'] . "</font></td>";
  echo "<td><font color='white'>" . $row['peso'] . "</font></td>";
  echo "<td><font color='white'>" . $row['descricao'] . "</font></td>";
  echo "<td><font color='white'>" . $row['time'] . "</font></td>";
  echo "</tr>";
}

echo "<table border='1'>
<tr>
<th><font color='white'>Medicamento</font></th>
<th><font color='white'>Timestamp</font></th>
</tr>";

while($row = mysql_fetch_array($resultq)) {
echo "<tr>";
  echo "<td><font color='white'>" . $row['medicamento'] . "</font></td>";
  echo "<td><font color='white'>" . $row['time'] . "</font></td>";
  echo "</tr>";
}

Answer 1

您可以将其作为url参数传递，并在page2.php使用$_GET ，例如：

...
"<td><a href='page2.php?username='".$row[1].">...</a>
...

并在page2.php

$username = $_GET['username'];

更新：：您的代码似乎引号有问题，请尝试更改为：

....
echo "<td><a href='page2.php?nome=" . $row['nome'] . "'><font color='white'>" . $row['nome'] . "</font></a></td>";
...