使用 "mysql_fetch_array();" 制作 table 无法正常工作

Question

我正在尝试制作一个从数据库中获取信息并将其发布到网页的页面。我尝试使用 "table border" 和 "table-height" 命令，但我认为我做错了。因为现在每当我回显某些内容时，它都会显示在 table 上方，即使代码位于 "table" 代码下方。看看我的代码（你不应该关心我设置的变量，因为它们可能彼此不匹配。不用担心这个。我唯一想修复的是 table）

<?php
include 'connection.php';


echo "<table border='2' style='border-collapse: collapse'>";  
echo "<th>ID</th><th>Name</th><th>Special</th><th>Size</th>";


while($elev = mysql_fetch_array($resultstudent)){
    echo "<h4> " . $student['Student']  .  $student['Room']  . " </h4>";
    //echo "<h4> has rom " . $ev['Rom'] . "</h4>";
}

echo "<h1> list of rooms </h1>";

while($rom = mysql_fetch_array($resultroom)){
    echo "<tr><td>"   .$rom['Room ID'] . ""   .$rom['Roomname'] .""  . $rom['Special'] ."</td><td>" . $rom['Size']  . "; 
}



?>

Answer 1

尝试更改此行：

echo "<th>ID</th><th>Name</th><th>Special</th><th>Size</th>";

有了这个：

echo "<tr><th>ID</th><th>Name</th><th>Special</th><th>Size</th></tr>";

然后不要忘记在最后关闭，但我认为也存在一些问题，因为它不在任何 table TAG

中

Answer 2

我在代码中添加了您需要的行，请参阅注释“// LINE ADDED”。试试这个代码：

<?php
include 'connection.php';

$queryroom = "SELECT * FROM rom";
$querystudent = "SELECT * FROM elev";
$queryspecial = "SELECT * FROM rom WHERE prosjektor = 1";

$resultroomm = mysql_query($queryRom);
$resultstudent = mysql_query($queryElev);
//$resultspecial = mysql_query($queryRomProsjektor);

echo "<table border='2' style='border-collapse: collapse'>";  
echo "<thead><tr>"; // LINE ADDED
echo "<th>ID</th><th>Name</th><th>Special</th><th>Size</th>";
echo "</tr></thead>"; // LINE ADDED

echo "<tbody>"; // LINE ADDED

while($elev = mysql_fetch_array($resultstudent)){
    echo "<tr><td colspan='4'>"; // LINE ADDED
    echo "<h4> " . $elev['Student'] . " has room nr: " .  $elev['Room']  . " </h4>";
    //echo "<h4> has rom " . $ev['Rom'] . "</h4>";
    echo "</td></tr>"; // LINE ADDED
}

echo "<tr><td colspan='4'>"; // LINE ADDED
echo "<h1> list of rooms </h1>";
echo "</td></tr>"; // LINE ADDED

while($rom = mysql_fetch_array($resultroom)){
    echo "<tr><td>" . $rom['Room ID'] . "</td><td>" . $rom['Roomname'] ."</td><td>"  . $rom['Special'] ."</td><td>" . $rom['Size']  . "</td></tr>"; 
}

echo "</tbody></table>"; // LINE ADDED 

?>

我认为你应该阅读这篇文章：

https://developer.mozilla.org/en-US/docs/Web/HTML/Element/table

使用 "mysql_fetch_array();" 制作 table 无法正常工作

Making a table with "mysql_fetch_array();" not working properly

html

php

mysql

datatable

html-table