我有两个我正在阅读的sql表,我想将每个表显示为自己的独立表。首先是内部,然后是外部。而且我希望每个人都有一个简单的标题。例如:
INTERNAL
Table here
EXTERNAL
Table here
发生的事情是INTERNAL和EXTERNAL标题不断出现 在同一条线上,或两者都在桌子前面。像这样:
INTERNAL
EXTERNAL
Internal Table here
External Table here
我已经尝试将标题包含在php标记中。这些是目前注释掉的行。而且我也试过在php标签之外添加html,如下所示:
<br><br>
<b>INTERNAL</b>
<br><br>
<?php Internal Table here ?>
然后:
<br><br>
<b>EXTERNAL</b>
<br><br>
<?php External Table here ?>
但是这仍然导致两个标题首先出现在表格之前。是否在PHP之前处理html?我必须过于线性地解释这个,就像脚本一样,因为它肯定没有按照我在页面上编码的顺序进行处理。
这是我的代码,其标题已注释掉。这些表格夹在一起,告诉我它们不会被解释为页面上的两个不同元素。我是否需要将每个标签放在自己的标签中?
<?php
$internal = getInternalNetworkTable();
if ($internal->num_rows > 0){
//echo " <b>INTERNAL</b>";
echo "<table cellspacing=\"0\">";
echo "<tr>";
echo " <th><b>Device</b></th>";
echo " <th><b>Label</b></th>";
echo " <th><b>Address</b></th>";
echo " <th><b>Type</b></th>";
echo " <th><b>DisplayNumber</b></th>";
echo " <th><b>Wiki</b></th>";
echo "</tr>";
while ($row = $internal->fetch_assoc()){
echo "<tr>";
echo ("<td>" . $row["Device"] ."</td>");
echo ("<td>" . $row["Label"]."</td>");
echo ("<td>" . $row["Address"]."</td>");
echo ("<td>" . $row["Type"]."</td>");
echo ("<td>" . $row["DisplayNumber"]."</td>");
echo ("<td>" . $row["Wiki"]."</td>");
echo "</tr>";
}
} else {
echo "No results founds";
}
?>
<?php
$external = getExternalNetworkTable();
if ($external->num_rows > 0){
//echo " <b>EXTERNAL</b>";
echo "<table cellspacing=\"0\">";
echo "<tr>";
echo " <th><b>Device</b></th>";
echo " <th><b>Label</b></th>";
echo " <th><b>Address</b></th>";
echo " <th><b>Type</b></th>";
echo " <th><b>DisplayNumber</b></th>";
echo " <th><b>Wiki</b></th>";
echo "</tr>";
while ($row = $external->fetch_assoc()){
echo "<tr>";
echo ("<td>" . $row["Device"] ."</td>");
echo ("<td>" . $row["Label"]."</td>");
echo ("<td>" . $row["Address"]."</td>");
echo ("<td>" . $row["Type"]."</td>");
echo ("<td>" . $row["DisplayNumber"]."</td>");
echo ("<td>" . $row["Wiki"]."</td>");
echo "</tr>";
}
} else {
echo "No results founds";
}
$conn->close();
?>
编辑: 需要补充:
echo "</table>";
新代码:
<?php
$internal = getNovatoInternalNetworkTable();
if ($internal->num_rows > 0){
echo " <b>INTERNAL</b>";
echo "<table cellspacing=\"0\">";
echo "<tr>";
echo " <th><b>Device</b></th>";
echo " <th><b>Label</b></th>";
echo " <th><b>Address</b></th>";
echo " <th><b>Type</b></th>";
echo " <th><b>DisplayNumber</b></th>";
echo " <th><b>Wiki</b></th>";
echo "</tr>";
while ($row = $internal->fetch_assoc()){
echo "<tr>";
echo ("<td>" . $row["Device"] ."</td>");
echo ("<td>" . $row["Label"]."</td>");
echo ("<td>" . $row["Address"]."</td>");
echo ("<td>" . $row["Type"]."</td>");
echo ("<td>" . $row["DisplayNumber"]."</td>");
echo ("<td>" . $row["Wiki"]."</td>");
echo "</tr>";
}
echo "</table>";
} else {
echo "No results founds";
}
?>
<?php
$external = getNovatoExternalNetworkTable();
if ($external->num_rows > 0){
echo " <b>EXTERNAL</b>";
echo "<table cellspacing=\"0\">";
echo "<tr>";
echo " <th><b>Device</b></th>";
echo " <th><b>Label</b></th>";
echo " <th><b>Address</b></th>";
echo " <th><b>Type</b></th>";
echo " <th><b>DisplayNumber</b></th>";
echo " <th><b>Wiki</b></th>";
echo "</tr>";
while ($row = $external->fetch_assoc()){
echo "<tr>";
echo ("<td>" . $row["Device"] ."</td>");
echo ("<td>" . $row["Label"]."</td>");
echo ("<td>" . $row["Address"]."</td>");
echo ("<td>" . $row["Type"]."</td>");
echo ("<td>" . $row["DisplayNumber"]."</td>");
echo ("<td>" . $row["Wiki"]."</td>");
echo "</tr>";
}
echo "</table>";
} else {
echo "No results founds";
}
$conn->close();
?>
答案 0 :(得分:1)
您在两个表格中都缺少</table>
。
因此,您的表将相互合并。
echo '</table>';
在你的两个} else {
之前应该做的伎俩