如何在if语句中回显PHP和HTML [英] How to echo PHP and HTML in if statement
问题描述
我有一个if语句，仅当$ result37 = ncrteam的值时才需要显示一些代码，但是如何显示HTML和PHP代码呢?回声";无法正常工作并回显'';也行不通.
I have an if statement that only has to show some code if the value of $result37 = ncrteam... but how can I echo that HTML and PHP code? echo" "; is not working and echo ' '; Is also not working.
这是我的代码:
<?php
if ($result37 === "ncrteam") {
?>
Status:<br>
<select class="form-control" name="status" style="width: 300px">
<?php
while ($row15 = mysqli_fetch_assoc($result15)):; ?>
<option selected value=\"<?php echo $row15['status'];?>\"><?php echo $row15['status'];?></option>
<?php endwhile;?>
<?php while ($row16 = mysqli_fetch_assoc($result16)):; ?>
<option value=\"<?php echo $row16['statusname'];?>\"><?php echo $row16['statusname'];?></option>
<?php endwhile;?>
</select>
<?php
}
?>
推荐答案
这不是一个完整答案，因为目前尚不清楚您到底想做什么，但是您的问题看起来像是How do I build html based on values of my php variables，我会尽力回答
This is not a complete answer, because it is unclear what exactly you are trying to do, but your question looks something like How do I build html based on values of my php variables and I will try to answer it as best I can
鉴于我从您的问题中获得的信息有限，我认为这是您要尝试做的事情:
With the limited information I have from your question, I think this is the kind of thing you are trying to do:
<?php
...
$html = "";
if ($result37 === "ncrteam") {
while ($row15 = mysqli_fetch_assoc($result15))
{
$html .= "<option value=".$row15['status'].">";
}
}
?>
<html>
<body>
...
<select>
<?php echo $html; ?>
</select>
...
</body>
</html>
这篇关于如何在if语句中回显PHP和HTML的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
