PHP完结脚本但是不影响后面的HTML输出
时间：2021-07-01 10:21:17
帮助过：1人阅读
PHP 结束脚本 但是不影响后面的HTML输出
可能是比较小白的问题。
比如：
if(!isset($_GET["id"])||!isset($_GET["code"]))
{
echo "注册验证";
exit();
}
else
{
echo "注册验证1";
}
......
?>
比如这段脚本在本身HTML中
的位置，判定是否有GET提交的数据来</pre>输出不同的title<br />
后面是有脚本的，但是如果GET没有数据传来，后面连接数据库什么的就没有必要，所以想要结束。<br />
但是结束后，使用exit或者return，后面HTML中<body>的内容等也就不</pre>输出了，该怎么办？
</p><br><font color='#FF8000'>------解决方案--------------------</font><br>不能 exit();<br />
而应将连接数据库什么的放在 else 分支中<br><font color='#FF8000'>------解决方案--------------------</font><br>你也可以把这些php代码放在单独的文件中 include 进来<br />
只是要将 exit(); 改为 return;<br /><br><font color='#FF8000'>------解决方案--------------------</font><br>本帖最后由 xuzuning 于 2013-02-21 12:57:32 编辑</p>
你还可以写成函数<br />
作为结构化编程，无非就是<br />
if else<br />
if elseif else<br />
switch case default<br />
另类点就是<br />
while break<br />
<br />
真的不喜欢结构化编程时<br />
php5.3已经恢复了对 goto 的支持，适当的使用也还是蛮不错的<br />
<pre class="brush: php">goto a;<br />
echo 123; //这里不会被执行<br />
a:<br />
echo 'abcd';<br />
</pre> </div>
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