Php 只有else语句正在执行

您需要学会调试代码：

<div id="register" class="animate form">
<form  action="signup_ac.php" autocomplete="on" method="POST"> 
<h1> Sign up </h1> 
<p> 
<label for="usernamesignup" class="uname" data-icon="u">Your username</label>
<input id="usernamesignup" name="usernamesignup" required type="text"   placeholder="mysuperusername690" />
</p>
<p> 
<label for="emailsignup" class="youmail" data-icon="e" > Your email</label>
<input id="emailsignup" name="emailsignup" required type="email"   placeholder="mysupermail@mail.com"/> 
</p>
<p> 
<label for="passwordsignup" class="youpasswd" data-icon="p">Your password </label>
<input id="passwordsignup" name="passwordsignup" required type="password" placeholder="eg. X8df!90EO"/>
</p>
<p> 
<label for="passwordsignup_confirm" class="youpasswd" data-icon="p">Please confirm your   password </label>
<input id="passwordsignup_confirm" name="passwordsignup_confirm" required  type="password" placeholder="eg. X8df!90EO"/>
</p>
<p class="signin button">
<input type="submit" value="Sign up"/> 
                            </p>
                            <p class="change_link">  
                                Already a member ?
                                <a href="#tologin" class="to_register"> Go and log in </a>
                            </p>
                        </form>
                       </div>

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您需要学习调试代码：

<div id="register" class="animate form">
<form  action="signup_ac.php" autocomplete="on" method="POST"> 
<h1> Sign up </h1> 
<p> 
<label for="usernamesignup" class="uname" data-icon="u">Your username</label>
<input id="usernamesignup" name="usernamesignup" required type="text"   placeholder="mysuperusername690" />
</p>
<p> 
<label for="emailsignup" class="youmail" data-icon="e" > Your email</label>
<input id="emailsignup" name="emailsignup" required type="email"   placeholder="mysupermail@mail.com"/> 
</p>
<p> 
<label for="passwordsignup" class="youpasswd" data-icon="p">Your password </label>
<input id="passwordsignup" name="passwordsignup" required type="password" placeholder="eg. X8df!90EO"/>
</p>
<p> 
<label for="passwordsignup_confirm" class="youpasswd" data-icon="p">Please confirm your   password </label>
<input id="passwordsignup_confirm" name="passwordsignup_confirm" required  type="password" placeholder="eg. X8df!90EO"/>
</p>
<p class="signin button">
<input type="submit" value="Sign up"/> 
                            </p>
                            <p class="change_link">  
                                Already a member ?
                                <a href="#tologin" class="to_register"> Go and log in </a>
                            </p>
                        </form>
                       </div>

---

可爱的漏洞，

mysql.*

函数已经被弃用了很长一段时间。请查看MySQLi:或PDO MySQL:以帮助您安全防范SQL注入。可爱的漏洞…MySQL的

MySQL.*

函数已经被弃用了很长一段时间。请看一下MySQLi:或PDO MySQL:以帮助您安全防范SQL注入。我刚刚看到并编辑了它，但它仍然显示相同的内容。我现在有点迷茫。因此，您需要从这里开始学习如何调试。开始检查脚本的输入。您试图访问的所有$\u POST值是否都存在？也许你的html表单中有一个输入错误。我刚才看到并编辑了它，但它仍然显示相同的内容。我现在有点迷路了。所以你需要从这里开始学习如何调试。开始检查脚本的输入。您试图访问的所有$\u POST值是否都存在？也许你的html表单有输入错误。

if [...snip...] $POST_['passwordsignup_confirm'])){
                 ^^^^^--- dyslexic moment?