값 설정 후 PHP 변수가 null로 보입니다

-1

null 변수를 값으로 설정하고 페이지에 게시 한 후 이러한 PHP 값이 null 데이터를 출력합니다. 아무도 이유를 아나요?값 설정 후 PHP 변수가 null로 보입니다

PHP :

<?php 
$name = $_POST['query']; 
$cap = null; 
$ugly = "http://25.media.tumblr.com/tumblr_m3v5kj4ERf1qijcxeo1_500.gif"; 
$pretty = "http://upload.wikimedia.org/wikipedia/commons/e/e8/Bebe_Phoque_de_Weddell_Baby_Weddell_Seal.gif"; 
$hot = "http://hosting01.hotchyx.com/adult-image-hosting-42/466laugh.gif"; 
$weird = "http://d2tq98mqfjyz2l.cloudfront.net/image_cache/1308355337507544.gif"; 
$choice = null; 
$little = null; 
$adj1 = "walrus devouring"; 
$adj2 = "cat canoodling"; 
$adj3 = "voluptuous machine of ghettoness and stupidity"; 
$adjch = null; 
$i = rand(1,4); 
$w = rand(1,3); 

if(w == 1) 
{ 
    $adjch = $adj1; 
} 
elseif (w == 2) 
{ 
    $adjch = $adj2; 
} 
elseif (w == 3) 
{ 
    $adjch = $adj3; 
} 

if(i == 1) 
{ 
    $choice = $ugly; 
    $little = "You should probably see a doctor."; 
    $cap = "$name, You're uglier than all of One Direction * 2 + 100. That's pretty bad."; 

} 
elseif (i == 2) 
{ 
    $choice = $weird; 
    $cap = "$name, you're weird."; 
    $little = "I wish I didn't know you and you didn't exist you creepy, $adjch diddly do gooder."; 
} 
elseif (i == 3) 
{ 
    $choice = $pretty; 
    $cap = "$name, you're the prettiest $adjch person on earth."; 
    $little = "ilu so much"; 
} 
else if(i == 4) 
{ 
    $choice = $hot; 
    $cap = "$name, help the homeless. Take me home with you."; 
    $little = "You're hot. Really hot."; 
} 

?>

HTML :

<html> 
<link href="css/bootstrap.css" rel="stylesheet"> 
<link href="css/style.css" rel="stylesheet"> 
<body> 
<div class="navbar navbar-inverse navbar-fixed-top"> 
     <div class="navbar-inner"> 
     <div class="container-fluid"> 
      <button type="button" class="btn btn-navbar" data-toggle="collapse" data-target=".nav-collapse"> 
      <span class="icon-bar"></span> 
      <span class="icon-bar"></span> 
      <span class="icon-bar"></span> 
      </button> 
      <div class="nav-collapse collapse"> 
      <ul class="nav"> 
       <li class="active"><a href="index.html">Home</a></li> 
       <li><a href="blog.html">Blog</a></li> 
       <li><a href="about.html">About</a></li> 
      </ul> 
      </div><!--/.nav-collapse --> 
     </div> 
     </div> 
    </div> 

<div class='push'></div> 
<p><img src="<?php echo $choice;?>"</p> 
<h2 align='center'><strong><?php echo $cap;?></strong></h2> 
<p><?php echo $little;?></p> 
<p><a href="index.html" class='btn'><i class="icon-search"></i> Go back </a></p> 

</body> 
<div class='footerpush'></div> 
<hr> 
<div align='center'> 
<a href='about.html'>about</a> 
| 
<a href='blog.html'>blog</a> 
| 
<a href='twitter.com'>twitter</a> 
| 
<a href='notice.html'>our policy</a> 
</div> 
<p>&copy bleh blah</p> 

</html>

감사합니다 모두 (PS, 나는 정말 죄송 PHP는 멍청한 놈이야)

내 코드입니다

출처

2013-10-15 Fixie

여기에 코딩을 넣으십시오. –

방금 규칙을 준수하도록 코드를 포맷했습니다. 빠른 질문 : 확장자가 .php 인 HTML 파일의 이름을 지정 했습니까? 그리고 이것을 서버에서 실행하고 있습니까? –

문제가 해결되면 삭제하지 마십시오. – michaelb958

if(w == 1)을 사용하는 방법을

if ($w == ...) 및 if ($i==...) 물론 의지? if(i == 1)? w은 무엇입니까? i은 무엇입니까? $w 및 $i 만 신고하셨습니다.

배열을 사용하면이 코드를 단순화 할 수 있습니다.

<?php 

    $name = $_POST['query']; 

    $adjs = array(
     1 => "walrus devouring", 
     2 => "cat canoodling", 
     3 => "voluptuous machine of ghettoness and stupidity" 
    ); 

    $adjch = $adjs[rand(1, 3)]; 

    $choices = array(
     1 => "http://25.media.tumblr.com/tumblr_m3v5kj4ERf1qijcxeo1_500.gif", 
     2 => "http://d2tq98mqfjyz2l.cloudfront.net/image_cache/1308355337507544.gif", 
     3 => "http://upload.wikimedia.org/wikipedia/commons/e/e8/Bebe_Phoque_de_Weddell_Baby_Weddell_Seal.gif", 
     4 => "http://hosting01.hotchyx.com/adult-image-hosting-42/466laugh.gif" 
    ); 

    $littles = array(
     1 => "You should probably see a doctor.", 
     2 => "I wish I didn't know you and you didn't exist you creepy, $adjch diddly do gooder.", 
     3 => "http://upload.wikimedia.org/wikipedia/commons/e/e8/Bebe_Phoque_de_Weddell_Baby_Weddell_Seal.gif", 
     4 => "You're hot. Really hot." 
    ); 

    $caps = array(
     1 => "$name, You're uglier than all of One Direction * 2 + 100. That's pretty bad.", 
     2 => "$name, you're weird.", 
     3 => "ilu so much", 
     4 => "$name, help the homeless. Take me home with you." 
    ); 

    $i = rand(1, 4); 

    $choice = $choices[$i]; 
    $cap = $caps[$i]; 
    $little = $littles[$i]; 

?>

출처

2013-10-15 00:18:18 Havenard

좋은 답변입니다. +1 – Fixie

변경 모든 IF를 가진 : switch statement

출처

2013-10-15 00:18:06

값 설정 후 PHP 변수가 null로 보입니다

답변

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