如果声明半工作

时间:2014-12-07 00:38:38

标签: php if-statement

我正在为提交的信息制作页面。当这个人没有填写一个方框(在这种情况下是名字)时,它应该改为$ okay为FALSE。现在,当我运行它时,这个代码只有一半有效 它会打印出句子,但不会将$ ok改为false。

<?php
   $FirstName = $_POST['firstname'];
   $LastName = $_POST['lastname'];
   $Email = $_POST['email'];
   $CompanyName = $_POST['companyname'];
   $Description = $_POST['description'];
   $okay = TRUE;


   if (empty($FirstName)) {
       echo '<p class="error"> Please enter first name</p>';
       $okay = FALSE;
   }


   if ($okay = TRUE) {
       echo "<p> Thank you $FirstName $LastName. We are looking foward to doing business with you. Please look for an email at $Email from
               BrandingYouSmall@gmail.com for '$CompanyName' </p>";
       echo "<p> Your Description: <br /> $Description </p>";
   }
?>

1 个答案:

答案 0 :(得分:4)

if($okay = TRUE){

这是一个问题。单=是赋值的运算符。所以你告诉&#34;将$ okay设置为TRUE&#34;。但你想要检查。所以你需要使用相等运算符 - ==。解决它:

if ($okay == true) {

或更短:

if ($okay) {