setcookie（）PHP无法使用变量

Question

我需要在作业中设置Cookie。当我使用字符串设置它们时，即setcookie('Name', 'John', time() + 86400)它可以工作，但是当使用PHP变量时，即setcookie('Name', $name, time() + 86400)未设置cookie。我不确定自己在做什么错。我已经在页面上回显了$ name变量，它显示了出来，所以我知道这已经设置了。当我使用显示Cookie时$_COOKIE['Name']，它在接收到字符串时才起作用，但在时则不起作用PHP variable。order01.php<form action="order02.php" class="formLayout"> <div class="formGroup">  <label>First name:</label>  <input type="text" name="fname" class="textbox" autofocus   required placeholder="First name" title="first name"   maxlength="20" pattern="[A-Za-z'-]{2,20}"> </div> <div class="formGroup">  <label> Car model:</label>  <div class="formElements">   <input type="radio" name="model" required value="Mustang">Ford Mustang<br>   <input type="radio" name="model" required value="Subaru">Subaru WRX STI<br>   <input type="radio" name="model" required value="Corvette">Corvette<br>  </div> </div><?php $name = $_GET['fname']; $model = $_GET['model']; //write cookies for name and model for 1 day setcookie('Name', $name, time() + 86400); setcookie('Model', $model, time() + 86400);?>order02.phpif(isset($_COOKIE['Name'])){ echo "Cookie ".$_COOKIE['Name']." is set";}else{ echo "<div class='pageContainer'>"; echo "<h2 class='containerText, centerText'>Failed to validate inputs"; echo "<br><br>"; echo "<a href='order01.php'><button>Go Back</button></a>"; echo "</div>"; exit();}

Answer 1

不得在Cookie行（如echo，print_r（）和HTML标记）之前输出

您可以在任何输出之前编写cookie代码

 <?php

  $name = $_GET['fname'];

  $model = $_GET['model'];

 //write cookies for name and model for 1 day

  setcookie('Name', $name, time() + 86400);

  setcookie('Model', $model, time() + 86400);

       ?>

<form action="order01.php" class="formLayout">

 <div class="formGroup">

  <label>First name:</label>

  <input type="text" name="fname" class="textbox" autofocus required placeholder="First name" title="first name" maxlength="20" pattern="[A-Za-z'-]{2,20}">

</div>

<div class="formGroup">

   <label> Car model:</label>

   <div class="formElements">

    <input type="radio" name="model" required value="Mustang">Ford Mustang<br>

    <input type="radio" name="model" required value="Subaru">Subaru WRX STI<br>

    <input type="radio" name="model" required value="Corvette">Corvette<br>

   </div>

   <input type="submit" value="submit">

  </div>

 </form>

Answer 2

您的代码中几乎没有问题，首先，没有结束</form>标记，其次，您必须将表单提交到order01.php保存位置，cookie因此请尝试这样做，它将起作用

Order01.php

<form action="order01.php" class="formLayout">

<div class="formGroup">

 <label>First name:</label>

 <input type="text" name="fname" class="textbox" autofocus 

 required placeholder="First name" title="first name" 

 maxlength="20" pattern="[A-Za-z'-]{2,20}">

</div>

<div class="formGroup">

 <label> Car model:</label>

 <div class="formElements">

  <input type="radio" name="model" required value="Mustang">Ford Mustang<br>

 <input type="radio" name="model" required value="Subaru">Subaru WRX STI<br>

 <input type="radio" name="model" required value="Corvette">Corvette<br>

</div>

 <input type="submit" value="submit">

</div>

</form>

Order02.php

  <?php

 $name = $_GET['fname'];

 $model = $_GET['model'];

 //write cookies for name and model for 1 day

 setcookie('Name', $name, time() + 86400);

 setcookie('Model', $model, time() + 86400);

if(isset($_COOKIE['Name'])){

  echo "Cookie ".$_COOKIE['Name']." is set";

 }

 else{

   echo "<div class='pageContainer'>";

   echo "<h2 class='containerText, centerText'>Failed to validate inputs";

   echo "<br><br>";

   echo "<a href='order01.php'><button>Go Back</button></a>";

   echo "</div>";

   exit();

  }

？>