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.php

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();

}

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&nbsp;<?php&nbsp; $name = $_GET['fname'];&nbsp; $model = $_GET['model'];&nbsp;//write cookies for name and model for 1 day&nbsp; setcookie('Name', $name, time() + 86400);&nbsp; setcookie('Model', $model, time() + 86400);&nbsp; &nbsp; &nbsp; &nbsp;?><form action="order01.php" class="formLayout">&nbsp;<div class="formGroup">&nbsp; <label>First name:</label>&nbsp; <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">&nbsp; &nbsp;<label> Car model:</label>&nbsp; &nbsp;<div class="formElements">&nbsp; &nbsp; <input type="radio" name="model" required value="Mustang">Ford Mustang<br>&nbsp; &nbsp; <input type="radio" name="model" required value="Subaru">Subaru WRX STI<br>&nbsp; &nbsp; <input type="radio" name="model" required value="Corvette">Corvette<br>&nbsp; &nbsp;</div>&nbsp; &nbsp;<input type="submit" value="submit">&nbsp; </div>&nbsp;</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">&nbsp;<label>First name:</label>&nbsp;<input type="text" name="fname" class="textbox" autofocus&nbsp;&nbsp;required placeholder="First name" title="first name"&nbsp;&nbsp;maxlength="20" pattern="[A-Za-z'-]{2,20}"></div><div class="formGroup">&nbsp;<label> Car model:</label>&nbsp;<div class="formElements">&nbsp; <input type="radio" name="model" required value="Mustang">Ford Mustang<br>&nbsp;<input type="radio" name="model" required value="Subaru">Subaru WRX STI<br>&nbsp;<input type="radio" name="model" required value="Corvette">Corvette<br></div>&nbsp;<input type="submit" value="submit"></div></form>Order02.php&nbsp; <?php&nbsp;$name = $_GET['fname'];&nbsp;$model = $_GET['model'];&nbsp;//write cookies for name and model for 1 day&nbsp;setcookie('Name', $name, time() + 86400);&nbsp;setcookie('Model', $model, time() + 86400);if(isset($_COOKIE['Name'])){&nbsp; echo "Cookie ".$_COOKIE['Name']." is set";&nbsp;}&nbsp;else{&nbsp; &nbsp;echo "<div class='pageContainer'>";&nbsp; &nbsp;echo "<h2 class='containerText, centerText'>Failed to validate inputs";&nbsp; &nbsp;echo "<br><br>";&nbsp; &nbsp;echo "<a href='order01.php'><button>Go Back</button></a>";&nbsp; &nbsp;echo "</div>";&nbsp; &nbsp;exit();&nbsp; }？>