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setcookie()PHP無法使用變量

setcookie()PHP無法使用變量

PHP
ibeautiful 2021-05-07 15:14:56
我需要在作業(yè)中設置Cookie。當我使用字符串設置它們時,即setcookie('Name', 'John', time() + 86400)它可以工作,但是當使用PHP變量時,即setcookie('Name', $name, time() + 86400)未設置cookie。我不確定自己在做什么錯。我已經(jīng)在頁面上回顯了$ name變量,它顯示了出來,所以我知道這已經(jīng)設置了。當我使用顯示Cookie時$_COOKIE['Name'],它在接收到字符串時才起作用,但在時則不起作用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();}
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2 回答

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繁星淼淼

TA貢獻1775條經(jīng)驗 獲得超11個贊

不得在Cookie行(如echo,print_r()和HTML標記)之前輸出


您可以在任何輸出之前編寫cookie代碼


 <?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>


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?
智慧大石

TA貢獻1946條經(jīng)驗 獲得超3個贊

您的代碼中幾乎沒有問題,首先,沒有結(jié)束</form>標記,其次,您必須將表單提交到order01.php保存位置,cookie因此請嘗試這樣做,它將起作用


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();

  }

?>


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