價值$ _POST從其他頁面不能通過isset形式

我有一個問題，通過值$result從不同的頁面到另一個頁面。

當我POST值$result進入此頁面。它工作正常，我回應出來。

但是，如果我嘗試將此$result轉換爲isset()表格並將其張貼回自我。
該值將成爲未定義索引。第2頁

PAGE1

<?PHP 
$connect=mysql_connect('localhost', 'root', ''); 
$db=mysql_select_db('survey', $connect); 
$SQL="SELECT * FROM question"; 
$data=mysql_query($SQL); 
$num=mysql_num_rows($data); 

$start=0; 
?> 


<style type="text/css"> 
#survey{ 
    width:100%; height:50px; 
    background:#e6e6e6; border-bottom:solid 1px white; 
    } 

#survey form{ 
    position:relative; 
    left:320px; top:15px; 
    float:left; 
    } 

#survey form a{ 
    text-decoration:none; 
    color:black; 
    } 

.submit{ 
background-color: transparent; 
border: none; 
cursor: pointer; 
} 

</style> 

<html> 
<div id="container"> 

<?PHP 
if($num<3){ 
    echo " 
    <div id='create'><a href='question.php'>+ Create a Survey</a></div> 
    "; 
} 
?> 


<?PHP 
while($start<$num){ 
    $result=mysql_result($data, $start, 'ID'); 
    $list_num=$start+1; 

     $ID=$result['ID']; 

     $list=" 
      <div id='survey'> 
       <span style='font-size:24px; color:gray; position:relative; float:left; left:10px; top:13px;'> 
       SURVEY 0$list_num 
       </span> 

       <form action='survey.php' method='POST'> 
        <input class='submit' type='submit' name='submit1' value='View' /> 
        <input type='hidden' name='result' value='$ID'/> 
       </form> 

       <form action='update.php' method='POST'> 
        <input class='submit' type='submit' name='submit2' value='Edit' /> 
        <input type='hidden' name='result' value='$ID'/> 
       </form>    

       <form action='delete.php' method='POST'> 
        <input class='submit' type='submit' name='submit3' value='&#10005 Delete' /> 
        <input type='hidden' name='result' value='$ID'/> 
       </form> 

      </div> 
     "; 

    echo $list; 
    $start++; 
    } 
?> 

</div> 
</html>

PAGE2

<?PHP 
    $result=$_POST['result']; 

    $error=""; 
    if(isset($_POST['submit'])){ 
     $question=$_POST['question']; 
     $answer_a=$_POST['a']; 
     $answer_b=$_POST['b']; 
     $answer_c=$_POST['c']; 


     if($question && $answer_a && $answer_b && $answer_c){ 
      $connect=mysql_connect("localhost", "root", ""); 
      $db=mysql_select_db("survey", $connect) or die(""); 

      $SQL="UPDATE question SET question='$question', answer_a='$answer_a', answer_b='$answer_b', answer_c='$answer_c' 
      WHERE ID='$result'"; 
      mysql_query($SQL); 
      $SQL="UPDATE answer SET question='$question', answer_a='0', answer_b='0', answer_c='0' 
      WHERE ID='$result' "; 
      mysql_query($SQL); 
      header("location:create.php"); 
     } 
     else{$error="<div style='color:white; background:gray; width:200px; height:20px; text-align:center; line-height:20px;'>" 
     . "Please fill all the field." . "</div>";} 
    } 


    ?> 
<style> 
body{ 
    font-family:Arial, Helvetica, sans-serif; 
    font-size:14px; 
    } 

</style> 

<html> 
<form action="update.php" method="POST"> 
Question<br /><input type="text" name="question" placeholder=" Enter your question here" maxlength="50" /><br/><br/>Answer<br/> 
a<input type="text" name="a" placeholder=" Answer a" maxlength="50"/> 
b<input type="text" name="b" placeholder=" Answer b" maxlength="50"/> 
c<input type="text" name="c" placeholder=" Answer c" maxlength="50"/> 
<input type="submit" name="submit" value="SAVE"/> 
</form> 
<?PHP echo $error; ?> 
</html>

來源

2012-12-24 chien pin wang

這是什麼意思？ **但如果我嘗試將此$結果傳遞給isset表單並將其發回自我。** – samayo

嗨，非常感謝你的幫助我找到了解決方案我認爲問題是當我將值傳遞給第二種形式時，我需要重新發送值。這是我做的。如果（isset（$ _ POST ['result']））{ $ result = $ _ POST ['result'];第一個來自上一頁，第二個是表單發佈到自身 ; echo $ result; }

if(isset($_POST['data'])){ 
    $ID=$_POST['data']; 
    echo $ID; 
    }

添加另一個隱藏在形式和重新發送值

<input type="hidden" name="data" value="<?PHP echo $result;?>"/>

更改MySQL值

WHERE... ID='$ID'";

來源

2012-12-24 23:51:11

嘗試編輯這個

<html> 
<form action="update.php" method="POST">

對此

<html> 
<form action="" method="POST">

來源

2012-12-24 13:15:24 kingW3

價值$ _POST從其他頁面不能通過isset形式

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