未定義的變量$ url，不確定爲什麼我得到這個錯誤

Question

任何人都可以告訴我爲什麼我得到這個腳本的以下未定義變量（$ url）錯誤。 錯誤是我的if語句if ($url)。我是一個新手。由於腳本 'testnew/edit_your_sites.php' 發生在28行

錯誤： 未定義的變量：URL

<div class="text"> 
    <?php 
    $page_title = 'Edit Your Account'; 
    include ('includes/header.html'); 
    include ('includes/functions.php'); 
    include ('includes/config.inc.php'); 
    if (isset($_GET['id']) && is_numeric($_GET['id'])){ 
     $id = $_GET['id']; 
    } elseif (isset($_POST['id']) && is_numeric($_POST['id'])) { 
     $id = $_POST['id']; 
    } else { 
     echo '<p class="error">This page has been accessed in error.</p>'; 
     include ('includes/footer.html'); 
     exit(); 
    } 
    if (isset($_SESSION['UserID'])){ 
    require (MYSQL); 
    $scrubbed = array_map('spam_scrubber', $_POST); 


    if ($_SERVER['REQUEST_METHOD'] == 'POST'){ 
    if (empty($scrubbed['url'])){ 
      echo '<p class="error">Please enter a url</p>'; 
     } else { 
      $url = mysqli_real_escape_string($dbc, $scrubbed['url']); 
     } 
    } 
    if ($url){ 
    $p = "SELECT UserID FROM sites WHERE UserID=$id"; 
    $q = mysqli_query($dbc, $p); 
    if (mysqli_num_rows($q) == 1){ 

    $i = "INSERT INTO sites (UserID, url, entry) VALUES('{$_SESSION['UserID']}', '$url', NOW())"; 
    $r = mysqli_query($dbc, $i); 
     if (mysqli_affected_rows($dbc) == 1){ 
     echo '<p>Your website was added succesfully.</p>'; 
     } else { 
      echo '<p class="error">Due to a system error your website could not be added.</p>'; 
     } 

    } else { 
     echo '<p class="error">Please try again'; 
    } 

    } 
    ?> 
    <form action="edit_your_sites.php" method="post"> 
    <?php $sql = "SELECT * FROM sitetypes"; 
    $f = mysqli_query($dbc, $sql) or trigger_error("Query: $sql\n<br />Mysqli Error: " . mysqli_error($dbc)); 
    echo '<select name="SiteType" selected="selected">'; 
    while($row2 = mysqli_fetch_array($f, MYSQLI_ASSOC)){ 
     if ($row2['SiteType'] == 'selected'){ 
      echo "<option Selected='selected' value=" . $row2["SiteTypeID"] . ">" . $row2['SiteType'] . "</option>"; 
     } else { 
      echo "<option value=" . $row2["SiteTypeID"] . ">" . $row2['SiteType'] . "</option>"; 
     } 
    } 

    echo '</select>'; 
    ?> 
    <p>Url:<input type="text" name="url" size="30" maxlength="60" value="<?php if(isset($scrubbed['url'])) echo $scrubbed['url']; ?>" /></p> 
    <?php 
    echo '</fieldset> 
    <p><input type="submit" name="submit" value="Edit Account!" /><input type="reset" name="reset" value="Clear Form" /> 
    <input type="hidden" name="id" value="' . $id . '" /> 
    </form>'; 
    } else { 
     $url = BASE_URL . 'index.php'; 
     header("Location: $url"); 
    } 
    ?> 

有可能在這個腳本，如果你看到一些 更多的錯誤任何它將不勝感激，如果你 讓我知道他們。 感謝你的幫助，你們是真的有很大的幫助

Answer 1

這是因爲您在else塊定義變量$url只。但是，如果empty($scrubbed['url'])爲true，則永遠不會定義變量，因爲else塊永遠不會執行。

if (empty($scrubbed['url'])) { 
    echo '<p class="error">Please enter a url</p>'; 
} 
else { 
    $url = mysqli_real_escape_string($dbc, $scrubbed['url']); 
} 

爲了避免這種錯誤，就在上述代碼段之前定義$url：

$url = null; 

Answer 2

變量只有一個if語句中定義的，你需要先定義它到...嘗試定義它在腳本的開頭有一個空白值。

$page_title = 'Edit Your Account'; 
$url=""; 
include ('includes/header.html'); 

Answer 3

這是因爲你曾經說過

if($url) 

但這不是定義。

嘗試

if(isset($url))