如何刪除php電子郵件中的空行

Question

我有一個php表單，當發佈時發送確認電子郵件。表格有幾個下拉列表，當填充時顯示零件的數量和名稱。我有一切工作正常，在我的$ usermessage中，我列出了所有對應於下拉列表的變量。我只想顯示已經實現的下拉列表和已填充的名稱。我無法弄清楚如何在變量未被填充時刪除多餘的空格。

我的確認郵件看起來是這樣的，當我有前兩個下拉菜單和一個第八填充：

用戶信息
名稱
李四

電子郵件地址
email@email.com

地址
12345 street

US
鎮州郵政編碼

電話號碼

數量
編號

CH171181

CH171182








CH171363

正如你所看到的，有空間補償無人居住變量的巨大差距。

這裏是我的代碼：

$email = $_POST["email"]; 
$myemail ="$email"; 

$CH171181 = $_POST["CH171181"]; 
$CH171182 = $_POST["CH171182"]; 
$CH171183 = $_POST["CH171183"]; 
$CH171184 = $_POST["CH171184"]; 
$CH171360 = $_POST["CH171360"]; 
$CH171361 = $_POST["CH171361"]; 
$CH171362 = $_POST["CH171362"]; 
$CH171363 = $_POST["CH171363"]; 
$CH171364 = $_POST["CH171364"]; 
$CH171404 = $_POST["CH171404"];  
$CH171408 = $_POST["CH171408"]; 
if (empty($CH171181)) {$var1 = '';} else {$var1 = 'CH171181';} 
if (empty($CH171182)) {$var2 = '';} else {$var2 = 'CH171182';} 
if (empty($CH171183)) {$var3 = '';} else {$var3 = 'CH171183';} 
if (empty($CH171184)) {$var4 = '';} else {$var4 = 'CH171184';} 
if (empty($CH171360)) {$var5 = '';} else {$var5 = 'CH171360';} 
if (empty($CH171361)) {$var6 = '';} else {$var6 = 'CH171361';} 
if (empty($CH171362)) {$var7 = '';} else {$var7 = 'CH171362';} 
if (empty($CH171363)) {$var8 = '';} else {$var8 = 'CH171363';} 
if (empty($CH171364)) {$var9 = '';} else {$var9 = 'CH171364';} 
if (empty($CH171404)) {$var10 = '';} else {$var10 = 'CH171404';} 
if (empty($CH171408)) {$var11 = '';} else {$var11 = 'CH171408';} 

$first_name = $_POST["first_name"]; 
$last_name = $_POST["last_name"]; 
$address = $_POST["address"]; 
$address2 = $_POST["address2"]; 
$country = $_POST["country"]; 
$city = $_POST["city"]; 
$state = $_POST["state"]; 
$zip = $_POST["zip"]; 
$phone = $_POST["phone"]; 

$user = "$myemail"; 
$usersubject = "Subject"; 
$userheaders = "From: support@support.com\n"; 
$usermessage = " 

Users Information 
Name 
$first_name $last_name 

Email address 
$email 

Address 
$address 
$address2 
$country 
$city $state $zip 

Phone Number 
$phone 

Qty 
Item # 

$CH171181 
$var1 

$CH171182 
$var2 

$CH171183 
$var3 

$CH171184 
$var4 

$CH171360 
$var5 

$CH171361 
$var6 

$CH171362 
$var7 

$CH171363 
$var8 

$CH171364 
$var9 

$CH171404 
$var10 

$CH171408 
$var11 
"; 

$headers = "MIME-Version: 1.0" . "\r\n"; 
$headers .= "Content-type:text/html;charset=UTF-8" . "\r\n";  

mail($user,$usersubject,$usermessage,$userheaders,$headers); 
} 

Answer 1

取而代之1個連續字符串內，以書面的輸出，你需要做一個檢查，看看是否變量包含一個字符串。如果它是空的，那麼甚至不要將它包含在$usermessage中。

例如：

if ($CH171183 !== ''): 
    $usermessage .= $CH171183 . "\n" . $var3; 
endif; 

這意味着如果$CH171183包含字符串，將包含在輸出（$usermessage）。但是，如果它不包含任何內容，那麼將會出現在輸出中。換句話說，你只是在做有數據顯示的地方創建輸出。

原始代碼的問題是$usermessage沒有檢查變量是否爲空。所以如果它是空的，你最終會在輸出中留下空白的空間。

Answer 2

試試這個

str_replace(chr(13), "", $usermessage) 

Answer 3

嘗試沿着下面的代碼線的東西：

$email = $_POST["email"]; 
$myemail ="$email"; 

$vars = Array("CH171181", 
       "CH171182", 
       "CH171183", 
       "CH171184", 
       ... etc 
     ); 

$first_name = $_POST["first_name"]; 
$last_name = $_POST["last_name"]; 
$address = $_POST["address"]; 
$address2 = $_POST["address2"]; 
$country = $_POST["country"]; 
$city = $_POST["city"]; 
$state = $_POST["state"]; 
$zip = $_POST["zip"]; 
$phone = $_POST["phone"]; 

$user = "$myemail"; 
$usersubject = "Subject"; 
$userheaders = "From: support@support.com\n"; 

$usermessage = " 

Users Information 
Name 
$first_name $last_name 

Email address 
$email 

Address 
$address 
$address2 
$country 
$city $state $zip 

Phone Number 
$phone 

Qty 
Item #"; 

foreach($vars as $idx => $v){ 
    if($_POST[$v]){ 
     $usermessage .= $idx . "\n" . $v . "\n"; 
    } 
} 

$headers = "MIME-Version: 1.0" . "\r\n"; 
$headers .= "Content-type:text/html;charset=UTF-8" . "\r\n";  

mail($user,$usersubject,$usermessage,$userheaders,$headers); 
}