PHP代碼在測試代碼時不顯示

Question

我試圖嘗試根據重要的信息片斷（例如姓名，祈禱請求，電話號碼等）來保存用戶輸入。這個想法是一個人導航到網頁並且然後填寫信息，然後單擊提交按鈕，然後將數據保存到數據庫。每當我嘗試測試頁面時，沒有任何東西拉上來。我多次查看了我的代碼並找不到問題。我有一個粗略的模板來基於我的PHP。在它內部有一個「全部顯示」按鈕，根本不需要在頁面上。任何建議將不勝感激。我的代碼（包括「全部顯示」按鈕的代碼）可以發現如下：

<?php 
$Req_F_Name = $_POST["Req_F_Name"]; 
$Reg_L_Name = $_POST["Reg_L_Name"]; 
$Reg_Phone = $_POST["Reg_Phone"]; 
$Reg_Email= $_POST["Reg_Email"]; 
$Reg_Request = $_POST["Reg_Request"]; 
$Reg_Address_1= $_POST["Reg_Address_1"]; 

if ($Submit) { 
$conn = mysql_connect('XXXXXX','XXXX','XXXXXX','prayer')    or 
die("Could not connect: " . mysql_error()); 
//select the database 
$db = mysql_select_db("prayer"); 

$query = "INSERT INTO Request VALUES  ('".$Req_F_Name."','".$Reg_L_Name."','".$Reg_Phone."','".$Reg_Email."','".$Reg_Request."',' ".$Reg_Address_1."')"or die(mysql_error()); 
echo "Your Query was successfully stored in the database :)"; 

mysql_close($conn); 
} 

if ($Show_All_Records) { 

//establish connection to mysql 
$conn = mysql_connect('XXXXXX','XXXXXX','XXXXXX','prayer') 
or 
die("Could not connect: " . mysql_error()); 
echo "Connected to MySql".'<br>'.'<br>'; 

//select the database 
$db = mysql_select_db("prayer"); 
$result = mysql_query("SELECT * FROM 'Request'",$conn); 

//display the results 
echo 'First Name:', $Req_F_Name ,'<br>' 
. 'Last Name:', $Reg_L_Name ,'<br>' 
. 'Phone Number:', $Reg_Phone ,'<br>' 
. 'Email:', $Reg_Email ,'<br>' 
. 'Prayer Request:', $Reg_Request ,'<br>' 
. 'Address:', $Reg_Address_1 ,'<br>'.'<br>'; 

//grabbing all data from the table 
echo 'Your Query was succesfully stored in the database :)'; 

while ($myrow = mysql_fetch_array($result)) { 
echo ''.$myrow["Req_F_Name"].' <br> '.$myrow["Reg_L_Name"].' <br>'.$myrow["Reg_Phone"].' <br> '.$myrow["Reg_Email"].' <br> '.$myrow["Reg_Request"].' <br> '.$myrow["Reg_Address_1"].' <br>'; 
} 
mysql_close($conn); 
} 
?> 

Answer 1

這種嘗試連接

$conn = mysql_connect('XXXXXX','XXXX','XXXXXX')    or 
die("Could not connect: " . mysql_error()); 
//select the database 
$db = mysql_select_db("prayer",$conn);