PHP代码在测试代码时不显示

-1

我试图尝试根据重要的信息片断（例如姓名，祈祷请求，电话号码等）来保存用户输入。这个想法是一个人导航到网页并且然后填写信息，然后单击提交按钮，然后将数据保存到数据库。每当我尝试测试页面时，没有任何东西拉上来。我多次查看了我的代码并找不到问题。我有一个粗略的模板来基于我的PHP。在它内部有一个“全部显示”按钮，根本不需要在页面上。任何建议将不胜感激。我的代码（包括“全部显示”按钮的代码）可以发现如下：PHP代码在测试代码时不显示

<?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); 
} 
?>

来源

2014-04-12 user3521756

我不知道，如果你是通过教程或课程学习PHP，但你正在使用不推荐使用的'mysql（）'函数。请使用['PDO']（http://be2.php.net/manual/en/book.pdo.php）（首选）或['mysqli（）']（http://be2.php.net/手册/ en/book.mysqli.php）。并检查最近的教程或将其提交给您的教练。 – KarelG

这种尝试连接

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

来源

2014-04-12 07:39:00 Sadikhasan

PHP代码在测试代码时不显示

回答

相关问题