测试代码时不显示PHP代码

时间:2014-04-12 07:37:44

标签: php sql database

我试图根据名称,祷告请求,电话号码等重要信息以表格保存用户输入。想法是一个人导航到网页,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,字面上都没有任何结果。我已多次查看我的代码,无法找到问题。我有一个粗略的模板来基于我的PHP。在其中有一个“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”按钮代码)可以在下面找到:

<?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); 
} 
?>

1 个答案:

答案 0 :(得分:0)

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);