PHP无法访问服务器上的数据库

Question

I am having difficulty getting my PHP code to reference my database and display certain information within it. The error I am getting is "Unable to access database." I have poured over this code numerous times and cannot seem to find the solution. The first file (Approve Deny Prayer Request) references another file (Prayer Request) in order to draw the data. The idea is that I would be able to edit prayer requests and then click an "Approved" button that would edit the entry and re-save it into the database. Below please find both CGI files.

Thanks for the help.

Approve Deny Prayer Request

<table cellpadding="10">
<tr>
<td>First Name</td>
<td>Last Name</td>
<td>Prayer Request</td>
</tr>

<?php

$username="XXXXXX";
$password="XXXXXXXXX";
$database="prayer";

mysqli_connect('fbcaltusprayerorg.ipagemysql.com',$username,$password,$database);
@mysqli_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM Request";
$result=mysqli_query($query);
mysqli_close();

while ($row=mysqli_fetch_array($result)){
echo ("<tr><td>$row[Reg_F_Name]</td>");
echo ("<td>$row[Reg_L_Name]</td>");
echo ("<td>$row[Reg_Request]</td>");
echo ("<td><a href=\"cgi-bin/PrayerRequest.php?id=$row[id]\">Edit</a></td></tr>");
}
echo "</table>";

?>

Prayer Request

<?php

$username="XXXXX";
$password="XXXXXXXX";
$database="prayer";

mysqli_connect('fbcaltusprayerorg.ipagemysql.com',$username,$password,$database);
@mysqli_select_db($database) or die( "Unable to select database");
$query = "SELECT * FROM Request"; 
$result = mysqli_query($query);
$row = mysqli_fetch_array($result);
?>

<form method="post" action="cgi-bin/ApproveDenyPrayerRequest.php" />

<table>

<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[Reg_F_Name]" ?>"></td>
</tr>

<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[Reg_L_Name]" ?>"></td>
</tr>

<tr>
<td>Prayer Request</td>
<td><input type="text" name="phone" value="<? echo "$row[Reg_Request]" ?>"></td>
</tr>

</table>

</form>

Answer 1

Don't use error suppression operator @ , it will hide the warnings and you will not be able to figure where the problem is..

Rewrite your code like this for establishing the connection..

<?php

$username="XXXXX";
$password="XXXXXXXX";
$database="prayer";

$link = mysqli_connect('fbcaltusprayerorg.ipagemysql.com', $username, $password, $database);

if (!$link) {
    die('Connect Error (' . mysqli_connect_errno() . ') '
        . mysqli_connect_error());
}

echo 'Success... ' . mysqli_get_host_info($link) . "
";

$query = "SELECT * FROM Request";
$result = mysqli_query($link,$query); //<----- Added link
$row = mysqli_fetch_array($result);
?>

<form method="post" action="cgi-bin/ApproveDenyPrayerRequest.php" />

<table>

    <tr>
        <td>First Name:</td>
        <td><input type="text" name="first" value="<? echo "$row[Reg_F_Name]" ?>"></td>
    </tr>

    <tr>
        <td>Last Name:</td>
        <td><input type="text" name="last" value="<? echo "$row[Reg_L_Name]" ?>"></td>
    </tr>

    <tr>
        <td>Prayer Request</td>
        <td><input type="text" name="phone" value="<? echo "$row[Reg_Request]" ?>"></td>
    </tr>

</table>

</form>

Read more here from the PHP manual.