PHP Forum

**djab** · 01-18-2011, 02:00 PM

This code produces no PHP fatal errors or warnings, but gives us an Apache 500 Internal Server error.
Why? I commented out pretty much the entire loop. If I comment out the first while too, it doesn't give a server error. But just with the while statement (and technically nothing happening inside it), it freaks out. What's goin on?

PHP Code:


			
<?php
 // Connects to your Database 
 require("libraries/db/dbconnect.php"); 
 // Encrypt Page
 require("libraries/auth/membersarea.php"…

$username = $_COOKIE['ID_my_site'];
$userID = mysql_result(mysql_query("select ID from users where username = '$username'"), 0);

$result = mysql_query("SELECT * FROM payroll ORDER BY PID DESC");
 
 if (!$result) {
     echo "Could not successfully run query from DB: " . mysql_error();
     exit;
     }
 if (mysql_num_rows($result) == 0) {
     echo "No past payments available";
     exit;
     }
    
// New code by Josh
// Gonna annotate this like a boss to keep track

// Open our table up
    echo "<table align=\"center\" width=\"400\">\r\n";
    
while($row = mysql_fetch_array(mysql_query("SELECT * FROM payroll ORDER BY PID DESC")))
    {
    // Table is opened. Let's bring in our Timestamp:
    //echo "<tr><td>Timestamp</td><td>".$row2['time…
    // Configure the columns of the table
    //echo "<tr><td></td><td><strong>Unit</strong><…
    // Bring in our variables now    
    /*while($row2 = mysql_fetch_array(mysql_query("SELECT * FROM vars ORDER BY ID ASC")))
        {
        // Determine variables
        $aliasName = $row2['varAlias'];
        $aliasRate = $row2['varAlias']."_rate";
        // Parse into table
        echo "<tr><td>".$aliasName."</td><td>".$row[$…
        }*/
    }

    echo "</table>\r\n";
    // End table

    
 // Footer Template
 require("template/footer.php");
 ?>

**lynxus** · 01-18-2011, 02:05 PM

Before i even look over your code.

What errors do you get in your apache / webserver error.log?

**lynxus** · 01-18-2011, 02:07 PM

Is this meant to be like this?

Not having a ); at the end?
// Encrypt Page
require("libraries/auth/membersarea.php"…

**mckain** · 01-18-2011, 02:07 PM

what if you test section by section,
in order to split the code in several sets of code, try it!

**djab** · 01-18-2011, 02:22 PM

PHP Code:


			
<?php
 // Connects to your Database 
 require("libraries/db/dbconnect.php"); 
 // Encrypt Page
 require("libraries/auth/membersarea.php");

$username = $_COOKIE['ID_my_site'];
$userID = mysql_result(mysql_query("select ID from users where username = ".$username.""), 0);

$result = mysql_query("SELECT * FROM payroll ORDER BY PID DESC");
 
 if (!$result) {
     echo "Could not successfully run query from DB: " . mysql_error();
     exit;
     }
 if (mysql_num_rows($result) == 0) {
     echo "No past payments available";
     exit;
     }
    
// New code by Josh
// Gonna annotate this like a boss to keep track

// Open our table up
    echo "<table align=\"center\" width=\"400\">\r\n";
    
while($row = mysql_fetch_array(mysql_query("SELECT * FROM payroll ORDER BY PID DESC")))
    {
    // Table is opened. Let's bring in our Timestamp:
    echo "<tr><td>Timestamp</td><td>".$row2['timestamp']."</td><td><i>N/A</i></td></tr>";
    // Configure the columns of the table
    echo "<tr><td></td><td><strong>Unit</strong></td><td><strong>Rate</strong></td></tr>\r\n";
    // Bring in our variables now    
    while($row2 = mysql_fetch_array(mysql_query("SELECT * FROM vars where ID = ".$userID." ORDER BY ID ASC")))
        {
        // Determine variables
        $aliasName = $row2['varAlias'];
        $aliasRate = $row2['varAlias']."_rate";
        // Parse into table
        echo "<tr><td>".$aliasName."</td><td>".$row[$aliasName]."</td><td>".$row[$aliasRate]."</td></tr>\r\n";
        }
    }

    echo "</table>\r\n";
    // End table

    
 // Footer Template
 require("template/footer.php");
 ?>

That is my code now. I have this in my error log about 1000 times.

Code:

[18-Jan-2011 14:18:25] PHP Warning:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/area51/public_html/RAI/secure/viewpastpayments.php on line 34

**mgraphic** · 01-18-2011, 02:29 PM

I would suggest avoiding putting the sql query inside the while evaluation. It would be better to refer to a resource handler var

**lynxus** · 01-18-2011, 03:01 PM

What happens if you simply change

while($row2 = mysql_fetch_array(mysql_query("SELECT * FROM vars where ID = ".$userID." ORDER BY ID ASC"))

to ( removing the ". and ." )

while($row2 = mysql_fetch_array(mysql_query("SELECT * FROM vars where ID = '$userID' ORDER BY ID ASC"))

**djab** · 01-18-2011, 03:04 PM

I tried this instead.

PHP Code:


			
<?php
 // Connects to your Database 
 require("libraries/db/dbconnect.php"); 
 // Encrypt Page
 require("libraries/auth/membersarea.php");

$username = $_COOKIE['ID_my_site'];
$userID = mysql_result(mysql_query("select ID from users where username = '$username'"), 0);

// New code by Josh
// Gonna annotate this like a boss to keep track

// Open our table up
    echo "<table align=\"center\" width=\"400\">\r\n";
// Configure the columns of the table
    echo "<tr><td></td><td><strong>Unit</strong></td><td><strong>Rate</strong></td></tr>\r\n";
//SQL Query for the first while loop
    $sql1 = mysql_query("SELECT * FROM payroll where ID = ".$userID." and confirmed = '1' ORDER BY PID ASC");
//SQL Query for the second while loop
    $sql2 = mysql_query("SELECT * FROM vars ORDER BY ID ASC");
//If user has no payrolls    
 if (mysql_num_rows($sql1) == 0) {
     echo "No past payments available";
     exit;
     }    
//First while loop
while($row = mysql_fetch_array($sql1))
    {
    // Table is opened. Let's bring in our Timestamp:
    // We'll put this on hold for now - echo "<tr><td>Timestamp</td><td>".$row2['timestamp']."</td><td><i>N/A</i></td></tr>";

    // Bring in our variables now    
    while($row2 = mysql_fetch_array($sql2))
        {
        // Determine variables
        $aliasName = $row2['varAlias'];
        $aliasRate = $row2['varAlias']."_rate";
        // Parse into table
        echo "<tr><td>".$aliasName."</td><td>".$row[$aliasName]."</td><td>".$row[$aliasRate]."</td></tr>\r\n";
        }

    }
    echo "</table>\r\n";
    // End table

    
 // Footer Template
 require("template/footer.php");
 ?>

It does not give me any errors but it only shows the first row of the payroll table when in fact it needs to show every table that matches the $userID and has confirmed set to 1. Any ideas?

**mgraphic** · 01-18-2011, 03:11 PM

You are reusing the same resource $sql2 for mysql_fetch_array() without redoing the query on it. I don't think that is allowed as the resource can only be looped once, I don't think it can be reset.

**djab** · 01-18-2011, 03:13 PM

How would i fix it? What would the code look like?

**mgraphic** · 01-18-2011, 03:16 PM

PHP Code:


			
<?php
 // Connects to your Database 
 require("libraries/db/dbconnect.php"); 
 // Encrypt Page
 require("libraries/auth/membersarea.php");
$username = $_COOKIE['ID_my_site'];
$userID = mysql_result(mysql_query("select ID from users where username = '$username'"), 0);
// New code by Josh
// Gonna annotate this like a boss to keep track
// Open our table up
    echo "<table align=\"center\" width=\"400\">\r\n";
// Configure the columns of the table
    echo "<tr><td></td><td><strong>Unit</strong></td><td><strong>Rate</strong></td></tr>\r\n";
//SQL Query for the first while loop
    $sql1 = mysql_query("SELECT * FROM payroll where ID = ".$userID." and confirmed = '1' ORDER BY PID ASC");
//If user has no payrolls    
 if (mysql_num_rows($sql1) == 0) {
     echo "No past payments available";
     exit;
     }    
//First while loop
while($row = mysql_fetch_array($sql1))
    {
    // Table is opened. Let's bring in our Timestamp:
    // We'll put this on hold for now - echo "<tr><td>Timestamp</td><td>".$row2['timestamp']."</td><td><i>N/A</i></td></tr>";
    // Bring in our variables now    
//SQL Query for the second while loop
    $sql2 = mysql_query("SELECT * FROM vars ORDER BY ID ASC");
    while($row2 = mysql_fetch_array($sql2))
        {
        // Determine variables
        $aliasName = $row2['varAlias'];
        $aliasRate = $row2['varAlias']."_rate";
        // Parse into table
        echo "<tr><td>".$aliasName."</td><td>".$row[$aliasName]."</td><td>".$row[$aliasRate]."</td></tr>\r\n";
        }
    }
    echo "</table>\r\n";
    // End table
    
 // Footer Template
 require("template/footer.php");
 ?>

**dagaul102** · 01-22-2011, 09:35 AM

It could simply be that the host is down

**djab** · 01-22-2011, 09:51 AM

Thanks everyone. I tested it out. Mgrapgic's code works and doesn't give any errors. Thanks everyone again.