PHP - Mysql_fetch_row()

Hello!

I've been trying to work on a small project of mine but I keep encountering this:
Quote

Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\book-suggestion\test.php on line 21


In test.php, this is the only PHP code I seem to have and I haven't found anything wrong with it. For additional assurance, I copy-pasted the SQL statement directly from phpmyAdmin but it continues to give me an error so I know for a fact that the SQL query works and there is a result for such a query.

Code: [Select]
$db = mysql_connect("127.0.0.1",$db_user,$db_password, $db_name);
if(!isset($db)) {
echo 'Error connecting to database...';
} else {
echo 'Success.';
}
$query = mysql_query("SELECT * FROM `book-listing` WHERE 1");
$row = mysql_fetch_row($query);
Any help with this?
I've tried var_dump() and it returns boolean(false). Can't seem to figure out what the mistake is.

I have a nagging question about loops when using mysql_fetch_row().

In other OOP languages whenever you use a for loop you have to use the variable you are incrementing to point to the right element in the array, but when you use mysql_fetch_row() you don't need to.

Why is that?

For example:
in js/php/as you write:
Code: [Select]
for(i=0; i<length; ++i) { doSomething [i]row; }
but when calling mysql_fetch_row you just write:
Code: [Select]
for(i=0; i<length; ++i) { doSomething row; }
I'm just curious about how the loop finds its way to next row. Is it built in the mysql_fetch_row function already?

Hi. I have a code and I'm trying to see if an e-mail address is stored in our database.

Code: [Select]
$query="SELECT email FROM users WHERE email='$email'";
    $row=mysql_query($query);
    $result=mysql_fetch_row($row);
    if ($email==$result[0])
    {
        return 1;
    }
    else
    {
        return 0;
    }
I get this error:
Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given.
On this line: $result=mysql_fetch_row($row);

Anyone knows why's broken? Can I do this in a better way ?

Hi

I am looking to be able to pass an array to mysql_fetch_row() not sure where i am going wrong.

works fine like this.

Code: [Select]
mysql_select_db($DB_NAME);

$new = mysql_query("SELECT * FROM {$DB_TABLE}");

if (!$new) {
   echo 'MySQL Error: ' . mysql_error();
   exit;
}

$row = mysql_fetch_array($new);

while ($row = mysql_fetch_array($new)) {

 echo $row['username'] . $DELIMITER . $row['firstname'] . $DELIMITER . $row['lastname'] . $DELIMITER . $row['password'] . $DELIMITER. "example@gmail.com" . $DELIMITER . "author";
  echo "<br />";
 
}

But really need too be able to do it this way.

Code: [Select]
mysql_select_db($DB_NAME);

$new = mysql_query("SELECT * FROM {$DB_TABLE}");

if (!$new) {
   echo 'MySQL Error: ' . mysql_error();
   exit;
}

$row = mysql_fetch_array($new);

echo "<pre>";
//print_r($row);
echo "</pre>";

$sam = array('username','firstname','lastname');

while ($row = mysql_fetch_row($new)) {

 echo $row[$sam] . $DELIMITER;
  echo "<br />";
 
}
So it will just loop through the array so i can keep adding to it by just adding an extra parameter to the array.

Any Help Stuck????

this Nofications appear..

Notice: Undefined index: id in C:\wamp\www\WAR\up3\view.php on line 4

Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\wamp\www\WAR\up3\view.php on line 9
invalid id

this is my code.

<?php
    require ('dbconnect.php'); // Connect to database

    $id = $_GET['id']; // ID of entry you wish to view.  To use this enter "view.php?id=x" where x is the entry you wish to view.

    $query = "SELECT data, filetype FROM uploads where uploadid=$id"; //Find the file, pull the filecontents and the filetype
    $result = MYSQL_QUERY($query);    // run the query

    if($row=mysql_fetch_row($result)) // pull the first row of the result into an array(there will only be one)
    {
        $data = $row[0];    // First bit is the data
        $type = $row[1];    // second is the filename

        Header( "Content-type: $type"); // Send the header of the approptiate file type, if it's' a image you want it to show as one
        print $data; // Send the data.
    }
    else // the id was invalid
    {
        echo "invalid id";
    }
?>