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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
When I run my php page, I get this error and do not know what's wrong, can anyone help? If anyone needs more infomation, I'll post the whole code.
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in
H:\Program Files\EasyPHP 2.0b1\www\test\info.php on line 16
<?PHP
$user_name = "root";
$password = "";
$database = "addressbook";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM tb_address_book";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['ID'] . "<BR>";
print $db_field['First_Name'] . "<BR>";
print $db_field['Surname'] . "<BR>";
print $db_field['Address'] . "<BR>";
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
It generally means that you've got an error in your SQL.
$sql = "SELECT * FROM myTable"; // table name only do not add tb
$result = mysql_query($sql);
var_dump($result); // bool(false)
Obviously, false is not a MySQL resource, hence you get that error.
EDIT with the code pasted now:
On the line before your while loop, add this:
if (!$result) {
echo "Error. " . mysql_error();
} else {
while ( ... ) {
...
}
}
Make sure that the tb_address_book table actually exists and that you've connected to the DB properly.
<?PHP
$user_name = "root";
$password = "";
$database = "addressbook";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM tb_address_book";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['ID'] . "<BR>";
print $db_field['First_Name'] . "<BR>";
print $db_field['Surname'] . "<BR>";
print $db_field['Address'] . "<BR>";
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
Related
I'm trying to create an upvote system where, after checking to see if you're logged in and after getting your userid, it checks if the table has your userid with the postid already in it and if it does then it means it was already upvoted. I just want to know what's wrong in my code, this is being used for learning, I don't need any complex thing.
Code:
if (isset($_GET['upvote'])) {
if ($_SESSION["loggedin"] == true) {
$upvoteid = $_GET["upvote"];
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id FROM users WHERE username=".$_SESSION["loggedinusername"];
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$userid = $row["id"];
}
$sql2 = "SELECT userid, postid FROM upvotedposts WHERE userid='".$userid."' AND postid='".$upvoteid."'";
$result2 = $conn->query($sql);
if (!$result2->numrows > 0) {
$sql = "UPDATE posts SET upvotes = upvotes + 1 WHERE id = ".$upvoteid;
if ($conn->query($sql) === TRUE) {
echo "Sucessfully upvoted";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
} else {
echo "Failed;
}
$conn->close();
}
}
When I do click the upvote button, it simply does nothing. The issue here is that as far as I know it looks like it would work but I may be forgetting something that I am unaware of or incorrectly using something.
You are missing a closing "
echo "Failed;
should be:
echo "Failed";
I have a fairly simple PHP page which displays some fields from a database.
For some reason I get:
Notice: Undefined index: entries.id in /var/www/html/originalprices.php on line 24
I can't see whats wrong, could anyone help? Thanks
<?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "zxdb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT entries.id, entries.title, entrytypes.text,entries.original_price, entries.budget_price,labels.name FROM entries,publishers, labels, entrytypes
where entries.entrytype_id = entrytypes.id
and publishers.entry_id = entries.id
and publishers.label_id = labels.id
and labels.id = '1371'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["entries.id"]. " - Name: " . $row["entries.title"]. " " . $row["entrytypes.text"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The Name of the returned field will be "id" not entries.id.
You can check that by using var_dump()
I'm a newbie at PHP. I'm simply trying to access my database and here is the code. I'm using the object oriented style for accessing the database.
It says "Trying to get property of non-object on line 25".
What am I missing?
<?php
$servername = "localhost";
//$username = "";
//$password = "";
$database = "MyPetProject";
$conn = new mysqli ($servername, $database);
if ($conn-> connect_error)
{
die ("Connection failed : " . $conn->connect_error);
}
else
{
echo "Connection Successfull";
}
$sql = "SELECT * FROM User";
$result = $conn->query($sql);
if($result->num_rows > 0)
{
while( $row = $result->fetch_assoc())
{
echo $row["Sr.no"] . " " . $row["Name"] . " " . $row["Online"];
}
}
else
{
echo "0 results";
}
echo ("hello world");
$conn->close();
?>
Edit the following:
while( $row = $result->fetch_assoc($result))
And it should work
Answering my own question here.
I had commented $username and $password.
These are mandatory parameters for mysqli. The rest of the code is perfectly fine.
You should need a username and password for accessing mysql databases .
$mysqli = new mysqli("localhost", "user", "password", "database");
which is the second and third parameters respectively.
You could find all related documentation in this link.
http://php.net/manual/en/mysqli.quickstart.connections.php
I want to have the print statements printed out on the browser. I tried but the code is not printing out. Please assist me with the code.
<?php
$user_name = "root";
$password = "";
$database = "addressbook";
$server = "127.0.0.1";
$db_handle = #mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
print("Connection to the server opened"."<br>");
if ($db_found) {
print "Database found";
$SQL = "SELECT * FROM tb_address_book";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['ID']."<br>";
print $db_field['First_Name']."<br>";
print $db_field['Surname']."<br>";
print $db_field['Address']."<br>";
}
mysql_close($db_handle);
}else {
print "Database not found";
mysql_close($db_handle);
}
?>
I created a members table on my database and entered the username row as user and the password row as password. Then I wrote a script that has to display the password and the username in a database. This is it:
<?PHP
$user_name = "root";
$password = "Hunter123";
$database = "adventure_of_dragons";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM members";
$result = mysql_query($SQL);
while ( $db_field = mysql_fetch_assoc($result) ) {
$id = array($db_field['member_id']); "<BR>";
$username = array($db_field['username']); "<BR>";
$password = array($db_field['password']); "<BR>";
$rank = array($db_field['rank']); "<BR>";
print_r($username);
print_r($password);
}
mysql_close($db_handle);
}
else {
print "Database NOT Found " . $db_handle;
}
?>
but when i run the code it displays this:
Array ( [0] => user ) Array ( [0] => password )
how do I make it display the text like this:
-User -Password
Please help.
That's simple. Just don't make arrays of them in the first place, and use regular echo.
Other bugs in the code
print_r is a debug function (just like var_dump), it is not used for printing out data to user.
Also, this statement: "<BR>"; simply means nothing.
You must echo it for it to have any effect at all.
Another thing is that you've overwritten the DB connection variables in your fetching loop. It's better to use constants for this, like shown below.
Here's your code, fixed
<?php
define("DB_USERNAME", "root");
define("DB_PASSWORD", "Hunter123");
define("DB_DATABASE", "adventure_of_dragons");
define("DB_SERVER", "127.0.0.1");
$db_handle = mysql_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD);
$db_found = mysql_select_db(DB_DATABASE, $db_handle);
if ($db_found || true) {
$SQL = "SELECT * FROM members";
$result = mysql_query($SQL) or die(mysql_error());
while ( $row = mysql_fetch_assoc($result) ) {
$id = $row['member_id'];
$username = $row['username'];
$password = $row['password'];
$rank = $row['rank'];
echo 'ID = ' . $id . '<br>';
echo 'RANK = ' . $rank . '<br>';
echo 'USERNAME = ' . $username . '<br>';
echo 'PASSWORD = ' . $password . '<br><br>';
// two <br>'s, so we get an empty line between users
}
mysql_close($db_handle);
} else {
echo "Database NOT Found " . $db_handle;
}