I have query, and i have database. When I query in dbforge(mysql) i got result. When from php, i got nothing. Can't understand why? If I write query simple like "select * from table where id = 1", its working.
Below my code:
<?php
$conn = mysqli_connect("localhost", "username", "pass", "db");
ini_set('max_execution_time', 300);
if (mysqli_connect_errno())
{
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
mysqli_query($conn, "SET CHARACTER SET utf8 ");
$result = mysqli_query($conn, "Select distinct level_two FROM reportls._report_sales_report Where level_one = 'Музыка'");
if($result == false)
{
die('INVALID QUERY: '.mysqli_error($conn));
echo "error";
}
while ($data = mysqli_fetch_row($result))
{
if(!empty($data[0]))
$sendBack = $sendBack."<option value=\"$data[0]\">$data[0]</option>";
}
echo ($sendBack);
$sendBack = '';
mysqli_free_result($result);
?>
This code return nothing. But if I paste my query into my dbforge, it returns rows.
Could you help me, or write some hits? What i wrote not correct?
Related
I am making a status update page for practicing PHP and MySQL.
I am currently trying to echo the saved status on the page.
The statuses are saved in a row called "Status" and they are all given a certain ID in a row called "statusID."
The problem I am having is which fetch I want to use because converting it into a string using (string)$var doesn't work. ($var is an example).
Also, the $idNum variable is something for later use, shouldn't have anything to do with this.
Here is the code: (Obviously the first variables are censored so none tries to connect to the database, the connection working in the actual code.)
The problem lies in the $fetchRes I believe.
<?php
$idNum = 1;
$servername = "censored";
$username = "censored";
$password = "censored";
$db_name = "censored";
$conn = mysqli_connect($servername, $username, $password, $db_name);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Status FROM SavedStatuses WHERE statusID=1;";
$statusQuery = mysqli_query($conn, $sql);
$fetchRes = mysqli_fetch_assoc($statusQuery);
if($conn->query($sql) == TRUE)
{
echo $fetchRes;
} else {
echo "Failed to retrieve status, error: " . $conn->error;
}
?>
As I mentioned in comments, you are querying twice and not looping over (successful) results.
A "loop" is to use either a while or a foreach. I used a while loop for this example.
From the "official" manual:
http://php.net/manual/en/mysqli-result.fetch-assoc.php
Example:
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}
/* free result set */
$result->free();
}
/* close connection */
$mysqli->close();
?>
So in your case, your code would read as:
Sidenote: Status and status are two different animals, so make sure the letter case matches (in the loop).
<?php
$mysqli = new mysqli("censored", "censored", "censored", "censored");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$query = "SELECT Status FROM SavedStatuses WHERE statusID=1";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
echo $row["Status"];
}
/* free result set */
$result->free();
}else{
echo "The query failed: " . mysqli_error($mysqli);
}
/* close connection */
$mysqli->close();
?>
I'm trying to list all my databases. But I only return information schema and one other table.. I checked my user settings/privileges in mysql and I have access to everything.. How can I return all databases
here is the code i used:
$set = mysql_query('SHOW DATABASES;');
$dbs = array();
while($db = mysql_fetch_row($set)) $dbs[] = $db[0]; echo implode('<br/>', $dbs);
As pointed out in the comments, you really should start using mysqli instead of mysql.
This should solve your problem though:
<?php
$link = mysqli_connect("localhost", "mysql_username", "mysql_password");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$res = mysqli_query($link, "SHOW DATABASES");
while ($row = mysqli_fetch_assoc($res)) {
var_dump($row['Database']);
}
I got this error while adding this code. Would appreciate some help. It's for a CS jackpot site.
$sitename = "website.com"; // YOUR DOMAIN
$link = mysql_connect("localhost", "db_user", "db_pass"); // MYSQL , LOCALHOOST , USERNAME , PASSWORD
$db_selected = mysql_select_db('db_name', $link); // MYSQL DATABASE
mysql_query("SET NAMES utf8");
function fetchinfo($rowname,$tablename,$finder,$findervalue) {
if($finder == "1") $result = mysql_query("SELECT $rowname FROM $tablename");
else $result = mysql_query("SELECT $rowname FROM $tablename WHERE `$finder`='$findervalue'");
while($row = mysql_fetch_assoc($query))
return $row[$rowname];
}
Some tips:
Use mysqli better than mysql
Split the vars in the query, like "SELECT ".$rowname." FROM ".$tablename;
Hope this help...
<?php
//Open a new connection to the MySQL server
$mysqli = new mysqli('host','username','password','database_name');
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}
//MySqli Select Query
$result = $mysqli->query("SELECT id, product_name FROM products");
while($row = $results->fetch_assoc()) {
echo $row["id"].' - '.$row["product_name"].'<br>';
}
$results->free();
$mysqli->close();
I'm wondering why it is not working for mysqli eventhou mysql_num_row is working.
if (mysql_num_rows($rows) > 0) {
echo "<p>That name has been taken </p>";
}
That is mysql. But, Im trying convert it to mysqli.
if (mysqli_num_rows($rows) > 0) {
echo "<p>That name has been taken </p>";
}
It supposed to be displayed on the screen but it's not. And there is nothing error message displayed. Or am I missing something? Any ideas?
First, a quick tutorial on some of the differences between mysql_* and mysqli_* functions.
In mysql_* you would have 3 parameters for your DB connection, then have a seperate line for your DB selection.
For example:
$db = mysql_connect("host","username", "password");
$db_selected = mysql_select_db('db_name', $db);
if (!$db_selected) {
die ('Can\'t use this : ' . mysql_error());
}
Your query would come first, followed by your DB connection.
For example:
mysql_query($query,$db);
But in mysqli_* things have changed including the parameters location. You now put all 4 parameters, for example (if you haven't done so yet):
$db = new mysqli("host","username", "password", "db_name");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
then the DB connection would come first, followed by the query instead of the other way around:
mysqli_query($db,$query);
A sample query:
$email = mysqli_real_escape_string($db,$_POST['email']);
$query = mysqli_query($db, "SELECT * FROM table_name WHERE email='".$email."'");
if(mysqli_num_rows($query) > 0){
echo "email already exists";
}else{
$sql="INSERT INTO table_name (email) VALUES ('$email')";
if (!mysqli_query($db,$sql))
{
die('Error: ' . mysqli_error($db));
}
}
You could try this code:
$connect = new mysqli("localhost", "user", "password", "database");
$query = "SELECT name FROM table";
$statement= mysqli_prepare($connect, $query)
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
if (mysqli_stmt_num_rows($statement) > 0) {
echo "<p>That name has been taken </p>";
}
mysqli_stmt_close($statement);
mysqli_close($connect);
I couldn't figure out why it has some problem like this. The error displayed mysqli_fetch_assoc() expects parameter 1 to be mysqli_result.
My code is like this. I don't know which one is wrong.
Have a look at my con.php.
<?php
$conn = mysqli_connect('localhost', 'username', 'password');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
?>
and this is my page.php
<?php
include 'con.php';
$link = mysqli_connect('localhost', 'username', 'password', 'username') or die("Error " . mysqli_error($link));
$query = " SELECT thread_id, thread_name, thread_date
FROM forum_thread
ORDER BY thread_date";
$result = mysqli_query($link,$query);
while ($row = mysqli_fetch_assoc($result)) {
$thread_id = $row ['thread_id'];
$thread_name = $row['thread_name'];
$thread_date = $row['thread_date'];
echo "$thread_id, $thread_name, $thread_date";
Any ideas? Appreciate any answers from you. Cheers!
I think I figured it out:
Change $result = mysqli_query($link,$query);
to $result = mysqli_query($conn,$query); since $conn is your DB connection, not $link
as per what you posted above
<?php
$conn = mysqli_connect('localhost', 'username', 'password', 'database_name');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
?>
and remove:
$link = mysqli_connect('localhost', 'username', 'password', 'username') or die("Error " . mysqli_error($link));
since you're already loading your DB con with include 'con.php';
Or this way:
<?php
$conn = mysqli_connect('localhost', 'username', 'password', 'database_name') or die("Error " . mysqli_error($conn));
$query = " SELECT thread_id, thread_name, thread_date
FROM forum_thread
ORDER BY thread_date";
$result = mysqli_query($conn,$query);
while ($row = mysqli_fetch_assoc($result)) {
$thread_id = $row ['thread_id'];
$thread_name = $row['thread_name'];
$thread_date = $row['thread_date'];
echo "$thread_id, $thread_name, $thread_date";
The error message indicates the query did fail for some reason. Instead of just using $result test to see if it's valid and if not output the error to the screen or a log file:
if ($result) {
// continue processing $result
}
else {
echo mysqli_error($link);
}
Edit:
The 4th parameter in the mysqli_connect should be the name of the database, "abc" or whatever.
Try to use following code
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli- >connect_error;
}
Here when you try to get connection object itself you can identify the problem if something goes wrong.