This my PHP URL for fetching the data from MySQL. I need to make mysqli_fetch_array code saying if filled uid in the app-data table is the same with uid in users table fetch the data from all row in app-data to the uid like every user show his items from app-data table.
$host = "";
$user = "";
$pwd = "";
$db = "";
$con = mysqli_connect($host, $user, $pwd, $db);
if(mysqli_connect_errno($con)) {
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
$sql = "SELECT * FROM app_data ORDER By id";
$result = mysqli_query($con, $sql);
$rows = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$rows[] = $row;
}
mysqli_close($con);
echo json_encode($rows);
I think this is the correct answer:
<?php
$host = "";
$user = "";
$pwd = "";
$db = "";
$con = mysqli_connect($host, $user, $pwd, $db);
if(mysqli_connect_errno($con)) {
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
$sql = "SELECT * FROM app_data WHERE u_id=".$_POST['posted_uid']." ORDER By id";
$result = mysqli_query($con, $sql);
$rows = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$rows[] = $row;
}
mysqli_close($con);
echo json_encode($rows);
Little tip for the future:
Don't search exactly what you want, only search in parts.
Related
I am Using Below Code to First Fetch all 'id' from MySql table.
<?php
$servername = "localhost";
$username = "**";
$password = "**";
$dbname = "**";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$slug = $_GET["category"];
$sql = "SELECT * FROM table WHERE category = '1'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$dealid = $row["dealid"];
}} else {}
$conn->close();
?>
$dealid should return with all id's but it is returning with only 1.
Now below code is to show data with those id's:-
<?php
$num_rec_per_page=52;
mysql_connect('localhost','**','**');
mysql_select_db('**');
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * $num_rec_per_page;
$sql = "SELECT * FROM table2 WHERE id = $dealid ORDER BY id DESC LIMIT $start_from, $num_rec_per_page";
$rs_result = mysql_query ($sql);
while ($row = mysql_fetch_assoc($rs_result)) {
?>
<?php include( $_SERVER['DOCUMENT_ROOT'] . '/includes/dealbox.php' ); ?>
<?php
};
?>
But its only showing 1 data because returning id is only 1. I am not able to understand what the issue is. Any help is appreciable and will gift if someone help me out with this issue.
Dear friend all Id's are set into one array and return that array,
eg:
while ($row = mysql_fetch_assoc($rs_result))
{
$dealid[]= $row['id']; //array creation
}
and
return ($dealid);
<?php
$username = "root";
$password = "password";
$database = "xxxxxx";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('xxxxxx', $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
mysql_close();
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
?>
This is my code i want to display the roll number of the currently logged in user and
when i run this code i get no database selected.
First of all, you should code properly, means database connection and database selection should be on top:
<?php
$username = "root";
$password = "password";
$database = "xxxx";
$link = mysql_connect("localhost", $username, $password);
mysql_select_db('xxxx', $link);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
mysql_close();
?>
Also moved mysql_close(); on last
One other main point was, now mysql_ is deprecated, please use mysqli_
Remove this statement from line number 10
mysql_close();
just remove these two lines before while that will solve ur problem
$result = mysql_query($query) or die(mysql_error());
$rows = array();
Use this code and use mysql_close function in the last.
<?php
$username = "root";
$password = "password";
$database = "xxxxxx";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db($database, $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
mysql_close();
?>
You have closed the connection from MySQL before the mysql_query mysql_close();
try this
<?php
$username = "root";
$password = "password";
$database = "dfsdftwsdgdfgdfsgsdf";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('xxxxxxx', $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
mysql_close();
print_r($rows);
?>
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
The order should be like this
mysql_select_db('meipolytechnic', $link);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('meipolytechnic', $link);
<?php
$username = "username";
$password = "password";
$hostname = "localhost";
$database = "database";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
$selected = mysql_select_db($database,$dbhandle)
or die("Could not select database");
$id = 0;
if(isset($_GET['Day'])){ $id = (int)$_GET['Day']; }
if(!$id){
$query = "SELECT * FROM `TimeTable`";
} else {
$query = "SELECT * FROM `TimeTable` WHERE `Day`='".$id."'";
}
$result = mysql_query($query);
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$rows[] = $r;
}
or die(mysql_error());
print json_encode($rows);
?>
This code worked previously, but has now stopped, and is producing Parse error: syntax error, unexpected T_LOGICAL_OR in /Directory/TimeTable.php on line 27
I am also looking to add more parameters, (eg: Where Day = $id and Year = $Year )
while($r = mysql_fetch_assoc($result)) {
$rows[] = $r;
}
or die(mysql_error());
This is a syntax error, doesn't know what to do with that or die() statement. Change to this:
$result = mysql_query($query);
if (!$result) {
die('Error');
}
while(...) {
}
I have looked up the new mysql functions an have changed to mysqli.
<?php
$username = "username";
$password = "password";
$hostname = "localhost";
$database = "database";
$link = mysqli_connect($hostname, $username, $password, $database);
if(mysqli_connect_errno()){
echo mysqli_connect_error();
}
$id= 0;
if(isset($_GET['Day'])){ $id=(int)$_GET['Day']; }
$year = 0;
if(isset($_GET['Year'])){ $year=(int)$_GET['Year'];}
if(!$id){
$query = "SELECT * FROM TimeTable";
} else {
if (!year) {
$query = "Select * FROM TimeTable";
} else {
$query = "SELECT * FROM TimeTable WHERE Day=$id AND Year=$year";
}
}
$rows = array();
//Perform JSON encode
if($result = mysqli_query($link, $query)){
while($r = mysqli_fetch_assoc($result)){
$rows[] = $r;
}
}
print json_encode($rows);
?>
I am new in PHP so I have to apologize for such a dumb question, but I am not sure how to find the right answer. I should check if my final result is empty (if $sql found anything). If it didnt find I would like to get some notification example "The list is empty". That message will also be visible from Android app when I call the url?
<?php
$host = "";
$user = "";
$pwd = "";
$db = "";
$con = mysqli_connect($host, $user, $pwd, $db);
if(mysqli_connect_errno($con)) {
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
// query the application data
$sql = "SELECT * FROM lista WHERE Grad='".$_GET['grad']."' AND Predmet='".$_GET['predmet']."'";
$result = mysqli_query($con, $sql);
$rows = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$rows[] = $row;
}
mysqli_close($con);
echo json_encode($rows);
If mysqli_num_rows returns 0, you have no records.
$result = mysqli_query($con, $sql);
if (mysqli_num_rows($result) == 0) {
$rows = "no rows found";
} else {
$rows = array();
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$rows[] = $row;
}
}
mysqli_close($con);
echo json_encode($rows);
I recently started updating some code to MySQL improved extension, and have been successful up until this point:
// old code - works
$result = mysql_query($sql);
if(mysql_num_rows($result) == 1){
$row = mysql_fetch_array($result);
echo $row['data'];
}
// new code - doesn't work
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]");
if($result->num_rows == 1) {
$row = $result->fetch_array();
echo $row['data'];
}
As shown I am trying to use the object oriented style.
I get no mysqli error, and vardump says no data... but there definitely is data in the db table.
Try this:
<?php
// procedural style
$host = "host";
$user = "user";
$password = "password";
$database = "db";
$link = mysqli_connect($host, $user, $password, $database);
IF(!$link){
echo ('unable to connect to database');
}
ELSE {
$sql = "SELECT * FROM data_table LIMIT 1";
$result = mysqli_query($link,$sql);
if(mysqli_num_rows($result) == 1){
$row = mysqli_fetch_array($result, MYSQLI_BOTH);
echo $row['data'];
}
}
mysqli_close($link);
// OOP style
$mysqli = new mysqli($host,$user, $password, $database);
$sql = "SELECT * FROM data_table LIMIT 1";
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
if($result->num_rows == 1) {
$row = $result->fetch_array();
echo $row['data'];
}
$mysqli->close() ;
// In the OOP style if you want more than one row. Or if you query contains more rows.
$mysqli = new mysqli($host,$user, $password, $database);
$sql = "SELECT * FROM data_table";
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
while($row = $result->fetch_array()) {
echo $row['data']."<br>";
}
$mysqli->close() ;
?>
As it was said, you're not checking for the errors.
Run all your queries this way
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]");
if no errors displayed and var dumps are saying no data - then the answer is simple: your query returned no data. Check query and data in the table.
In PHP v 5.2 mysqli::num_rows is not set before fetching data rows from the query result:
$mysqli = new mysqli($host,$user, $password, $database);
if ($mysqli->connect_errno) {
trigger_error(sprintf(
'Cannot connect to database. Error %s (%s)',
$mysqli->connect_error,
$mysqli->connect_errno
));
}
$sql = "SELECT * FROM data_table";
$result = $mysqli->query($sql);
// a SELECT query will generate a mysqli_result
if ($result instanceof mysqli_result) {
$rows = array();
while($row = $result->fetch_assoc()) {
$rows[] = $row;
}
$num_rows = $result->num_rows; // or just count($rows);
$result->close();
// do something with $rows and $num_rows
} else {
//$result will be a boolean
}
$mysqli->close() ;