I am trying to retrieve multiple rows from the database and process it for response as service using php code, but I wouldn't retreive all values, only first row from the table is displaying, how can I make dis to work?
Here is my code:
$servername="localhost";
$username="root";
$conn= mysql_connect($servername,$username)or die(mysql_error());
mysql_select_db("testing",$conn);
$sql="insert into login (src,dest)values('$from','$tona')";
$result=mysql_query($sql,$conn) or die(mysql_error());
$res = mysql_query("SELECT * FROM login");
$numrows = mysql_num_rows($res);
setcookie('a',$numrows);
Note the use of _mysql is discouraged for new development ... please read this on selecting a new API
This is pretty basic but you need to loop the returned result like so :
$res = mysql_query("SELECT * FROM login");
while ($row = mysql_fetch_array($res, MYSQL_BOTH)) {
// your columns are accessible using
$row['columnname'];
// or
$row[columnnumber];
}
Docs for mysql_fetch_array are here
mysql_query returns a resource on success or false on failure
do like this
<?php
$servername="localhost";
$username="root";
$conn= mysql_connect($servername,$username)or die(mysql_error());
mysql_select_db("testing",$conn);
$sql="insert into login (src,dest)values('$from','$tona')";
$result=mysql_query($sql,$conn) or die(mysql_error());
$res = mysql_query("SELECT * FROM login");
$numrows = mysql_num_rows($res);
while ($result=mysql_fetch_array($res)){
echo $result['src']."<br/>";
}
setcookie('a',$numrows);
mysql_* is deprecated
use
$res = mysql_query("SELECT * FROM login");
$numrows = mysql_num_rows($res);
while($row = mysql_fetch_assoc($res))
{
print_r($row);
}
I see that you using mysql instead of mysqli. Try changing to mysqli instead, because this will become the new standard. As of version PHP 5.5.0 mysql will become deprecated.
Related
I have looked for an answer for ages now, lots of similar questions but found no solutions yet...
Anyway,
all I am trying to do is get the id of a user from the database using a mysqli_query, the query seems to work when I use it in phpmyadmin but doesn't when I use it as part of a php script.
$username = "bob";
$db = mysqli_connect("localhost", "username", "password", "user_data");
$sql1 = "select id from user_information where username='$username'";
$id = mysqli_query($db, $sql1) or die(mysql_error());
echo $id;
The database connection works fine, I am able to input data through php.
Any suggestions? (Anyone's help is greatly appreciated).
you can't print the result from mysqli_query, it is mysqli_resource and for dumping the error you need to change mysql_error() to mysqli_error()
$username = "bob";
$db = mysqli_connect("localhost", "username", "password", "user_data");
$sql1 = "select id from user_information where username='$username'";
$result = mysqli_query($db, $sql1) or die(mysqli_error());
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo $row['id'].'<br>';
}
I am having trouble selecting my data from a database and displaying it. I have looked at tutorials and i still get the same error. Some help would be appreciated. The error i am getting is couldnt fetch result.
$sql = "SELECT * FROM data";
$result = mysql_query($sql) or die("couldnt fetch result");
if($result > 0){
while ($rows = mysql_fetch_array($result)){
$username = $rows['username'];
echo $username;
}
}
Just do that (assuming got it right connecting to DB, first thing to check !)
$sql = "SELECT * FROM `data`"; // data is a reserved keyword, protect it !!!
$result = mysql_query($sql) or die("couldnt fetch result"); // potentially diying here
if($result){
while ($row = mysql_fetch_assoc($result)){
$username = $row['username'];
echo $username;
}
}
If what you're getting is literally 'couldnt fetch result' it means your mysql_query() fails, and die statement takes over. Check your database connection.
I think the very simple problem is that you check if the $result is greater then 0. But you get an resource.
$conn = mysql_connect.......
$sql = "SELECT * FROM data";
$result = mysql_query($sql) or die("couldnt fetch result");
if($result){
while ($rows = mysql_fetch_array($result)){
$username = $rows['username'];
echo $username;
}
}
And if you see your die statement you have an error in your SQL Syntax. Its very short but its possible that your table doesn't exist in that database you're trying to connect. I hope you have a connect before and its not your complete code.
You use the old mysql functions. Its better to use MySQLi or PDO.
And DATA is a reserved keyword its possible that you get problems if you use it in your query. Rename your table in prefix_data for example.
https://dev.mysql.com/doc/refman/5.7/en/keywords.html
Yes it connects to the database, everything else works fine. I cant seem to pull the pass from the db its showing no returned echo
<?php
$username="test";
include("db.php");
$con=mysql_connect($server, $db_user, $db_pwd) //connect to the database server
or die ("Could not connect to mysql because ".mysql_error());
mysql_select_db($db_name) //select the database
or die ("Could not select to mysql because ".mysql_error());
$query="select password from ".$table_name." where username='$username'";
$result=mysql_query($query,$con) or die('error');
while ($row = mysql_fetch_assoc($result));
$un_pass_s1=$row['password'];
echo $un_pass_s1;
?>
while ($row = mysql_fetch_assoc($result)); loops until $row is false. The loop body is a single empty statement, ;. You need to put your code which accesses $row inside the loop, not after it.
$sql=mysql_query("select password from ".$table_name." where username='$username'");
while($row=mysql_fetch_array($sql))
{
$un_pass_s1=$row['password'];
}
echo "value=".$un_pass_s1;
I am trying to display a record from my database, however the page appears blank and doesn't display the data I am expecting. The code follows below:
<?php
$mysqli = new mysqli(localhost, root, USERPASS, DBNAME);
$query = "SELECT * FROM usertable WHERE userID= '" . $_SESSION["sess_uid"] . "'";
$result = mysqli_query($mysqli, $query);
$row = mysqli_fetch_row($result);
echo $row['userQuestion'];
?>
Any help would be appreciated.
Thanks
<?php
// there need to be strings arguments here
$mysqli = new mysqli('localhost', 'root', USERPASS, DBNAME);
// sql injection friendly query
$query = "SELECT * FROM `usertable`
WHERE `userID`='{$_SESSION["sess_uid"]}' LIMIT 1;";
// do we have a result
if($result = mysqli_query($mysqli, $query)){
// fetch a single row
if($row = mysqli_fetch_row($result)){
// print the record
var_dump($row);
}
}
?>
You need to wrap 'localhost' and 'root' as strings.
mysqli_fetch_row returns a numerical array.
You can print the content of the record using var_dump or use mysqli_fetch_assoc instead.
Here is my error:
Warning: mysql_query() expects parameter 2 to be resource, null given...
This refers to line 23 of my code which is:
$result = mysql_query($sql, $connection)
My entire query code looks like this:
$query = "SELECT * from users WHERE userid='".intval( $_SESSION['SESS_USERID'] )."'";
$result = mysql_query($query, $connection)
or die ("Couldn't perform query $query <br />".mysql_error());
$row = mysql_fetch_array($result);
I don't have a clue what has happpened here. All I wanted to do was to have the value of the users 'fullname' displayed in the header section of my web page. So I am outputting this code immediately after to try and achieve this:
echo 'Hello '; echo $row['fullname'];
Before this change, I had it working perfectly, where the session variable of fullname was echoed $_SESSION['SESS_NAME']. However, because my user can update their information (including their name), I wanted the name displayed in the header to be updated accordingly, and not displaying the session value.
Your $connection variable is NULL that's what your error message is referring to.
Reason being is that you have not called mysql_connect. Once called it will assign you a resource where you can set it to the $connection variable, thus being non-null.
As an example:
$connection = mysql_connect('localhost', 'mysql_user', 'mysql_password');
// now $connection has a resource that you can pass to mysql_query
$query = "SELECT * from users WHERE userid='".
intval( $_SESSION['SESS_USERID'] )."'";
$result = mysql_query($query, $connection)
include the mysql connections on your class file, for example:
connections/mysql.php
<?
$hostname_MySQL = "localhost";
$database_MySQL = "database";
$username_MySQL = "user";
$password_MySQL = "password";
$MySQL = mysql_pconnect($hostname_MySQL, $username_MySQL, $password_MySQL) or trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_MySQL,$MySQL);
?>
class.php
<?
include "Connections/MySQL.php";
class utils {
public function myFunction()
{
global $MySQL;
$sql = "select * from table";
$rs = mysql_query($sql, $MySQL) or die(mysql_error());
$filas = mysql_fetch_assoc($rs);
$totalFilas = mysql_num_rows($rs);
...
}
}
?>
You have two ways of doing this, you need to use mysql_connect to connect to your database, you can pass this to mysql_query if you desire, if you don't pass anything to mysql_query PHP uses the last link opened from mysql_connect
$conn = mysql_connect("localhost", "mysql_user", "mysql_password");
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1";
$result = mysql_query($sql);
Have you connected to your database? If so please show this code too.
For now, just try removing the $connection variable, like this:
$result = mysql_query($query);
And see where that gets you.
$connection is assigned the value of the database connection resource id. You don't have that in your script, so the value of $connection is NULL, and that is why you are getting the error. You need to connect to the database before using mysql_query(). You should be okay after that.
You need to do:
$connection=mysql_connect('host','user','pass');
if($connection === false) {
echo "Error in connection mysql_error()";
}