<?php
include_once "Database.php";
$server="localhost";
$db="fantasy cricket";
$user="root";
$password="";
$db=new Database($server,$db,$user,$password);
$name=$_POST["name"];
$country=$_POST["country"];
$db->removePlayerFromTeam($name,$country);
This is the code that I am using to delete a row ,but the deletion is not working and database remains same .The removePlayerFromTeam($name,$country) is given below.When i use the query "delete from demo" ,that works and all rows get deleted,but when specifying column and value its not working.
By echoing i get the desired values passed as parameter to the function.Please help me out what is going wrong.
Thanks in advance.
public function removePlayerFromTeam($n,$cn)
{
echo "$n"."<br>"."$cn"."<br>";
$sql = sprintf("DELETE FROM demo WHERE Pname='%s' and country='%s'",$n,$cn);
$v=$this->connection->exec($sql);
if($v>0)echo "ok"."<br>";
else echo "wrong"."<br>";
}
Related
I am trying to delete an item from my database but it isnt working.
I thought I had it working but it was deleting the first item in the database but not the item selected.
Here is what I have.
A link to delete.php then I have this for delete.php
<?php
ob_start();
include_once('../mysql_connect.php');
// contact to database
$host = "localhost";
$username = "admin";
$password = "password";
$database="database";
$tbl_name="new_equip";
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$database")or die("cannot select DB");
// get value of id that sent from address bar
$id=$_GET['id'];
// Delete data in mysql from row that has this id
$sql="DELETE FROM $tbl_name WHERE id='$id'";
$result=mysql_query($sql);
// if successfully deleted
if($result){
echo "Deleted Successfully";
echo "<BR>";
echo "<a href='inventory.php'>Back to main page</a>";
}
else {
echo "ERROR";
}
?>
I know this is probably something simple and I have been searching and trying everything I can find, but I cannot seem to get it working. I believe the delete.php link needs to have the item number in it. Here is what the link is
"delete.php?id=<?php echo $eid; ?>"
I also have this on the top of the delete.php
<?php
$eid = (int) $_GET['id'];
if ($eid < 1)
?>
Do not put the GET directly in your variable that might cause SQL Injections
Do not use mysql... use mysqli instead!
Format your code more
Than back to the main problem:
Echo your $id to see if it is the correct one ;) I could not see some other problem at your code.
I have been trying to figure this out for hours, I have created a database ( MySql/PHPMyadmin) and i am trying to get user input stored to be able to call back up, however the info is not making it/ saving it to the database, everything shows up okay except this part of code:
$registered = mysqli_affected_rows ($dbc);
echo $registered. "Row is affected";
when run gives me a display of -1 row, I believe this to be a big part of the problem as everything else seems to work okay. I am a complete beginner so could you guys tell me how the best way of debugging this is.
$dbc = $dbc = mysqli_connect ($hostname, $username, $password, $dbname) OR die("Could not Connect");
To input the data to the db i have the following:
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$comments = $_POST ['Comments'];
if (!empty ($comments)){
include ('mysqldb.php');
mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");
$registered = mysqli_affected_rows ($dbc);
echo $registered. "Row is affected";
}else {
echo "Nothing Submitted";
}
}
The line:
mysqli_query ($dbc,"INSERT INTO 'User-Comments' (Comments) VALUES ('$comments')");
should be:
mysqli_query ($dbc,"INSERT INTO `User-Comments` (Comments) VALUES ('$comments')");
Notice the change in the apostrophe character ( ` ) around your table name.
An excerpt from the documentation for function mysqli_stmt_affected_rows(): -1 indicates that the query has returned an error.
You should check the value returned by mysqli_query(). If it returns FALSE then you can get details about the reason (error message) by using function mysqli_error().
I have a table with two value.
ID, Building(is the name of Building)
i write a code with jquery to insert or Update the name of Building (i take the ID value from list1 and the new name from text_build)
function saveBuilding()
{
alert(document.getElementById("list1").value)
alert(document.getElementById("text_build").value)
$.get("saveBuilding.php",{ID:document.getElementById("list1").value, val:document.getElementById("text_build").value},
function(ret) { alert(ret);});
}
where my saveBuilding is:
<?php
$idbuilding=$_GET['ID'];
$name=$_GET['val'];
require_once '../../../dbconnection.php';
$db_server = mysql_connect($db_hostname, $db_username, $db_password);
if (!db_server) die("Unable to connection_aborted to MySQL: " . mysql_error());
mysql_select_db($db_database) or die ("Unable to connection_aborted to MySQL: " . mysql_error());
$query = "UPDATE Building SET Name = '$name' WHERE ID_Building = '$idbuilding';";
$result = mysql_query($query);
if (mysql_error()) {
echo mysql_error();
}
mysql_close();
?>
Now, if i update the value with a new value, it works, if i update the value with a value already used previously, it said that the query is successfully but it dont change nothing.
I try to insert new value by changing the query. and the result is the same.
i also try to add this value directly from mysql and it works!
so which is the problem in my code?
Thanks
MySQL is working as designed (and other rdbms). If you try to change a value to its current value, no error is thrown but nothing happens.
You just need to run
SELECT Name FROM your_table WHERE idbuilding=$id
then compare Name to what is already stored. If it's different, run your UPDATE; if not don't do anything because no action is required.
I solved the problem..
first i verified that the php code works.
and then i verified that the problem was the js.
i dont know why but changing the jquery function $.get in $.post now it work!!
I'm having a problem trying to truncate the 'requestID' field from my requests table.
This is my code.
<?php
include 'mysql_connect.php';
USE fypmysqldb;
TRUNCATE TABLE requestID;
echo "Request ID table has been truncated";
?>
I'm using server side scripting so no idea what error is coming back.
Anyone got an idea?
You aren't executing queries, you're just putting SQL code inside PHP which is invalid. This assumes you are using the mysql_*() api (which I kind of suspect after viewing one of your earlier questions), but can be adjusted if you are using MySQLi or PDO.
// Assuming a successful connection was made in this inclusion:
include 'mysql_connect.php';
// Select the database
mysql_select_db('fypmysqldb');
// Execute the query.
$result = mysql_query('TRUNCATE TABLE requestID');
if ($result) {
echo "Request ID table has been truncated";
}
else echo "Something went wrong: " . mysql_error();
Take a look at the function mysql_query which performs the query execution. The code to execute a query should look something like this.
$link = mysql_connect('host', 'username', 'password') or die(mysql_error());
mysql_select_db("fypmysqldb", $link) or die(mysql_error());
mysql_query("TRUNCATE TABLE requestID", $link) or die(mysql_error());
mysql_close($link);
I want to retrieve or output data in the database but I kept on getting the error called "Resource ID".
Here is my code:
<?php
$host="localhost";
$username="root";
$password ="123192";
$db_name = "customers";
//Connecting to your Host
mysql_connect("$host","$username","$password") or die("Failed To Connect The server");
//Selecting your Database
mysql_select_db("$db_name") or die("Failed To Select The DB");
$name = $_REQUEST['customerName'];
echo 'WELCOME! <b>'.$name.'</b> We hope that you\'ll Enjoy your stay ';
$sql="SELECT Name FROM `people` WHERE id =2 && Name = 'Kyel'";
$rs=mysql_query($sql);
echo "$rs";
?>
If I need improvement regarding my code kindly tell me.
mysql_query() returns a resource. The to string (implicitly triggered by using echo to output it) of that is Resource ID # followed by the id.
A resource in PHP is only supposed to be used with other PHP functions. This includes but is not limited to file, curl, ftp handles, etc.
I could tell you to..
(a) use mysql_fetch_array() (or similar) or
(b) use PDO.
The latter is by far much better advice.
Try this instead of the echo statement:
$array = mysql_fetch_assoc($rs);
var_dump ($array);