I am wondering what mistakes has been made in this pretty simple update statement using old version of PHP. If I echo the statement it says update statement is getting form submitted properly.
Here is the code:
<?php
echo $q = "UPDATE notice SET FromDate = $notice_fromdate, ToDate = $notice_todate, VacType ='$notice_vactype',NoticeDetail ='$notice_detail',Status ='$notice_status' WHERE ID=$id";
if (mysql_query($link, $q)) {
echo "Record updated successfully";
} else {
echo "<h3>Error updating record</h3>". mysql_error($link)."-". mysql_errno($link). "\n";
}
?>
and the output returns this
UPDATE notice SET FromDate = 2017-01-08, ToDate = 2017-01-09, VacType ='May Day',NoticeDetail ='Testing',Status ='Enabled' WHERE ID=3
Error updating record
-0
I know its a pretty simple thing, I guess I have not made any mistake in the update statement but instead it is showing Error update record. I copied the output SQL statement and run at phpmyadmin, it has worked properly. It would be nice if you can help me. Thank in advance
Note: Clients website built on old version of PHP, I know that few functions got deprecated so it would be better if you do not discuss or criticize about the version.
Apply quotes to dates it will work
<?php
echo $q = "UPDATE notice SET FromDate = '$notice_fromdate', ToDate = '$notice_todate', VacType ='$notice_vactype',NoticeDetail ='$notice_detail',Status ='$notice_status' WHERE ID=$id";
if (mysql_query($link, $q)) {
echo "Record updated successfully";
} else {
echo "<h3>Error updating record</h3>". mysql_error($link)."-". mysql_errno($link). "\n";
}
?>
Related
<?php
header('Content-Type: application/json');
include '../../connection.php';
$id_e = $_POST['id'];
$location_e = $_POST['location'];
$state_e = $_POST['state'];
$notes_e = $_POST['notes'];
$sql="UPDATE chassis SET location='%$location_e%', state='%$state_e%', notes='%$notes_e%' WHERE id='%$id_e%'";
$query = mysqli_query($db,$sql);
if (!$query) {
echo json_encode(["message"=>mysqli_error($db)]);
}else {
echo json_encode(["message"=>"Success"]);
}
/* I used this code below to check if the data sent by ajax is received
switch($state_e){
case "New":
echo json_encode(["message"=>"hi"]);
break;
default:
echo json_encode(["message"=>"nope"]);
break;
}
}*/
?>
Hi, I'm not sure what the problem with my code is but the UPDATE statement isn't updating my database. I have an ajax jquery posting data which is received by this PHP file. I know that the data is being received by the PHP file because I tried a the piece of code to check it. Also, the query is returning success. I think the problem lies with the UPDATE statement however, I've tried different variations of the UPDATE statement like:
$sql="UPDATE chassis SET location='".$_POST['location']."', state='".$_POST['state']."', notes='".$_POST['notes']."' WHERE id='".$_POST['id']."";
It still doesn't work. I also checked the privileges on the SQL and I have All privileges. Please help, I'm a high school student and this is for my end of year project. Cheers :)
You should be using prepared statements. Also i do not see the reason for adding the % next to the variables as per my comment. Also you should be getting the real error and not just outputting Error
$sql="UPDATE chassis SET location=?, state=?, notes=? WHERE id=?";
$result = $db->prepare($sql);
$result->bind_param('sssi', $location_e, $state_e, $notes_e, $id_e);
echo $result->execute() === true ? 'Success' : 'Failed: '.$result->error;
I am using this code to update rows and the value of $result is coming true. The if condition is executing. But the data in the table isnt updated. So, No errors yet the updation isnt happening.
What is the problem? Any help appriecated. :) Happy Diwali
for($i=0;$i<$len-1;$i++){
$sqlquery="UPDATE items_data SET item_price='".$price[$i]."' WHERE item_name='".$items[$i]."'";
$result=mysqli_query($connection,$sqlquery);
if($result){
echo "Working";
}
else
{
echo mysqli_error($connection);
}
}
<?php
include "conn.php";
include "session.php";
$name_enterd=$_GET['Name'];
$sql = "DELETE FROM myDB.Mynew WHERE firstname='$name_enterd' OR lastname='$name_enterd'";
echo "<br>";
$result=$conn->query($sql);
if($result==1)
{
echo "<br> Data deleted successfully";
}
else
{
echo "No Data Found<br>";
}
?>
when I run this code 1st time it works properly by deleting the data. But when i run it again it still gives me the same answer" Data Deleted Successfully" even there is no data with that value exists.
i.e $result still gets value1.
Your code should look more like this:
<?php
include "conn.php";
include "session.php";
$name_enterd=$_GET['Name'];
$sql = "DELETE FROM myDB.Mynew WHERE firstname='$name_enterd' OR lastname='$name_enterd'";
echo "<br>";
$result=$conn->query($sql);
if($result->rowCount() > 0)
{
echo "<br> Data deleted successfully";
}
else
{
echo "No Data Found<br>";
}
?>
Specifying rowCount gives you just the number of rows affected by the query
Even when the query only affects 0 rows it has still completed successfully, so you would expect $result to be 1.
You are getting the correct output. When doing that query, you're asking the database to check if there is data with that firstname or lastname and delete it. Even if there is no data with that matches it, the query has still run successfully.
You need to do use
$result->rowCount() == 1
instead of
$result == 1
It really depends what you want to use the result for. If you simply want to tell the user it has been deleted, using what you have is fine. However, if you want to let the user knows if anything has actually been deleted, you need to use my suggestion above or an alternate method to determine if this is the case.
Actually it looks like you might be using mysqli in this code, so maybe you could try using affected_rows instead of rowCount:
see http://php.net/manual/en/mysqli.affected-rows.php.
What does
$result->affected_rows
give you?
(Sorry, I don't really know what I am doing.)
I have this Unity game in an iframe on Facebook calling a php file in the same directory, and that much is working. What I want it to do is update the player record if it is there and make one if it isn't.
This script runs but it always returns a "not here" and when I check the database, it is in fact creating the records each time, identical but for the datetime field. So I don't understand why affected_rows is never coming back as "1".
<?php
$db = #new mysqli('••.•••.•••.••', '•••••••••••', '••••••••','•••••••••••');
if ($db->connect_errno)
{
echo("Connect failed "+mysqli_connect_error());
exit();
}
$inIP = $_POST["ip"];
$playerIP = mysqli_real_escape_string($db, $inIP);
$inUN = $_POST["un"];
$playerUN = mysqli_real_escape_string($db, $inUN);
$query = "UPDATE lobby SET whens=NOW(), wherefores='$playerIP', whys=0 WHERE whos='$playerUN'";
mysqli_query($db, $query);
if (mysqli_affected_rows($db) > 0)
{
echo "here";
}
else
{
$query2 = "INSERT INTO lobby (whens,whos,wherefores,whys) values (NOW(),'$playerUN','$playerIP',0)";
mysqli_query($db, $query2);
echo "not here";
}
if ($db)
{
$db->close();
}
?>
You have a typo:
wherefores=$playerip
it should be
wherefores=$playerIP
because of that
mysqli_affected_rows($db)
returns
-1
Sounds like you're experiencing the same problem as me, especially if you are running your code through a debugger. I've investigated the issue with Netbeans and Xdebug and it seems this is a bug in the MySQLi extension itself. An according bug report has been made. In the meantime you can instead use another expression, e.g.:
if (mysqli_sqlstate($dbc) == 00000) {
//your code
}
to continue debugging your remaining code.
I have a code which kinda works, but not really i can't figure out why, what im trying to do is check inside the database if the URL is already there, if it is let the user know, if its not the go ahead and add it.
The code also makes sure that the field is not empty. However it seems like it checks to see if the url is already there, but if its not adding to the database anymore. Also the duplicate check seems like sometimes it works sometimes it doesn't so its kinda buggy. Any pointers would be great. Thank you.
if(isset($_GET['site_url']) ){
$url= $_GET['site_url'];
$dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'");
$num_rows = mysql_num_rows($dupe);
if ($num_rows) {
echo 'Error! Already on our database!';
}
else {
$insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
echo $url;
echo ' added to the database!';
}
}
else {
echo 'Error! Please fill all fileds!';
}
Instead of checking on the PHP side, you should make the field in MySQL UNIQUE. This way there is uniqueness checking on the database level (which will probably be much more efficient).
ALTER TABLE tbl ADD UNIQUE(URL);
Take note here that when a duplicate is INSERTed MySQL will complain. You should listen for errors returned by MySQL. With your current functions you should check if mysql_query() returns false and examine mysql_error(). However, you should really be using PDO. That way you can do:
try {
$db = new PDO('mysql:host=localhost;db=dbname', $user, $pass);
$stmt = $db->query('INSERT INTO tbl (URL) VALUES (:url)');
$stmt->execute(array(':url' => $url));
} catch (PDOException $e) {
if($e->getCode() == 1169) { //This is the code for a duplicate
// Handle duplicate
echo 'Error! Already in our database!';
}
}
Also, it is very important that you have a PRIMARY KEY in your table. You should really add one. There are a lot of reasons for it. You could do that with:
ALTER TABLE tbl ADD Id INT;
ALTER TABLE tbl ADD PRIMARY KEY(Id);
You should take PhpMyCoder's advice on the UNIQUE field type.
Also, you're not printing any errors.
Make sure you have or die (mysql_error()); at the end of your mysql_* function(s) to print errors.
You also shouldn't even be using mysql_* functions. Take a look at PDO or MySQLi instead.
You're also not executing the insert query...
Try this code:
if(isset($_GET['site_url']) ){
$url= $_GET['site_url'];
$dupe = mysql_query("SELECT * FROM $tbl_name WHERE URL='$url'") or die (mysql_error());
$num_rows = mysql_num_rows($dupe);
if ($num_rows > 0) {
echo 'Error! Already on our database!';
}
else {
$insertSite_sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
mysql_query($insertSite_sql) or die (mysql_error());
echo $url;
echo ' added to the database!';
}
}
else {
echo 'Error! Please fill all fileds!';
}
As PhpMyCoder said, you should add a unique index to the table.
To add to his answer, here is how you can do what you want to do with only one query.
After you add the unique index, if you try to "INSERT INTO" and it result in a duplicate, MySQL will produce an error.
You can use mysql_errno() to find out if there was a duplicate entry and tell the user.
e.g.
$sql = "INSERT INTO $tbl_name (URL) VALUES('$url')";
$result = mysql_query($sql);
if($result === false) {
if(mysql_errno() == $duplicate_key_error) {
echo 'Error! Already in our database!';
} else {
echo 'An error has occurred. MySQL said: ' . mysql_error();
}
}
mysql_error() will return the mysql error in plain english.
mysql_errno() returns just the numeric error code. So set $duplicate_key_error to whatever the code is (I don't know it off the top of my head) and you are all set.
Also note that you don't want to print any specific system errors to users in production. You don't want hackers to get all kinds of information about your server. You would only be printing MySQL errors in testing or in non-public programs.
ALSO! Important, the mysql functions are deprecated. If you go to any of their pages ( e.g. http://php.net/manual/en/function.mysql-errno.php) you will see recommendations for better alternatives. You would probably want to use PDO.
Anyone who wants to edit my answer to change mysql to PDO or add the PDO version, go ahead.