i got issue while if statement running. i want to update query on those user_id's who status changed from 0 to 1. but this query run on all status userid who already 1. but i want run on those who changed to 0 to 1. and run only one time don't many time. please help me to solve this. Thanks
if($status==1)
{
$query22= "UPDATE `user` SET `global` =$row[user_id] WHERE
`user_id` = '".$_SESSION['userid']."' ";
$current_id = $db_handle->insertquery($query22);
}
Try with below condition:
if($status=1)
If I understand correctly you want to take some action when a status is changed by user, Ignore answer if thats not the case.
First you need to check the initial status. This you most probably need to fetch from database.
Lets call this variable $initialStatus
$initialStatus = fetchUserStatus($userId); // need to implement fetchUserStatus
then you can put your check something like this,
if($status==1 && $initialStatus==0){
// do some action
}
Related
I have a table in a database called 'threads' and I want to delete a row where the id is equal to $forumid which is set by the URL. The URL looks something like this: domain.com/viewThread.php?forumid=1
I am getting this using
$forumid = $_GET['forumid'];
and I am sure that this is working because I use
echo $forumid;
and it works correctly. But when I go to delete a row using
$db->query("DELETE FROM threads WHERE id='$forumid'");
its not working for some reason.
Can somebody please help me with this? Is it possible that there is something wrong with my phpMyAdmin or mySQL database?
Try the below query
$db->query("DELETE FROM threads WHERE id='.$forumid.'");
Check the query
("DELETE FROM threads WHERE id= '.$forumid.'");
Try by using any of the below codes.
$db->query("DELETE FROM threads WHERE id='.$forumid.'");
OR
$db->query("DELETE FROM threads WHERE id=$forumid");
Basically what I am trying to do is when I delete a client with an ID of lets say 6 and I have 50 clients, I then want to update the client with the ID of 50 to 6.
This is my code, in PHP, but it won't execute 2 mysql_query-s at the same time at least I think that's the problem. Otherwise the SQL syntax works fine.
public function delete () {
$last=$this->numrow; //contains last ID works fine
if (isset ($_GET['x'])) {
mysql_query('DELETE FROM proba WHERE ID ='.$_GET['x']);
mysql_query('UPDATE proba SET ID='.(int)$_GET['x'].'WHERE ID='.(int)$last);
}
}
The $_GET['x'] contains the ID on which it was clicked . But only the first mysql_query gets executed how do i make it so the second one gets executed also ?
And another question is is it possible to get <a href="munka/index.php?x=5" > [-] </a> the x=5 with a $_POST ?
You might save yourself a lot of trouble by using mysql's replace query:
see http://dev.mysql.com/doc/refman/5.6/en/replace.html
for details.
most probably you are facing a php error on the first query. Check the php error log.
for the second question $_GET is used to take parameters from the URL for example
munka/index.php?x=5
$_POST is used to get parameters posted on http post (usually on form submits).
just change the update query with a space before the where clause
mysql_query('UPDATE proba SET ID='.(int)$_GET['x'].' WHERE ID='.(int)$last);
Better to use transactions support by using InnoDB Mysql DB Engine, so both delete and update execute together wuth COMMIT without miss , and in case anything goes wrong your delete changes get ROLLBACK
if (isset($_GET['x'])) {
mysql_query('DELETE FROM proba WHERE ID =' . $_GET['x']);
mysql_query('ALTER TABLE `proba` DROP `ID`;');
mysql_query('ALTER TABLE `proba` ADD `ID` INT NOT NULL AUTO_INCREMENT PRIMARY KEY FIRST');
}
Try in Phpmyadmin delete record no of 5 and drop id column and recreate id column.
It is work.
i use my own mvc framework.
and post action is :
function post($pid = 0 , $title = '')
{
$pid = $this->input->clean($pid);
$stm =$this->Post->query("UPDATE posts SET `show_count`=show_count+1 WHERE id = $pid");
$stm->execute();
$row = $this->Post->find()->where(array('id'=>$pid))->fetchOne();
$this->layout->set('comments' , $this->comments($pid));
$this->layout->set('row' , $row);
$this->side();
$this->layout->view('post');
echo $this->layout->render($row['title']);
}
i want to when a record fetch from database plus one show_count column .
i use this query :
UPDATE posts SET show_count = show_count + 1 WHERE id = $pid
this right in localhost but in my shared host when run query instead of one pluses ,2 plus show_count column.
how can i solve this problem?
It's working in your localhost but not your internet host. The host is probably doing something behind the scenes that makes it not work. Also, since you don't have access to all the configuration stuff on the host server, you probably won't be able to just fix it.
Simplest fix is to do it in the software layer (assuming $pid is unique).
First get the show_count.
Then:
$oldShowCount = the current show_count
$newShowCount = $oldShowCount + 1
UPDATE posts SET show_count = $newShowCount WHERE id = $pid AND show_count = $oldShowCount
At this point, if no race conditions occurred the row will update (rows update = 1). If a race condition occurred, the update will fail (rows updated = 0).
Then check the # rows updated to verify it worked and repeat until it does.
This is quite common case.
You are running your script twice. Probably because browser is calling it twice. Check your rewrite rules.
Thanks for asking me to make my question clearer.
I have a form which is used to save content in a database.
The user is also allowed to update any content by editing the fields in the form and submitting it.
The update query looks something like this:-
$res = mysql_query("UPDATE freight SET .... WHERE id='$id'") or die ( mysql_error());
if(mysql_affected_rows() > 0){
//do some stuff
return true;
}
else{
return false;
}
Now my problem is this: If the user changes some field while updating the form, mysql_affected_rows() is > 0 so I am able to "//do some stuff".
But if the user does not change any field in the form and simply submits, mysql_affected_rows return 0 and I reach the false block.
I was under the impression that UPDATE should overwrite every time thereby mysql_affected_rows() should return 1 irrespective of whether any fields have been changed or not.
I use the return value to display an appropriate success or error message in the screen. Hence this is a serious bug.
Can you suggest whats the right way to achieve the correct functionality?
Thanks,
Vish
mysql_affected rows() -1 will be returned if the query itself can not be issued to the server, possibly because of syntax error AND if the last query was not either an INSERT or UPDATE statement.
When using UPDATE, MySQL will not update columns where the new value
is the same as the old value. This creates the possibility that
mysql_affected_rows() may not actually equal the number of rows
matched, only the number of rows that were literally affected by the
query.
so your solution can be achieved by below logic ( but this is not safe)
if(mysql_affected_rows()!=-1){
}
if( $_POST['message'] != '' ) {
// add the field to your query
} else { // this is optional
// don't add it to your query
}
// do the mysql update here
I have a bit of an issue with my code.
I'm making an administrative panel for users to add things to the database. On occasion, they might try to save data without changing it (open a dialog, click save without changing anything). This, of course, will make mysql_affected_rows() return '0' when checking to see if the UPDATE query worked.
Is there another query to make that will always UPDATE regardless of whether the data is the same or not (or can I modify a simple UPDATE query to always update)?
EDIT
This is for users who don't have programming experience. Of course you wouldn't want to update if there's no reason to, but when a user tries to update and it doesn't happen I end up showing a failure message. Rather than there being something wrong, its just it doesn't need to be updated. I need a way to show the user that, instead of a generic 'failure' message. If it failed for another reason, I still need to know.
From the MySQL Documentation:
If you set a column to the value it currently has, MySQL notices this
and does not update it.
Instead of checking mysql_affected_rows, just check to see if the query was successful:
if(!mysql_query("UPDATE ..."))
{
//failure
}
else
{
$verification = mysql_query("SELECT ROW_COUNT() as rows_affected");
$row = mysql_fetch_row($verification);
$rows_affected = $row[0];
if ($rows_affected > 0)
{
//update was performed
}
else
{
//no update was needed
}
}