I'm trying to update my users table using php.
$update_query="
UPDATE `users` SET `name`='$addname',
`lastname`='$addlastname',
`password`='$addpsswrd',
`email`='$addemail'
where `username`='$modifyusername'
";
echo $update_query;
if( mysql_query($update_query) or die('Erreur SQL !'.$req.'<br>'.mysql_error()))
echo "Lignes modifiées : ", mysql_affected_rows();
But I always get :
UPDATE `users` SET `name`='Jolia ',`lastname`='roberta', `password`='password1234',`email`='roberta.joli#hotmail.fr' where `username`='user11'
Lignes modifiées : 0
How can I fix this? the user11 exist in my database and I tried to copy past this query as it is in the output echo message I get 0 modified line so how can I fixed on the php part?
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.
http://us1.php.net/manual/en/function.mysql-affected-rows.php
I am not sure , but my suggestion is try to remove the single quotation marks ['] from the column name. ps: this is my first time answer question on stackoverflow
like
$update_query="
UPDATE users SET name='$addname',
lastname='$addlastname',
password='$addpsswrd',
email='$addemail'
where username='$modifyusername'
";
echo $update_query;
if( mysql_query($update_query) or die('Erreur SQL !'.$req.'<br>'.mysql_error()))
echo "Lignes modifiées : ", mysql_affected_rows();
Related
I am trying to code a user system. I am having an issue with the activation part. I can select and insert data to my table but now I am trying to create an update statement and got stuck.
<?PHP
include "db_settings.php";
$stmt = $dbcon->prepare("UPDATE 'Kullanicilar' SET 'Aktivasyon' = ? WHERE 'KullaniciID'=?");
// execute the query
$stmt->execute(array('1','5'));
// echo a message to say the UPDATE succeeded
echo $stmt->rowCount() . " records UPDATED successfully";
?>
And I am getting error as:
"0 records UPDATED successfully".
This is my table; http://i.imgur.com/PL2eD80.png
I have tried by changing my 'Aktivasyon' type int to char but it also does not work.
EDIT:
I am trying to make this a function;
function dataUpdate($tableName, $updateRow, $updateValue, $conditonRow, $conditionValue)
{
include "db_settings.php";
$q = $dbcon->prepare("UPDATE $tableName SET $updateRow= ? WHERE $conditonRow= ?");
$q->execute(array($updateValue,$conditionValue));
}
I also try this :
...
$q = $dbcon->prepare("UPDATE `$tableName` SET `$updateRow`= ? WHERE `$conditonRow`= ?");
...
How can I make this statement work?
You are using wrong quotes. You are basically saying "update this table, set this string to something when the string KullaniciID equals the string 5" which of course never is true.
You should use backticks ` if you want to specify column names. Then your query would work. Usually you don't even need those, but for some reason MySQL world is always adding them.
So to clarify, this is a string: 'KullaniciID' and this is a column name: `KullaniciID`.
Also you should not send integers as strings. It causes extra conversions or even errors with more strict databases.
I have a query like this:
$query="INSERT IGNORE INTO mytable (key, word1, word2) VALUES (1, 'abc', 'def')";
And I'd like to somehow "catch" if the query was ignored because the primary key already exists.
Is there any was to make an "if" statement so my program could notify the user that the query was ignored?
I solved it by using mysql_affected_rows to check if no rows were affected by the query.
if($productquery&&mysql_affected_rows($conn)>0){
echo "Success.";
echo mysql_affected_rows($conn); //returns 1
}else{
echo "Some error." . mysql_error(). " ".mysql_affected_rows($conn); //returns 0 affected rows since it was ignored
}
The problem is when you especially ignore the errors then you don't get them. So you have to remove the IGNORE and catch the error instead.
You don't get any errors in your case.
I am trying to get the number of rows affected in a simple mysql update query. However, when I run this code below, PHP's mysql_affected_rows() always equals 0. No matter if foo=1 already (in which case the function should correctly return 0, since no rows were changed), or if foo currently equals some other integer (in which case the function should return 1).
$updateQuery = "UPDATE myTable SET foo=1 WHERE bar=2";
mysql_query($updateQuery);
if (mysql_affected_rows() > 0) {
echo "affected!";
}
else {
echo "not affected"; // always prints not affected
}
The UPDATE statement itself works. The INT gets changed in my database. I have also double-checked that the database connection isn't being closed beforehand or anything funky. Keep in mind, mysql_affected_rows doesn't necessarily require you to pass a connection link identifier, though I've tried that too.
Details on the function: mysql_affected_rows
Any ideas?
Newer versions of MySQL are clever enough to see if modification is done or not. Lets say you fired up an UPDATE Statement:
UPDATE tb_Employee_Stats SET lazy = 1 WHERE ep_id = 1234
Lets say if the Column's Value is already 1; then no update process occurs thus mysql_affected_rows() will return 0; else if Column lazy had some other value rather than 1, then 1 is returned. There is no other possibilities except for human errors.
The following notes will be helpful for you,
mysql_affected_rows() returns
+0: a row wasn't updated or inserted (likely because the row already existed,
but no field values were actually changed during the UPDATE).
+1: a row was inserted
+2: a row was updated
-1: in case of error.
mysqli affected rows developer notes
Have you tried using the MySQL function ROW_COUNT directly?
mysql_query('UPDATE myTable SET foo = 1 WHERE bar = 2');
if(mysql_result(mysql_query('SELECT ROW_COUNT()'), 0, 0)) {
print "updated";
}
else {
print "no updates made";
}
More information on the use of ROW_COUNT and the other MySQL information functions is at: http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_row-count
mysqli_affected_rows requires you to pass the reference to your database connection as the only parameter, instead of the reference to your mysqli query. eg.
$dbref=mysqli_connect("dbserver","dbusername","dbpassword","dbname");
$updateQuery = mysqli_query($dbref,"UPDATE myTable SET foo=1 WHERE bar=2");
echo mysqli_affected_rows($dbref);
NOT
echo mysqli_affected_rows($updateQuery);
Try connecting like this:
$connection = mysql_connect(...,...,...);
and then call like this
if(mysql_affected_rows($connection) > 0)
echo "affected";
} else { ...
I think you need to try something else in update then foo=1. Put something totaly different then you wil see is it updating or not without if loop. then if it does, your if loop should work.
You work this?
$timestamp=mktime();
$updateQuery = "UPDATE myTable SET foo=1, timestamp={$timestamp} WHERE bar=2";
mysql_query($updateQuery);
$updateQuery = "SELECT COUNT(*) FROM myTable WHERE timestamp={$timestamp}";
$res=mysql_query($updateQuery);
$row=mysql_fetch_row($res);
if ($row[0]>0) {
echo "affected!";
}
else {
echo "not affected";
}
This is because mySql is checking whether the field made any change or not,
To over come this, I created a new TINY field 'DIDUPDATE' in the table.
added this to your query 'DIDUPDATE=DIDUPDATE*-1'
it looks like.
$updateQuery = "UPDATE myTable SET foo=1, DIDUPDATE=DIDUPDATE*-1 WHERE bar=2";
mysql_query($updateQuery);
if (mysql_affected_rows() > 0)
{
echo "affected!";
}
else
{
echo "not affected";
}
it works fine!!!
Was My Tought !
I was just about to tell to check if the function's being called many times !
Just a little advice:
try using isset() & POST / GET or something like that;
if ( isset( $_POST['Update'] == 'yes' ) ) :
// your code goes here ...
endif;
Hope it was clear and useful, Ciao :)
I know there are a lot of topics with the same title. But mostly it's the query that's been inserted in the wrong place. But I think I placed it right.
So the problem is, that I still get 0 even when the data is inserted in the db.
Does someone knows an answer where I could be wrong?
here's my code:
mysql_query('SET NAMES utf8');
$this->arr_kolommen = $arr_kolommen;
$this->arr_waardes = $arr_waardes;
$this->tabel = $tabel;
$aantal = count($this->arr_kolommen);
//$sql="INSERT INTO `tbl_photo_lijst_zoekcriteria` ( `PLZ_FOTO` , `PLZ_ZOEKCRITERIA`,`PLZ_CATEGORIE`)VALUES ('$foto', '$zoekje','$afdeling');";
$insert = "INSERT INTO ".$this->tabel." ";
$kolommen = "(";
$waardes = " VALUES(";
for($i=0;$i<$aantal;$i++)
{
$kolommen .=$this->arr_kolommen[$i].",";
$waardes .="'".$this->arr_waardes[$i]."',";
}
$kolommen = substr($kolommen,0,-1).")";
$waardes = substr($waardes,0,-1).")";
$insert .=$kolommen.$waardes;
$result = mysql_query($insert,$this->db) or die ($this->sendErrorToMail(str_replace(" ","",str_replace("\r\n","\n",$insert))."\n\n".str_replace(" ","",str_replace("\r\n","\n",mysql_error()))));
$waarde = mysql_insert_id();
Thanks a lot in advance, because I have been breaking my head for this one for almost already a whole day. (and probably it's something small and stupid)
According to the manual mysql_insert_id returns:
The ID generated for an AUTO_INCREMENT column by the previous query on
success, 0 if the previous query does not generate an AUTO_INCREMENT
value, or FALSE if no MySQL connection was established.
Since it does not give you false and not the correct number it indicates that the queried table didn't generate an auto-increment value.
There are two possibilities I can think of:
Your table doesn't have an auto_increment field
Since you doesn't provide the link to the mysql_insert_id() but using a link with mysql_query() it might not be the correct table that's queried when retrieving the last inserted id.
Solution:
Make sure it has an auto_increment field
Provide the link aswell: $waarde = mysql_insert_id($this->db);
It is possible that your INSERT query was not successful - e.g., maybe you were trying to insert duplicate data on a column whose data must be unique?
If the id is indeed set to auto increment and still get '0' as your response do a column and value count i experienced this only later on I noticed a number of my column count did not match values count.
Codeigniter has an odd behaviourd when calling mysql_insert_id(). The function returns 0 after the first call. So calling it twice will return 0.
Use a variable instead of calling the function more times:
$id = mysql_insert_id();
i have written few codes to show time spent by users at site but when a users click on submit it should be stored in mysql but its not getting stored can you please tell where i have done mistake here is my code.
Your query seems to be wrong.
Either use INSERT without WHERE if you want to insert a new record. If however you want to update an already present record, use UPDATE instead of INSERT.
And it is always a good idea to check whether a query was successful:
if (mysql_query ("insert into jcow_accounts(Time_Spent) values ('{$Time}') where uid='{$client['id']}' ") === FALSE) {
echo 'MySQL error: ' . mysql_error() . "\n";
}
You need to use an UPDATE instead of an insert.
$dtime = getChangeInTime(); // this is the change in time from the last update
mysql_query( "UPDATE jcow_accounts SET `TIME_SPENT` = `TIME_SPENT` + $dtime ".
" where id='{$client['id']}'" );
Try
insert INTO `jcow_accounts` (`Time_Spent`) VALUES ('{$Time}') where uid='{$client['id']}' WHERE `uid` = '{$client['id']}'
Are you sure the uid is in the DB? try running it without the WHERE...