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.
Related
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();
i am using this query to execute the session, but it does not work on the website
$_SESSION['bericht'] = $stamVanStatus;
$status = $_SESSION['bericht'];
print_r($status);
$result = mysql_query("SELECT bericht,naam,datum FROM discussies WHERE bericht LIKE '$status'");
The Print_r shows me the message i send from the post function, which i want to compare to the column. in the column there are couple of messages i created to test.
It should compare it with my data in the database, but that does not seem to work, as i just see the printed message i send. When i change the LIKE function to for example '%Test%' It shows everything inside the column which has Test in the text.
It does not show any errors
Hopefully someone can help me with this
Thanks in Advance
$result = mysql_query
("
SELECT bericht,
naam,
datum
FROM discussies
WHERE bericht LIKE '%$status%'
");
Append two (%) percent signs to the query. One before and one after the variable.
If "$status" is set to Foo, the query would return; Foobar, if Foobar is a value in your Database, it would also return BarFoo and BarFooBar. So long the is present in the Database, a value will be returned.
Appending a percent sign in either the beginning or the end will rule out BarFooBar and only return the value if it's present, and any other information before(or after, your choice) the string itself.
You literally queried "$status". Meaning instead of the message, like 'hello this is a message', you got '$status'
change last line to
$result = mysql_query("SELECT bericht,naam,datum FROM discussies WHERE bericht LIKE '" . $status . "'");
Try assigning the query string to a variable, inspecting the variable with a echo or debugger, then query it. In my opinion this makes debugging SQL queries a lot easier. If this doesn't work, what does the evaluated string become?
I know this is late answer but I hope it may help someone.
Variables are connected to strings with .
$result = mysql_query("SELECT bericht,naam,datum FROM discussies WHERE bericht='".$status."'");
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();
HI everyone i tried for 3 days and i'm not able to solve this problem. This is the codes and i have went through it again and again but i found no errors. I tried at a blank page and it worked but when i put it inside the calendar it has the syntax error. Thanks a million for whoever who can assist.
/** QUERY THE DATABASE FOR AN ENTRY FOR THIS DAY !! IF MATCHES FOUND, PRINT THEM !! **/
$testquery = mysql_query("SELECT orgid FROM sub WHERE userid='$userid'");
while($row4 = mysql_fetch_assoc($testquery))
{
$org = $row4['orgid'];
echo "$org<br>";
$test2 = mysql_query("SELECT nameevent FROM event WHERE `userid`=$org AND EXTRACT(YEAR FROM startdate)='2010' AND EXTRACT(MONTH FROM startdate)='08' AND EXTRACT(DAY FROM startdate)='15'") or die(mysql_error());
while($row5=mysql_fetch_assoc($test2))
{
$namethis = $row5['nameevent'];
$calendar.=$namethis;
}
}
First question: what calendar are you talking about?
And here are my 2-cents: does the EXTRACT function returns a string or a number?
Are the "backticks" (userid) really in your query? Try to strip them off.
Bye!
It's a guess, given that you haven't provided the error message you're seeing, but I imagine that userid is a text field and so the value $org in the WHERE clause needs quotes around it. I say this as the commented out testquery has quotes around the userid field, although I appreciate that it works on a different table. Anyway try this:
SELECT nameevent FROM event WHERE userid='$org' AND EXTRACT(YEAR FROM startdate)='2010' AND EXTRACT(MONTH FROM startdate)='08' AND EXTRACT(DAY FROM startdate)='15'
In such cases it's often useful to echo the sql statement and run it using a database client
First step in debugging problems like this, is to print out the acutal statement you are running. I don't know PHP, but can you first build up the SQL and then print it before calling mysql_query()?
EXTRACT() returns a number not a character value, so you don't need the single quotes when comparing EXTRACT(YEAR FROM startdate) = 2010, but I doubt that this would throw an error (unlike in other databases) but there might be a system configuration that does this.
Another thing that looks a bit strange by just looking at the names of your columns/variables: you are first retrieving a column orgid from the user table. But you compare that to the userid column in the event table. Shouldn't you also be using $userid to retrieve from the event table?
Also in the first query you are putting single quotes around $userid while you are not doing that for the userid column in the event table. Is userid a number or a string? Numbers don't need single quotes.
Any of the mysql_* functions can fail. You have to test all the return values and if one of them indicates an error (usually when the function returns false) your script has to handle it somehow.
E.g. in your query
mysql_query("SELECT orgid FROM sub WHERE userid='$userid'")
you mix a parameter into the sql statement. Have you assured that this value (the value of $userid) is secure for this purpose? see http://en.wikipedia.org/wiki/SQL_injection
You can use a JOIN statement two combine your two sql queryies into one.
see also:
http://docs.php.net/mysql_error
http://docs.php.net/mysql_real_escape_string
http://www.w3schools.com/sql/sql_join.asp
Example of rudimentary error handling:
$mysql = mysql_connect('Fill in', 'the correct', 'values here');
if ( !$mysql ) { // some went wrong, error hanlding here
echo 'connection failed. ', mysql_error();
return;
}
$result = mysql_select_db('dbname', $mysql);
if (!$result ) {
echo 'select_db failed. ', mysql_error($mysql);
return;
}
// Is it safe to use $userid as a parmeter within an sql statement?
// see http://docs.php.net/mysql_real_escape_string
$sql = "SELECT orgid FROM sub WHERE userid='$userid'";
$testquery = mysql_query($sql, $mysql);
if (!$testquery ) {
echo 'query failed. ', mysql_error($mysql), "<br />\n";
echo 'query=<pre>', $sql, '</pre>';
return;
}