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.
Related
Apologies if this has been asked before, but I havent found an answer that suits my needs.
I have a prepared statement to delete data from a table using user input:
$delete = $handler->prepare("DELETE FROM screenings WHERE
time = :time ORDER BY screeningid LIMIT 1; ");
$delete- >execute(array( ':time' => $screeningtime ));
I'm trying to add an if statement before the delete statement to first check if the record exits, and if so executes the delete statement otherwise echos an error message.
so far i've been trialling:
if(empty($row['column'])) {
echo "error";
}else {
//delete statement here
}
but to no avail, the error message prints out in all instances
You'd need to do a SELECT to detect if the row exists.
However, the result of the DELETE would return the number of rows affected, so you could just run it regardless, and then check if 0 or 1 or >1 rows were deleted. That would be $delete->rowCount()
http://php.net/manual/en/pdostatement.rowcount.php
<? Php
$delete->execute();
if ($delete->rowCount() >= 1)
echo 'ok';
else
echo 'error or not found';
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 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( $_POST['Update'] == 'yes' ) ) :
// your code goes here ...
endif;
Hope it was clear and useful, Ciao :)
I am using UNIQUE Constraint in my table for two columns to prevent duplicate rows. If I try to insert a row with the same 2 columns I get
A Database Error Occurred Error
Number: 1062
Duplicate entry '62-88' for key
'subscription'
INSERT INTO subscriptions
(user_id, subscribed_to, date)
VALUES ('62', '88', '2011-07-11
19:15:13')
Filename:
C:\wamp\www\mysite\system\database\DB_driver.php
Line Number: 330
How can I return a php error, e..g Already subscribed! instead of displaying the mysql error?
Call mysql_errno() after your mysql_query() and check for 1062.
Note: It's a more common/intuitive solution to query the database first. See answer by Manocho.
I think you will need to run a query manually to check if the entry already exists in your database. so something like: SELECT COUNT(*) FROM subscriptions WHERE (user_id = '62' AND subscribed_to = '88'). If the count returned is > 0 then return the error, however you want it displayed.
http://php.net/manual/en/function.mysql-error.php I assume you are able to programatically obtain the MySQL error
http://www.php.net/manual/en/function.mysql-errno.php
http://dev.mysql.com/doc/refman/5.5/en/error-handling.html
Using the 3rd link, compare the mysql error or mysql errno to the list of errors and if the condition is met, provide your alternate error message.
To solve the error you can do this:
mysql_query($sql);
if (mysql_errno() == 1062) {
print "<script type=\"text/javascript\">";
print "alert('The informations are already inserted')";
print "</script>";
}
try this code :
$query = "INSERT INTO ".$table_name." ".$insertdata;
if(mysqli_query($conn,$query)){
echo "data inserted into DB<br>";
}else{
if(mysqli_errno($conn) == 1062)
echo "duplicate entry no need to insert into DB<br>";
else
echo "db insertion error:".$query."<br>";
}//else end
Just learning PHP and I'm having some trouble understanding mysql_query. My understanding is that mysql_query is supposed to return FALSE if the record is not found. However, it seems that it always returns true because "FOUND!" is always the result:
$q = "SELECT * FROM users WHERE username = 'doesnotexist'";
$r = mysql_query($q);
if (!$q) {
echo "<p>NOT FOUND!</p>";
} else {
echo "<p>FOUND!</p>";
}
mysql_close();
Thanks in advance for any light you can shed.
mysql_query returns false if there is an error, not if there are no results found. From the documentation:
For SELECT, SHOW, DESCRIBE, EXPLAIN
and other statements returning
resultset, mysql_query() returns a
resource on success, or FALSE on
error.
For other type of SQL statements,
INSERT, UPDATE, DELETE, DROP, etc,
mysql_query() returns TRUE on success
or FALSE on error.
mysql_query() will also fail and
return FALSE if the user does not have
permission to access the table(s)
referenced by the query.
If you want to check to see if there were results returned by your query, use mysql_num_rows(). See the documentation:
Use mysql_num_rows() to find out how
many rows were returned for a SELECT
statement or mysql_affected_rows() to
find out how many rows were affected
by a DELETE, INSERT, REPLACE, or
UPDATE statement.
You are checking the '$q' variable (your sql statement) instead of the '$r' variable (the mysql result)
if (empty($r)) {
echo "<p>NOT FOUND!</p>";
} else {
echo "<p>FOUND!</p>";
}
The following things are considered to be empty:
"" (an empty string)
0 (0 as an integer)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)
Ok i have worked out the answer to this.
This is the version for updating a database that tells you if a record was valid.
$updatequery = "update `mydb` set `userid` = '$arr[0]', `name` = '$arr[1]' where `age` = '$arr[2]'";
$updatequeryresult= mysql_query($updatequery);
$howmanyupdatedrecords = mysql_affected_rows();
if ($howmanyupdatedrecords == 0) {
echo("The update didn't update any records as no one matched an age of " .$arr[2]");
}
This will iterate through the DB updating all people with the specified age, if the age does not exist in the DB a message will be displayed showing you the age that does not exist. Also something to note, even if the mysql query matches a record, it won't update the record if the data already matches what it is being updated to. This causes the script to return "no one matched an age" even though there are people who did match. Can only attribute that to a bug in mysql. I told MySQL to update the information I don't see why it should take it upon itself to not bother doing as i told it. ;)
Your 'if (!$q)' should be 'if (!$r)' I think.
if that dosen't work, try this:
<?php
$q = "SELECT * FROM users WHERE username = 'doesnotexist'";
$r = mysql_query($q);
if (!mysql_num_rows($r) >= 1) {
echo "<p>NOT FOUND!</p>";
} else {
echo "<p>FOUND!</p>";
}
mysql_close();
?>