$entries = "INSERT INTO allowances (totalGrossPay) VALUES ('".$totalGrossPay."') WHERE (allowances.SSN = '".$SSN."')";
mysql_query ($entries) or die (mysql_error());
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE (allowances.SSN = '300497654')'
What do you think should happen? INSERT is unconditional, no WHERE clause is required or even allowed.
What you really want is an UPDATE.
Just do an update:
$entries = "UPDATE `allowances` SET `totalGrossPay` = '{$totalGrossPay}' WHERE `SSN` = '{$SSN}'";
mysql_query ($entries) or die (mysql_error());
And if you want to update a possibly existing record and otherwise insert the record, you can use the ON DUPLICATE KEY phrase of the INSERT statement to specify what happens if the record already exists.
Related
I'm not a newbie to PHP but I have encountered a [seemingly] simple problem which I cannot figure out how to resolve.
MySQL throws error that the syntax is wrong.
My Statement is this:
if($value){
$query = "UPDATE ".$preuploads." SET words = '$words_amount' WHERE id= $sn_id";
$db->sql_query( $query ) or die( mysql_error() );
}
And then $words_amount is an integer, $sn_id is also an integer. They are double checked.
The statement when printed before execution is as follows:
UPDATE SET uploads words = '250' WHERE id= 8081
// edited, with the name of table added since the problem primarily was
// with the encapsulation and the name of table just was dropped in this question
// and not in the app
however words value ('250') is tested with integer data-type as well, but no change occurs and the error lingers on.
And the error thrown is:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SET words = '250' WHERE id= 8081' at line 1
If I understand your question (and preuploads is a table), then
$query = "UPDATE ".$preuploads." SET words = '$words_amount' WHERE id= $sn_id";
should be
$query = "UPDATE ".$preuploads." SET words = '".$words_amount."' WHERE id=".$sn_id;
Or, even better prepare and use bind_param,
$stmt = $mysqli->prepare("UPDATE ? SET words=? WHERE id=?");
$stmt->bind_param($preuploads, $words_amount, $snd_id);
$stmt->execute();
check your string ($words_amount) has any single quotes ' if it is then remove it by using this option on php $words_amount=string_replace("'","/'",$your_string_variable);
I have found two errors:
First, not encapsulation of the data should occur, thus:
$words_count should be left as is, not to be encapsulated with '
And the table and fields name should be encapsulated with backtick
I think your having problem with name of table. The syntax for update query is
UPDATE table_name SET words = '250' WHERE id= 8081
I get the following error when running my .php file:
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near ''feeds_list' SET 'cron' = now() WHERE 'feed_id' = '1'' at line 1
Here is the code:
$sql = mysql_query("UPDATE 'feeds_list' SET 'cron' = now() WHERE 'feed_id' = '$feed_id'") or die (mysql_error());
Thanks for helping me out
Remove the '' around table names and column names, it should be:
UPDATE feeds_list SET cron = now() WHERE feed_id = '$feed_id'
or use a backticks (``).
$sql = mysql_query("UPDATE feeds_list SET cron = now() WHERE feed_id = '$feed_id'") or die (mysql_error());
USE cron Not 'cron'
if you are putting single quotes on the field name it will show error, don't get confused in single quotes and backticks they are completely different.
I am trying to insert data into Mysql table, but it is giving me an error as-
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Scoretab VALUES ('UX 345','22','0.8562675')' at line 1
This is the php-mysql snippet that im using :
if($value >= 0.70){
$mu_id = $ros['c_id'];
$moc_id = $ram['t_id'];
$query="INSERT INTO Scoretab VALUES ('$mu_id','$moc_id','$value')";
$op1 = mysql_query($query) or die(mysql_error());
}
This is my table structure:
CREATE TABLE IF NOT EXISTS `Scoretab` (
`mu_id` varchar(10) NOT NULL,
`moc_id` int(5) NOT NULL,
`score` decimal(5,4) NOT NULL,
UNIQUE KEY `mu_id` (`mu_id`)
)
There could potentially be a few problems with this query
$query="INSERT INTO Scoretab VALUES ('$mu_id','$moc_id','$value')";
Does the number of columns match the fields your trying to insert? Have you tried using using specific column identifier Scoretab (col,col,col) values (val, val, val)
Does any of your values contain an unescaped apostrophe? You might want to consider using mysql_real_escape_string for $mu_id and intval for $moc_id maybe!
$value is a float you don't need to ad apostrophes while inserting
Are you sure you are connected to the same database you have this table in?
this could be a possible working solution (edit)
if ($value >= 0.70)
{
$mu_id = mysql_real_escape_string($ros['c_id']);
$moc_id = intval($ram['t_id']);
$query = "INSERT INTO `Scoretab` VALUES ('$mu_id', $moc_id, $value)";
$op1 = mysql_query($query) or die(mysql_error());
}
try this
$query="INSERT INTO Scoretab (mu_id,moc_id,score) VALUES ('$mu_id','$moc_id','$value')";
The error seems to be before the table name Scoretab. Did you check your syntax carefully?
Sometimes we don't see what's right in front of our eyes! :D
Just replicated the example and everything worked for me.
I'm using MySQL 5.1 hosted at my ISP. This is my query
mysql_query("
IF EXISTS(SELECT * FROM licensing_active WHERE title_1='$title_1') THEN
BEGIN
UPDATE licensing_active SET time='$time' WHERE title_1='$title_1')
END ELSE BEGIN
INSERT INTO licensing_active(title_1) VALUES('$title_1')
END
") or die(mysql_error());
The error is
... check the manual that corresponds to your MySQL server version for the right syntax to use near 'IF EXISTS(SELECT * FROM licensing_active WHERE title_1='Title1') THEN ' at line 1
My actual task involves
WHERE title_1='$title_1' AND title_2='$title_2' AND version='$version' ...ETC...
but I have reduced it down to make things simpler for my problem solving
In my searches on this, I keep seeing references to 'ON DUPLICATE KEY UPDATE', but don't know what to do with that.
Here is a simple and easy solution, try it.
$result = mysql_query("SELECT * FROM licensing_active WHERE title_1 ='$title_1' ");
if( mysql_num_rows($result) > 0) {
mysql_query("UPDATE licensing_active SET time = '$time' WHERE title_1 = '$title_1' ");
}
else
{
mysql_query("INSERT INTO licensing_active (title_1) VALUES ('$title_1') ");
}
Note: Though this question is from 2012, keep in mind that mysql_* functions are no longer available since PHP 7.
This should do the trick for you:
insert into
licensing_active (title_1, time)
VALUES('$title_1', '$time')
on duplicate key
update set time='$time'
This is assuming that title_1 is a unique column (enforced by the database) in your table.
The way that insert... on duplicate works is it tries to insert a new row first, but if the insert is rejected because a key stops it, it will allow you to update certain fields instead.
The syntax of your query is wrong. Checkout http://dev.mysql.com/doc/refman/5.0/en/control-flow-functions.html
Use the on duplicate key syntax to achieve the result you want. See http://dev.mysql.com/doc/refman/5.0/en/insert-select.html
Another solution
$insertQuery = "INSERT INTO licensing_active (title_1) VALUES ('$title_1')";
if(!$link->query($insertQuery)){ // Insert fails, so update
$updateQuery = "UPDATE licensing_active SET time='$time' WHERE title_1='$title_1'";
$link->query($updateQuery);
}
Here is the example I tried and its works fine:
INSERT INTO user(id, name, address) VALUES(2, "Fadl", "essttt") ON DUPLICATE KEY UPDATE name = "kahn ajab", address = "Address is test"
I am amazed to see so many useless codes and answers...
Just replace INSERT with REPLACE.
¯\(ツ)/¯
im trying to update my table using the following query...
$query = mysql_query("UPDATE `outgoings` (id, user_id, bill, bill_name, bill_description, bill_colour ) VALUES ('$id', '$uid', '$bill', '$billname', '$billdescription', '$billcolour') WHERE id = '$id'") or die(mysql_error());
It returns...
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(id, user_id, bill, bill_name, bill_description, bill_colour ) VALUES ('', '8464' at line 1
Ive tried removing ' around my variables and googling for alternative methods but cant seem to figutre out what imdoing wrong?
Use this syntax for update statements:
UPDATE `outgoings` set id = '$id', user_id = '$uid' ... where ...
You got it mixed with insert statement I guess.
It looks like your ID is empty (...VALUES ('',...). Should there be an ID there?
Your $id seems to be empty or not defined yet. Read mysql.error() up to the end.
The update query has different syntax, something like that:
UPDATE `outgoings` SET user_id='$uid', bill='$bill' WHERE id = '$id'