hi and thank you in advance for any help you can give!
i'm iffy on syntax with variables in queries.
i have a database table with a column called 'iltr' and a column called 'refline'.
the 'refline' column is filled with text data like this: '1LINE1' '1LINE2' '1LINE3' '2LINE1' '2LINE2' '2LINE3' etc etc etc.
i want to use a variable that relates to the first digit of the data in the 'refline' column to update another column.
in the example below i'm trying to set 'iltr'= 'Y' where 'refline' is '2LINE3' by using a variable for the "2" digit in '2LINE3' instead of the "2" digit itself, and i don't even know if this kinda thing can be done.
this code syntax below does not work specifically in the WHERE segment where i try to mix the variable with the actual text in the column:
$slot=2;
$sql = "UPDATE overall SET iltr='Y' WHERE refline='$slotLINE3'";
i hope you can forgive my inexperience. thanks again and have an awesome day!
Your code is trying to substitute a variable named $slotLINE3. You need to use a delimiter to end the variable name.
$sql = "UPDATE overall SET iltr='Y' WHERE refline='{$slot}LINE3'";
Related
I'm trying to add +1 in custom row. Example:
UPDATE `users` SET `MVP` = `MVP` + 1 WHERE `steam_id` = `%s`;
But nothing. What's wrong? Syntax looks good i think.
%s is a value so remove `
You can usually omit that everywhere unless you are using some "unlucky" column names
UPDATE `users` SET `MVP` = `MVP` + 1 WHERE `steam_id` = %s;
Take the ' away from the MVP you are incrementing.
UPDATE `users` SET `MVP` = MVP+1 WHERE `steam_id` = `%s`;
Post the entire code snipped. Can´t really help you like that. What is steam_id? What is %s. When do you replace it with an actual value? You should also just use prepared statements and not DYI that. %s is not how they look like in PDO or MYSQLI, but sure looks like a placeholder for a string.
You also marked this as insert, yet you´re doing an update.
Those `` are kinda unnecessary, never used them for column/table names, but appears to work at least in mysql console.
The sql snipped looks right, assuming MPV is numeric and the id is a string/varchar and equals %s, or you´re replacing it with something.
EDIT: As some have said the %s is the problem use nothing if it´s a int. Use single or double quotes if it´s a string. And you don´t need to use anything anywhere else, but if you wish to do so you can.
Hi everybody and sorry for my english.
I have the column "example" that is a SET type.
I have to make a php page where you can add values to that column.
First of all I need to know what is just in "example", to prevent the adding of an existing value by a control. Second of all I need to add the new value.
Here's what I had thinked to do.
//I just made the connection to the db in PDO or MySQLi
$newValue=$_POST['value']; //I take the value to add in the possible values from a form
//Now I have to "extract" all the possible values. Can't think how.
//I think I can store the values into an array
$result=$sql->fetch(); //$sql is the query to extract all the possible values from "example"
//So now i can do a control with a foreach
foreach($result as $control){
if ($newValue == $control){
//error message, break the foreach loop
}
}
//Now, if the code arrives here there isn't erros, so the "$newValue" is different from any other values stored in "example", so I need to add it as a possible value
$sql=$conn->query("ALTER TABLE 'TableName' CHANGE 'example' 'example' SET('$result', '$newValue')"); //<- where $result is the all existing possible values of "example"
In PDO or MySQLi, it's indifferent
Thanks for the help
We can get the column definition with a query from information_schema.columns
Assuming the table is in the current database (and assuming we are cognizant of lower_case_table_names setting in choosing to use mixed case for table names)
SELECT c.column_type
FROM information_schema.columns c
WHERE c.table_schema = DATABASE()
WHERE c.table_name = 'TableName'
AND c.column_name = 'example'
Beware of the limit on the number of elements allowed in a SET definition.
Remove the closing paren from the end, and append ',newval').
Personally, I don't much care for the idea of running an ALTER TABLE as part of the application code. Doing that is going to do an implicit commit in a transaction, and also require an exclusive table / metadata lock while the operation is performed.
If you need a SET type - you should know what values you add. Otherwise, simply use VARCHAR type.
Answer found (syntax): The column name of my string had to be encased in backticks " ` " as they contained spaces. Note that this means that the majority of this post has no relevance to the issue. The code has been corrected in case someone wants to do something similar.
So, I am doing a foreach loop to assign a value (1/0) to non-static columns in my database (it needs to support addition/deletion/editing of columns). I am using $connectionvar->query($queryvar); to do my queries which worked fine up until now when I'm trying to use a custom built string as $queryvar in order to change the column name to a variable within the loop. I've been outputting this string through echo and it looks exactly like my functional queries but somehow doesn't run. I've attempted to use eval() to solve this but to no avail (I feel safe using eval() as the user input is radio buttons).
Here's the loop as well as my thought processes behind the code. If something seems incoherent or just plain stupid, refer to my username.
foreach($rdb as $x) { //$rdb is a variable retrieved from $_POST earlier in the code.
$pieces = explode("qqqppp", $x); //Splits the string in two (column name and value) (this is a workaround to radio buttons only sending 1 value)
$qualname = $pieces[0]; //Column name from exploded string
$qualbool = $pieces[1]; //desired row value from exploded string
$sql = 'UPDATE users SET '; //building the query string
$sql .= '`$qualname`';
$sql .= '=\'$qualbool\' WHERE username=\'$profilename\''; //$profilename is retrieved earlier to keep track of the profile I am editing.
eval("\$sql = \"$sql\";"); //This fills out the variables in the above string.
$conn->query($sql); //Runs the query (works)
echo ' '.$sql.' <br>'; //echoes the query strings on my page, they have the exact same output format as my regular queries have.
}
}}
Here's an example of what the echo of the string looks like:
UPDATE users SET Example Qualification 3='1' WHERE username='Admin2'
For comparison, echoing a similar (working) query variable outside of this loop (for static columns) looks like this:
UPDATE users SET profiletext='qqq' WHERE username='Admin2'
As you can see the string format is definitely as planned, yet somehow doesn't execute. What am I doing wrong?
PS. Yes I did research this to death before posting it, as I have hundreds of other issues since I started web developing a month ago. Somehow this one has left me stumped though, perhaps due to it being a god awful hack that nobody would even consider in the first place.
You need to use backticks when referring to column names which have spaces in them. So your first query from the loop is outputting as this:
UPDATE users SET Example Qualification 3='1' WHERE username='Admin2'
But it should be this:
UPDATE users SET `Example Qualification 3`='1' WHERE username='Admin2'
Change your PHP code to this:
$sql = 'UPDATE users SET `'; // I added an opening backtick around the column name
$sql .= '$qualname`'; // I added a closing backtick around the column name
$sql .= '=\'$qualbool\' WHERE username=\'$profilename\'';
Example Qualification 3 : Is that the name of your Mysql Column name ?
You shouldnt use spaces nor upper / lower case in your columnname.
Prefere : example_qualification_3
EDIT :
To get column name and Comment
SHOW FULL COLUMNS FROM users
I have a table which contains an field zoneShape of the type Polygon.
When inserting some geometry in this field, in phpmyadmin it says: affected 1 row. The row is added, but the geometry field says it's size is 0B.
This is the Insertion code:
SET #g = 'POLYGON(50.866753 5.686455, 50.859819 5.708942, 50.851475 5.722675, 50.841611 5.720615, 50.834023 5.708427, 50.840744 5.689373, 50.858735 5.673923, 50.866753 5.686455)';
INSERT INTO zone SET zoneShape = PolygonFromText(#g)
As you can see; the first and last point are the same, so the polygon is closed.
Can anyone help me with adding this polygon to the database?
Thanx
It turns out that you need to add some extra brackets around the Points; because the standard uses an outside and inside line aka holes or islands.
So:
SET #g = 'POLYGON((50.866753 5.686455, 50.859819 5.708942, 50.851475 5.722675, 50.841611 5.720615, 50.834023 5.708427, 50.840744 5.689373, 50.858735 5.673923, 50.866753 5.686455))';
^ outside ^
INSERT INTO zone SET zoneShape = PolygonFromText(#g) doesn't look like a valid INSERT statement to me. Did you mean INSERT INTO zone (zoneShape) VALUES (PolygonFromText(#g));?
First of all I'm a rookie to Programming, I created a PHP page to update a value from my mysql(myadmin) database, but the value is not updating. I also tried to retrieve values from database it's working just fine but this UPDATE code is not working! I don't know why, please check out my code below.
$qs=mysql_query("update staff set review=$newrate where name=$rateuser");
$resu=mysql_query($qs);
All variables are double defined, assigned with proper values, checked and I tested variables using echo, table name is also checked, it's all fine, but I think the problem is with Update query, I searched internet for the syntax but it's not different than mine. Please help me out
How are $newrate and $rateuser set?
mysql_query("UPDATE staff SET review = '".mysql_real_escape_string($newrate)."' WHERE name = '".mysql_real_escape_string($rateuser) ."'");
http://php.net/manual/en/function.mysql-real-escape-string.php
Try:
$qs=mysql_query("update staff set review='$newrate' where name='$rateuser'");
Do not use second line.
You probably just need some " around your values $newrate and $rateuser
But if you did an echo, why not actually echo for us what the query-string becomes?
You need single quotes around string values on your query:
$qs=mysql_query("update staff set review='$newrate' where name='$rateuser'");
(assuming both variables are strings)