PHP SQL Views counter using PDO prepare() - php

I have a variable $id which gives me the id of the current article and this can help me to make an update query in my database on current article.
This is my code:
$vizualizari = $current_views+1;
$sql1= "UPDATE detalii_cantari SET viz = viz WHERE id = {$id};";
$q1 = $dbh->prepare($sql1);
$q1->execute(array(':viz'=>$vizualizari));
I don't get any errors but my code is still not working...


Your correct code is here:
$vizualizari = $current_views+1;
$sql1= "UPDATE detalii_cantari SET viz = :viz WHERE id = {$id}";
$q1 = $dbh->prepare($sql1);
$q1->execute(array(':viz'=>$vizualizari));
; from the end of sql is not needed here and viz = viz must become viz = :viz because of PDO.


It seems you have to get rid of the previous query and make it in a single statement
$sql = "UPDATE detalii_cantari SET viz = viz + 1 WHERE id = ?";
$stm = $dbh->prepare($sql);
$stm->execute(array($id));

Related

I am trying update multiple rows using values is this correct?

$query = "UPDATE INTO Sanctions SET (idNumber, lastName,firstName, section,sanction,expireDate) VALUES('$idNumber','$lastName', '$firstName','$section','$sanction', '$dueDate') WHERE id= '$id'";

Wrong
$query = "UPDATE INTO Sanctions
SET (idNumber, lastName,firstName, section,sanction,expireDate)
VALUES('$idNumber','$lastName', '$firstName','$section','$sanction', '$dueDate')
WHERE id= '$id'";
Correct way:
$query = "UPDATE Sanctions
SET idNumber = '{$idNumber}',
lastName = '{$lastName}', ....
WHERE id = '{$id}'";
The INTO command is not valid for UPDATE query. You need to assign the table equals to (=) values for every column you want to edit.
Notes:
These query are not well secured, please use prepared statement insted. :)

update data in existing row in mysql database

I have a table called pack_details with 4 columns. I'm trying to insert new data into an existing table. Can somebody tell me what's wrong with my codes and why i have a parse error?
$sql_query = "UPDATE pack_details SET $delivery_date = $_POST["delivery_date"], $delivery_time = $_POST["delivery_time"]
WHERE $delivery_building = $_POST["delivery_building"]
AND $delivery_room = $_POST["delivery_room"]";

Try any from below options:
$sql_query = "UPDATE pack_details SET $delivery_date = '{$_POST['delivery_date']}', $delivery_time = '{$_POST['delivery_time']}' WHERE $delivery_building = '{$_POST['delivery_building']}' AND $delivery_room = '{$_POST['delivery_room']}'";
or
$sql_query = "UPDATE pack_details SET delivery_date = '".$_POST["delivery_date"]."', delivery_time = '".$_POST["delivery_time"]."' WHERE delivery_building = '".$_POST["delivery_building"]."' AND delivery_room = '".$_POST["delivery_room"]."'";
Note: If field name doesn't contain $, remove $ from field name in query. For eg. "$delivery_date" should be "delivery_date"
Suggestion: Instead of using string concatenation for building, You should use bind parameters to pass value to query. It helps to prevent SQL injection as well as code look well.

PHP MySQL Update Query not working with vars

$ID = trim($_GET["uid"]);
$Name = trim($_GET["name"]);
$result = $mysqli->query("UPDATE `Benutzer` SET `R_NAME`='$Name' WHERE `ID` = '$ID'");
The Result returns fine, but the Database is not updated. If I replace the vars with static values the Database IS updated.

Use mysqli prepare statement.
$stmt = $mysqli->prepare("UPDATE Benutzer SET R_NAME = ? WHERE ID = ?");
$stmt->bind_param($Name,$ID);
$stmt->execute();
$stmt->close();

Follow these steps:
Remove "trim" & use "mysql_escape_string".
Echo Check the values of Name & ID. Once you are getting them then follow up with the 3rd step.
Concatinate the sql string as shown by removing the tild operators:
$result = $mysqli->query("UPDATE Benutzer SET R_NAME ='".$Name."' WHERE ID = '".$ID."'");

What we pass in the query arguments is a string or we can say query in the form of string. you can change the query like below.
$result = $mysqli->query("UPDATE `Benutzer` SET `R_NAME`='".$Name."' WHERE `ID` = '".$ID."'");

SQL Multiple Statements

I am using php and mysql to update rows in my DB. I have 4 update statements in a row, yet only the last one works. I have confirmed that the statements work if they are used alone, but when I have them executed one after another only the last one executed works. I am receiving no error messages. Any help? Thanks!
$sql = "UPDATE comlog SET name='$name1', message='$message1' WHERE id=1";
$sql = "UPDATE comlog SET name='$name2', message='$message2' WHERE id=2";
$sql = "UPDATE comlog SET name='$name3', message='$message3' WHERE id=3";
$sql = "UPDATE comlog SET name='$name', message='$message' WHERE id=4";
In the above code, only the row with id 4 is being updated.

The answer is simple.
You are declaring the same variable for EACH sql string.
You need to declare it something like:
$sql1 = "";
$sql2 = "";
$sql3 = "";
$sql4 = "";

UPDATE and Increment column value POD

I'm trying to sum/add another value to the actual value in the database but this is not working. Any suggestions?
$suplies=15;
$user_id="100234";
$sql = "UPDATE table SET suplies=suplies+".$suplies." WHERE user_id=?";
$q = $conn->prepare($sql);
$q->execute(array(':suplies'=>$suplies,':user_id'=>$user_id));

Just use the named placeholder all through out.
$sql = "UPDATE table SET suplies = suplies + :suplies WHERE user_id = :user_id";

Categories