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I am a beginner in PHP and SQL. I have been trying to delete rows in SQL table using the following code but it doesn't work. Please help.
<?php
/*
DELETE.PHP
Deletes a specific entry from the 'db' table
*/
// connect to the database
include('connect-db.php');
// check if the 'id' variable is set in URL, and check that it is valid
// get id value
$id = $_GET['id'];
// delete the entry
$result = mysql_query("DELETE FROM db WHERE 'Report No.'= '$id'")
or die(mysql_error());
// redirect back to the view page
header("Location: view.php");
// if id isn't set, or isn't valid, redirect back to view page
{
header("Location: view.php");
}
?>
Apply backticks(`) around table field name "Report No." (its not standard way to define a table field name)
Try this
$result = mysql_query("DELETE FROM db WHERE `Report No.`= '$id'");
Fix your query by removing single quote of name table:
$result = mysql_query("DELETE FROM db WHERE `Report No.`= '$id'");
Make sure that you type right for Report No. column name. Actually for naming Report No. is not recommended.
Related
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When I put a number myself the database gets updated to Success but when I try to access the same number from $_POST It complete the transaction but doesn't affect the rows even thought it's using the same number.
Example of a query that works perfectly and updates the database
$sql = "UPDATE `transactions` SET `status` ='Success' WHERE `transactions`.`txn_id` = 65765756";
Example of a query that doesn't work
$sql = "UPDATE `transactions` SET `status` ='Success' WHERE `transactions`.`txn_id` = ".$_POST['m_payment_id'];
First, check what's in the value, and make sure it is the same as what you are manually entering.
var_dump($_POST['m_payment_id'])
Second, the code without any other checks is a SQL injection vulnerability.
You could convert the value into an integer to protect against this, e.g. intval($_POST['m_payment_id'])
Ideally though, you would be using bindings.
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I'm a begginer to PHP and I want to know how can I fetch some text from the corresponding ID and store it as a variable in PHP.
The table is like
ID----NAME----ACCOUNT----PASSWORD
1----name1----accont1----password2
2----name2----accont2----password2
3----name3----accont3----password3
Now if I want to get the account2 as text and save it in an variable (say acc2) then what should I do. Assuming that I have connection information in connect.php.
Edit: I want to select the account2 using the ID like from ID 2 select account.
Thanks In Advance!!!
Assuming you use MySQL, the table is named users and you are using PDO, this would get what you need:
$stmt = $conn->query("SELECT * FROM users WHERE ID = 2");
$row = $stmt->fetch()
$account = $row['ACCOUNT']
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I believe it has to be simple, but I'm totally green in SQL.
I will include pictures of the database so it will be easier to imagine.
I use it in my PHP code that is accessed by Flash to enter data into the database.
This is my code which doesn't work:
$sql = "SELECT * FROM users
WHERE username = '$username' AND password = '$password'
INSERT INTO users (contactlist) VALUES ('$xmlcontactlist1')";
I want the data from variable $xmlcontaclist1 were entered to 'contactlist' column but to specific User (based on their Name and Password). Somehow when the code was doing something it was creating NEW empty space in the database with just contactlist instead of adding it for each user.
Database Screenshot
Try this one:
$sql = "UPDATE users
SET contactlist = '$xmlcontactlist1'
WHERE username = '$username'
AND password = '$password'";
But this is a bad practice. You can get SQL injections with this code. Read this post here to prevent this: How can I prevent SQL injection in PHP?
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I'm trying to do in-place editing way in my site.
Now I set up all the things I need.
When the user clicks Submit, it will send the id of the div element (what kind of content) and the new value to update.
Here's my code:
if($pedit = $mysqli->prepare("UPDATE `accounts` SET ? = ? WHERE `id`= ? ")){
$pedit->bind_param("sss", $id, $value, $_SESSION["user_id"]);
$pedit->execute();
$pedit->free_result();
$pedit->close();
}
I don't know why it doesn't update the information.
$id = the row that change: description, fullname, email etc.
$value = the new information about $id. User can update his profile information.
The code doesn't show me any kind of error but still doesn't update.
You can't specify a column name as a parameter in a prepared statement. You'll instead have to substitute column names into the statement before preparing it. Don't forget to whitelist editable column names to make sure no unwanted SQL gets executed.
<?php
$accounts_editable_cols = array(
'name'=>true, 'street'=>true, 'city'=>true,
'region'=>true, 'postal'=>true, 'phone'=>true
);
// prevent SQL injection by whitelisting column names
if (!array_key_exists($id, $accounts_editable_cols)) return false;
$pedit = $mysqli->prepare("UPDATE `accounts` SET $id = ? WHERE `id`= ? ")
if ($pedit) {
$pedit->bind_param("ss", $value, $_SESSION["user_id"]);
$pedit->execute();
$pedit->free_result();
$pedit->close();
}
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I have problem only if I use select * but if I select exact field from my database it is working fine
$sql = "SELECT * FROM `product id`;";
$resutl = mysql_query($sql);
while ($row = mysql_fetch_assoc($result))
{
echo $row["product name"];
};
It is working if I use
SELECT `product name` FROM `product id`
Thank you
$sql = "SELECT * FROM `product id`";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result))
{
print_r($row);
}
and check your array and traverse it perfectly as when you call all rows it will not be same as fetching one row.
Try this it should work fine, and you will get more idea.
only 2 errors I found is $result variable was not correct and semicolon in query!
Table names e column names that include white-space are not a good idea because may be in conflict with mysql sintax (when mysql parse the query). You can use var_dump($row).
Use mysql_fetch_array() instead of mysql_fetch_assoc()
While the two are similar mysql_fetch_assoc() only returns an associative array.
Also you should think of moving from mysql to mysqli or PDO. mysql is being removed as of PHP6 and already depreciated as of PHP5.5.