I am new to PHP. And I'm trying to prevent duplicates through php script. I use wamp server. here's the code,
$query1="select COUNT(email) from customer_info where email=$eml";
$res1=mysqli_query($db_Con,$query1);
$row=mysqli_fetch_assoc($res1);
I wanted to know if $row here has the count value. i.e If xyz#abc.com already exists(one occurance) in "customer_info", wil $row be having the value "1"??
And what exactly should be passed to mysqli_fetch_assoc() ? I got a warning saying it expects "mysqli_result" , not a boolean.
P.S: I don't want to make the column email in my table as UNIQUE.
Thanks!
try this solution
$query = "SELECT COUNT(email) AS res FROM customer_info WHERE email = $eml"; // res is a alias... can be any unreserved word
$st = mysqli_query($db_con, $query);
$row = mysqli_fetch_object($st); //mysqli_fetch_object bind the resultset as object into the variable as described here http://php.net/mysqli_fetch_object
// $row->res is a object with property "res" from your query
if($row->res > 0)
{
//has an e-mail
}
else
{
///
}
Related
My connection & sql queries have already been tested and they work. I am just trying to print one column of data into my browser for output so I can test the rest of the database. I am just trying to print one row and when I run the code, nothing shows up in the browser as output. The database column name is "type" and there is a "1" in there for the user I am logging in with. The login works but I am trying to read the integer in the database in order to direct it to a specific profile. I can't get anything to printout from my table into the browser.
$role = "SELECT type FROM fp.user WHERE usname ='$username' AND pswd = '$password'";
$access = mysqli_query($link, $role);
$row = mysqli_fetch_row($access);
echo $row['type'];
There may be a problem with your query. You should check to see if $access is first false (Which would indicate a query failure), and echo out the error associated with the query.
$role = "SELECT type FROM fp.user WHERE usname ='$username' AND pswd = '$password'";
$access = mysqli_query($link, $role);
if ( !$access )
{
echo 'There was a problem running this query: ', mysqli_error($link);
exit;
}
$row = mysqli_fetch_row($access);
Also, according to the docs:
Fetches one row of data from the result set and returns it as an enumerated array, where each column is stored in an array offset starting from 0 (zero). Each subsequent call to this function will return the next row within the result set, or NULL if there are no more rows.
Check to make sure the user/password combo in the database is correct. Another issue could be that no user exists for that $username/$password combo.
I'm trying to create an update function in PHP but the records don't seem to be changing as per the update. I've created a JSON object to hold the values being passed over to this file and according to the Firebug Lite console I've running these values are outputted just fine so it's prob something wrong with the sql side. Can anyone spot a problem? I'd appreciate the help!
<?php
$var1 = $_REQUEST['action']; // We dont need action for this tutorial, but in a complex code you need a way to determine ajax action nature
$jsonObject = json_decode($_REQUEST['outputJSON']); // Decode JSON object into readable PHP object
$name = $jsonObject->{'name'}; // Get name from object
$desc = $jsonObject->{'desc'}; // Get desc from object
$did = $jsonObject->{'did'};// Get id object
mysql_connect("localhost","root",""); // Conect to mysql, first parameter is location, second is mysql username and a third one is a mysql password
#mysql_select_db("findadeal") or die( "Unable to select database"); // Connect to database called test
$query = "UPDATE deal SET dname = {'$name'}, desc={'$desc'} WHERE dealid = {'$did'}";
$add = mysql_query($query);
$num = mysql_num_rows($add);
if($num != 0) {
echo "true";
} else {
echo "false";
}
?>
I believe you are misusing the curly braces. The single quote should go on the outside of them.:
"UPDATE deal SET dname = {'$name'}, desc={'$desc'} WHERE dealid = {'$did'}"
Becomes
"UPDATE deal SET dname = '{$name}', desc='{$desc}' WHERE dealid = '{$did}'"
On a side note, using any mysql_* functions isn't really good security-wise. I would recommend looking into php's mysqli or pdo extensions.
You need to escape reserved words in MySQL like desc with backticks
UPDATE deal
SET dname = {'$name'}, `desc`= {'$desc'} ....
^----^--------------------------here
you need to use mysql_affected_rows() after update not mysql_num_rows
I have a script that follows that is supposed to collect data from a field"UserID" in my sql table, submit all data into an array, and then compare a variable to whats in the array. If the value of the variable is already in the array, tell the user that that value is invalid.
$sql = "SELECT *" //User info
. " FROM Users" ;
$result = mysql_query($sql);
//insert where line for assessorid
$users = array();
while(($user = mysql_fetch_assoc($result))) {
$users[] = $user;
}
foreach($users as $user){
$user['UserID'];
}
I need the output of $users to be equivalent to array('user1','user2','user3');
Whats happening is data comes in from a form as $user_name. I want to use this in a statement like follows:
if(in_array($user_name,$users)){
echo "username available"
}
else{
echo "not available"}
I tried using the extract function, but that just created a big mess.
Im not sure what is incorrect about what I'm doing, unless the format of $users as an array cannot be parsed in the in_array() function as it is formatted currently. Any advice is much appreciated. Thanks!
$sql = "SELECT USERID FROM Users" ;
$result = mysql_query($sql);
$users = array();
while(($user = mysql_fetch_assoc($result))) {
$users[] = $user['USERID'];
}
When you are saying
$users[] = $user;
You are not specifying which column in the result set to be appended to the array.
Maybe I am missing something... Why not do it like this:
SELECT UserID FROM Users WHERE Username = 'username'
Then just use mysql_num_rows() to check if the username already exists or not. This should be both faster and more efficient (memory-wise).
In that case, you collect all data from the database and need to do some inefficient processing in PHP as well. It is better to query for that value to see if it is in the database, so:
$username = mysql_real_escape_string($username);
$query = "
select
count('x') as usercount
from
users u
where
u.username = '$username'";
The, if the 'usercount' is 0, the username does not exist. If > 0, the username does exist. This way, you let the database do the work it is designed to do, and the only value that is actually retreived is that single number.
Have you tried modifying your query? Currently you are getting all of the values for every user, but you just seen to need UserID. You could do this:
$sql = "SELECT UserID FROM Users";
$result = mysql_query($sql);
$users = array();
while(($user = mysql_fetch_assoc($result)))
{
$users[] = $user['UserID'];
}
// ...
if (in_array($user_name, $users))
{
echo 'Username not available';
}
else
{
echo 'Username available';
}
Or you could just look up in the database for the given username:
$sql = 'SELECT count(*) FROM Users WHERE UserID = '.mysql_escape_string($user_name);
$result = mysql_query($sql);
// and then just check if the resulting row is equal to 0
Are you attempting to write a script that will check if a username is taken?
If so, it may be easier (and more efficient) to structure the actual query towards this end rather than relying on the programmatic approach.
$sql = "SELECT COUNT(*) FROM Users WHERE Username = '$username'";
Then you could apply this result to a count and allow the user to register or not based on whether a value greater than zero (a user has already taken that name) or not (its free) is returned.
As has been mentioned, that is a rather inefficient way to check for an existing username. The suggestions for modifying your query are good advice.
However, to address the problem with the code you provided:
in_array() will not detect the presence of a value in a multi-dimensional array. Your $users array probably looks something like this:
$users = array(
array('userID', 'foo', 'bar'),
array('userID', 'foo', 'bar'),
array('userID', 'foo', 'bar')
)
and in_array will not search below the first set of indexes. If this is really what you want to do, see this question: in_array() and multidimensional array
I'm new to PHP and SQL, but I need a way to store the result of an SQL Query into a variable.
The query is like this:
$q = "SELECT type FROM users WHERE username='foo user'";
$result = pg_query($q);
The query will only return one string; the user's account type, and I just need to store that in a variable so I can check to see if the user has permission to view a page.
I know I could probably just do this query:
"SELECT * FROM users WHERE username='foo user' and type='admin'";
if(pg_num_rows($result) == 1) {
//...
}
But it seems like a bad practice to me.
Either way, it would be good to know how to store it as a variable for future reference.
You can pass the result to pg_fetch_assoc() and then store the value, or did you want to get the value without the extra step?
$result = pg_query($q);
$row = pg_fetch_assoc($result);
$account_type = $row['type'];
Is that what you are looking for?
Use pg_fetch_result:
$result = pg_query($q);
$account_type = pg_fetch_result($result, 0, 0);
But on the other hand it's always good idea to check if you got any results so I'll keep the pg_num_rows check.
Let say that we have the following query:
SELECT DISTINCT COUNT(`users_id`) FROM `users_table`;
this query will return the number of the users from a table. I need to pass this value to a PHP variable. I'm using this:
$sql_result = mysql_query($the_query_from_above) or die(mysql_error());
if($sql_result)
{
$nr_of_users = mysql_fetch_array($sql_result);
}
else
{
$nr_of_users = 0;
}
please correct my code where you think is necessary.
Which is the best approach. How do you recommend to do this ?
Like this:
// Changed the query - there's no need for DISTINCT
// and aliased the count as "num"
$data = mysql_query('SELECT COUNT(`users_id`) AS num FROM `users_table`') or die(mysql_error());
// A COUNT query will always return 1 row
// (unless it fails, in which case we die above)
// Use fetch_assoc for a nice associative array - much easier to use
$row = mysql_fetch_assoc($data);
// Get the number of uses from the array
// 'num' is what we aliased the column as above
$numUsers = $row['num'];
Also, an alternative using mysqli, which you should be using anyway for parameter interpolation:
$statement = $connection->prepare($the_query_from_above);
$statement->execute();
$statement->bind_result($nr_of_users);
$statement->fetch();