i have tried alot and this is my last code. Any ideas how i can get this work?
$id_name = $this->name->Text;
$finder = prdtblRecord::finder();
$result = $finder->findAllByname($id_name);
$row_cnt = $result->num_rows;
You can create a query that checks if the id you have exists or not
$sql = "SELECT EXISTS(SELECT * FROM YOUR_TABLE_NAME WHERE id = TO_CHECK_ID )"
then you could do something like
$id = mysql_query($sql);
if($id)
echo "success"; //your code upon success
else
echo "failed"! //your code upon failure
hope this helps
Related
In the below script I want to fetch data from mysql using a explode function and also a variable within an explode function.
Here's how I want to get
<?php
include ('config.php');
$track = "1,2,3";
$i = 1
$trackcount = explode(",",$track);
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
This is the code
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
I want sql to fetch data from tracks table where id = $trackcount[$i]
Whatever the value of $trackcount[$i] mysql should fetch but it shows a blank screen.
If I put this
$sql = "SELECT * FROM tracks WHERE id='$trackcount[1]'";
It works perfectly
save your $trackcount[$i] in one variable and then pass it in the query as given below
<?php
include ('config.php');
$track = "1,2,3";
$i = 1;
$trackcount = explode(",",$track);
$id=$trackcount[$i];
$sql = "SELECT * FROM tracks WHERE id='$id'";
$retval = mysql_query($sql, $conn);
while ($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "{$row['track_name']}";
}
mysql_free_result($retval);
?>
and one more thing check your previous code with echo of your query and see what is passing ok.
echo $sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//like this
problem is with your query
$sql = "SELECT * FROM tracks WHERE id='$trackcount['$i']'";//change
to
$sql = "SELECT * FROM tracks WHERE id='$trackcount[$i]'";
Generally you would want to use the IN operator with this type of query, so for you this would be:-
$sql="SELECT * FROM `tracks` WHERE `id` in (".$track.");";
or, if the $ids are in an array,
$sql="SELECT * FROM `tracks` WHERE `id` in (".implode( ',', $array ) .");";
Basically, I have been having some trouble with sending a request to a MySQL server and receiving the data back and checking if a user is an Admin or just a User.
Admin = 1
User = 0
<?php
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`";
$checkAdmin = $checkAdminQuery
mysqli_query = $checkAdmin;
if ($checkAdmin == 1) {
echo '<h1>Working!</h1>';
}else {
echo '<h1>Not working!</h1>';
}
?>
Sorry that this may not be as much info needed, I am currently new to Stack Overflow.
Firstly, your SQL query is wrong
SELECT * FROM `users` WHERE `admin`
It's missing the rest of the WHERE clause
SELECT * FROM `users` WHERE `admin` = 1
Then you're going to need fetch the result from the query results. You're not even running the query
$resultSet = mysqli_query($checkAdminQuery)
Then from there, you'll want to extract the value.
while($row = mysqli_fetch_assoc($resultSet))
{
//do stuff
}
These are the initial problems I see, I'll continue to analyze and find more if needed.
See the documentation here
http://php.net/manual/en/book.mysqli.php
You need to have something like user id if you want to check someone in database. For example if you have user id stored in session
<?php
// 1. start session
session_start();
// 2. connect to db
$link = mysqli_connect('host', 'user', 'pass', 'database');
// 3. get user
$checkAdminQuery = mysqli_query($link, "SELECT * FROM `users` WHERE `id_user` = " . $_SESSION['id_user'] );
// 4. fetch from result
$result = mysqli_fetch_assoc($checkAdminQuery);
// 5. if column in database is called admin test it like this
if ($result['admin'] == 1) {
echo '<h1>Is admin!</h1>';
}else {
echo '<h1>Not working!</h1>';
}
?>
// get all admin users (assumes database already connected)
$rtn = array();
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`=1";
$result = mysqli_query($dbcon,$checkAdminQuery) or die(mysqli_error($dbconn));
while($row = mysqli_fetch_array($result)){
$rtn[] = $row;
}
$checkAdminQuery = "SELECT * FROM `users` WHERE `admin`"; !!!!
where what ? you need to specify where job = 'admin' or where name ='admin'
you need to specify the column name where you are adding the admin string
I am using if else condition with foreach loop to check and insert new tags.
but both the conditions(if and alse) are being applied at the same time irrespective of wether the mysql found id is equal or not equal to the foreach posted ID. Plz help
$new_tags = $_POST['new_tags']; //forget the mysl security for the time being
foreach ($new_tags as $fnew_tags)
{
$sqlq = mysqli_query($db3->connection, "select * from o4_tags limit 1");
while($rowq = mysqli_fetch_array($sqlq)) {
$id = $rowq['id'];
if($id == $fnew_tags) { //if ID of the tag is matched then do not insert the new tags but only add the user refrence to that ID
mysqli_query($db3->connection, "insert into user_interests(uid,tag_name,exp_tags) values('$session->userid','$fnew_tags','1')");
}
else
{ //if ID of the tag is not matched then insert the new tags as well as add the user refrence to that ID
$r = mysqli_query($db3->connection, "insert into o4_tags(tag_name,ug_tags,exp_tags) values('$fnew_tags','1','1')");
$mid_ne = mysqli_insert_id($db3->connection);
mysqli_query($db3->connection, "insert into user_interests(uid,tag_name,exp_tags) values('$session->userid','$mid_ne','1')");
}
}
}
i think you are inserting
$r = mysqli_query($db3->connection, "insert into o4_tags(tag_name,ug_tags,exp_tags)
values('$fnew_tags','1','1')");$mid_ne = mysqli_insert_id($db3->connection);
and then you are using while($rowq = mysqli_fetch_array($sqlq))
which now has records you just inserted therefore your if is executed
I'm pretty sure the select query below will always return the same record.
$sqlq = mysqli_query($db3->connection, "select * from o4_tags limit 1");
I think most of the time it will goes to the else, which execute the 2 insert.
Shouldn't you write the query like below?
select * from o4_tags where id = $fnew_tags limit 1
Is it possible to re-write the code below, maybe even with an if (result > 0) statement, in just one line (or simply shorter)?
// a simple query that ALWAYS gets ONE table row as result
$query = $this->db->query("SELECT id FROM mytable WHERE this = that;");
$result = $query->fetch_object();
$id = $result->id;
I've seen awesome, extremely reduced constructs like Ternary Operators (here and here - btw see the comments for even more reduced lines) putting 4-5 lines in one, so maybe there's something for single result SQL queries like the above.
You could shorten
$query = $this->db->query("SELECT id FROM mytable WHERE this = that;");
$result = $query->fetch_object();
$id = $result->id;
to
$id = $this->db->query("SELECT id FROM mytable WHERE this = that")->fetch_object()->id;
but this, and the original code will emit errors, if any of the functions returns an unexpected response. Better to write:
$query = $this->db->query("SELECT id FROM mytable WHERE this = that");
if (!$query) {
error_log('query() failed');
return false;
}
$result = $query->fetch_object();
if (!$result) {
error_log('fetch_object() failed');
return false;
}
$id = $result->id;
I have the following code and it should return just one value (id) from mysql table. The following code doesnt work. How can I output it without creating arrays and all this stuff, just a simple output of one value.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = map_query($query);
echo $result;
I do something like this:
<?php
$data = mysql_fetch_object($result);
echo $data->foo();
?>
You have to do some form of object creation. There's no real way around that.
You can try:
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
//$result = map_query($query);
//echo $result;
$result = mysql_query($query); // run the query and get the result object.
if (!$result) { // check for errors.
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result); // get the single row.
echo $row['id']; // display the value.
all you have is a resource, you would still have to make it construct a result array if you want the output.
Check out ADO if you want to write less.
Not sure I exactly understood, what you want, but you could just do
$result = mysql_query('SELECT id FROM table WHERE area = "foo" LIMIT 1');
list($data) = mysql_fetch_assoc($result);
if you wish to execute only one row you can do like this.
$query = "SELECT id FROM users_entity WHERE username = 'Admin' ";
$result = mysql_query($query);
$row = mysql_fetch_row($result);
echo $row[0];
there have been many ways as answered above and this is just my simple example. it will echo the first row that have been executed, you can also use another option like limit clause to do the same result as answered by others above.