Hello I have a Database named as "admin" in which i have two tables
Table 1 Name = "register"
Table 2 Name = "noti"
In Register Table i've approx more than 10+ User entries which comes through Registration Page
In Noti Table, its empty at this time (Column Name is also "noti")
I want to perform this thing
First I want to count the total no. of records in "register" table
and it checks, if the records are greater than ZERO then it runs the INSERT query otherwise it runs the UPDATE Query
And i want to INSERT and UPDATE that count value into "noti" table
Here's my code
<?php
include('config.php');
$sql2 = "SELECT count(*) as count FROM register";
$result2 = mysqli_query($con, $sql2);
if($result2->num_rows>0)
{
while($rw1=$result2->fetch_array())
{
$value1 = $rw1['count'];
$result = mysqli_query($con, "SELECT count(*) as count FROM register ");
if(!empty($value1)) {
mysqli_query($con, "UPDATE noti SET noti = '$value1' ");
}
else
{
mysqli_query($con, "INSERT INTO noti(noti) VALUES ('$value1') ");
}
}
}
?>
Use this:
include('config.php');
$sql2 = "SELECT count(*) as count FROM register";
$result2 = mysqli_query($con, $sql2);
$count = $result->num_rows;
if($count != 0)
{
mysqli_query($con, "UPDATE noti SET noti = '$count' ");
}
else
{
mysqli_query($con, "INSERT INTO noti(noti) VALUES ('$value1') ");
}
Related
I did query to the database where I need the results from it and then store it in a variable. Then I will pass the variable to the INSERT INTO statement but for some reason my code does not work. This is my code/
$query = "SELECT * from animals where old= 1 AND user_id=".$_SESSION['user_id'];
$result = mysqli_query(mysqli_connect("","","", ""), $query);
while ($row = mysqli_fetch_array($result))
{
$variable[] = $row['number'];
}
//now I will pass the $variable to the INSERT INTO statement
if(isset($_POST['submit_d']))
{
foreach($variable as $var)
{
$query="INSERT INTO selectedanimals(number) VALUES ({$var},2)";
mysqli_query($con, $query) or die (mysql_error());
}
?>
<script>
alert("Animal added.");
self.location="chooseAnimals.php";
</script>
<?php
}
?>
You can use INSERT INTO ... SELECT for this purpose in a single query:
INSERT INTO selectedanimals (number)
SELECT number
FROM animals
WHERE old = 1 AND user_id = some_id
PHP code:
$query = "INSERT INTO selectedanimals (number) ";
$query.= "SELECT number FROM animals WHERE old = 1 AND user_id = ".$_SESSION['user_id'];
mysqli_query($con, $query) or die (mysql_error());
Ok. I ran into an issue. My code is able to insert records into my merchandise table. I truncated the table and a record is still inserted into the table but with an error "Undefined variable last_id". I assume that this is because when the table was truncate, there isn't a previous id since the record being inserted is the FIRST. Can someone help me resolve this issue. Thanks!
$sql = "SELECT m_id FROM merchandise";
$result = mysqli_query($connection, $sql);
while($row = mysqli_fetch_assoc($result)) {
$last_id = $row["m_id"];
}
$next_id = $last_id+1;
$conc = number_format($next_id/100,2,'-','');
$query = "INSERT INTO merchandise (mfr,type,description,mer_sku,price,qty) ";
$query .="VALUES ('$mfr','$type','$desc','MR{$mfr}{$conc}','$price','$qty')";
$add_sku_query = mysqli_query($connection, $query);
Declare last id as zero in case there are no rows.
$last_id = 0;
$sql = "SELECT m_id FROM merchandise";
$result = mysqli_query($connection, $sql);
while($row = mysqli_fetch_assoc($result)) {
$last_id = $row["m_id"];
}
$next_id = $last_id+1;
I am looking to count the number of times 'yes' in present for a user in a table, then post the result into anther table for that same user. Both tables have the username. I would like this done for each user. I have the following but it is not working.
$sql = $item_count = "SELECT SUM(if(strike='yes',1,0)) AS strike_total FROM weekpicks WHERE username = 'username'";
// execute SQL query and get result
$sql_result = mysql_query($sql) or die (mysql_error());
if (!$sql_result) {
echo "Something has gone wrong!";
}
else {
//loop through record and get values
while ($row = mysql_fetch_array($sql_result)) {
$item_result = ($row = #mysql_query($item_count)) or die(mysql_error());
$strike_total = ($row = #mysql_result($item_result,"strike_total"));
$strikes = ($row = $strike_total ['strike_total']);
$username = $row["username"];
// the following will insert number of strikes into table for each user.
$sql = "UPDATE authorize SET strikes = '($strikes)' WHERE username='$username'";
//mysql_query(" UPDATE authorize SET " . "strikes = '" . ($strikes) . "' WHERE username='$username' ");
$result = mysql_query($sql) or die (mysql_error());
Just one query should be enough
Update for single user..
UPDATE authorize SET strikes = (select count(*) from weekpicks WHERE username = '$username' and strike='yes') WHERE username='$username';
For bulk update all users
UPDATE authorize as A SET strikes = (select count(*) from weekpicks B WHERE strike='yes' and A.username=B.username group by B.username)
Isn't that simple.
I have a table with a column called 'status'. The defaut value of 'status' is 0. I want to update the value to '1' after using it.
I basically want to check if the status is 0, if it is, do an operation and then change the value to 1.
Here is the code. All works perfectly except that the value of 0 is not changed to 1.
I am a novice so maybe is a very basic mistake :(
<?php
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$sql = "UPDATE summonerid SET status='1' WHERE status='0';"; // THIS IS THE PART THAT DOES NOT WORK WELL
}
}
}
?>
Any help would be highly appreciated
Try this.. you are not executing the sql statements
$sql = "SELECT number, status FROM summonerid";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$SummonerID = $row["number"];
$status = $row["status"];
if($status=='0'){
$recentgames=$lol->getRecentGames($SummonerID);
$MatchID1=$recentgames->games[0]->gameId;
$sql = "INSERT INTO matchid (number) SELECT * FROM (SELECT '$MatchID1') AS tmp WHERE NOT EXISTS (SELECT number FROM matchid WHERE number = '$MatchID1') LIMIT 1;";
$result1 = $conn->query($sql);
$sql = "UPDATE summonerid SET status='1' WHERE status='0';";
$result1 = $conn->query($sql);
}
}
}
?>
I have sql + php query and i need inform user when update fail exmpl:
$sql = "UPDATE db SET
date = GetDate(),
...
...
...
WHERE name = '$name1' and code = '$code' and value1 = '$value1' and value2='$value2'
";
sqlsrv_query( $con, $sql);
And now if php variables values not 100% match values in db update fails but users cant see that. He can check records and try again. I would like inform him when query update nothing.
Like GOB commented, you can use the PHP sqlsrv_rows_affected function to retrieve the number of affected rows. For example:
$stmt = sqlsrv_query( $conn, $sql , $params, $options );
$row_count = sqlsrv_rows_affected( $stmt );
if ($row_count === false)
echo "Error in retrieving row count.";
else
echo $row_count;
Before directly executing update query,check whether condition in update query exists or not. This can be done by selecting count of that condition.
Try below code:
$sql = "select count(*) as count from db WHERE name = '$name1' and code = '$code' and value1 = '$value1' and value2='$value2' ";
while($row = mysqli_fetch_array($sql))
{
$count = $row['count'];
}
if($count == 0)
{
echo 'update will fail';
}
else
{
$sql = "UPDATE db SET
date = GetDate(),
...
...
...
WHERE name = '$name1' and code = '$code' and value1 = '$value1' and value2='$value2'
";
}