how to change mysql_result($res, 0, "url"); to mysqli - php

I don't know how to use mysqli_result like past.
is this ok? how is correct?
$res = mysqli_query($con,"SELECT `id`,`usrid`,`url` FROM site WHERE state = 'popup' AND creditsp >= 1 AND (cth < cph || cph=0) order by rand() limit 1");
$urll = mysqli_result($res, 0, "url");
$ownerid = mysqli_result($res, 0, "usrid");
$siteidd = mysqli_result($res, 0, "id");

PHP 5.4 now supports function array dereferencing:
http://php.net/manual/en/migration54.new-features.php
so, in order to get the results from sql you need to use mysqli_fetch_assoc() function.
Usage can be found here:
http://php.net/manual/en/mysqli-result.fetch-assoc.php
http://www.w3schools.com/php/func_mysqli_fetch_assoc.asp
________Small Example________
Suppose we have a DB:
Id Name Age Occupation
1. William 11 Student
2. Uname 14 Student
3. Yem 22 Teacher
$query = "SELECT * FROM persons";
$res = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($res)); \\returns an associative array to the row variable. You can use foreach loop as well to loop throgh the various data items.
{
echo $row["Id"] . "<br>";
echo $row["Name"] . "<br>";
echo $row["Age"] . "<br>";
echo $row["Occupation"] . "<br>";
}

Use fetch_assoc on the result:
$row = $res->fetch_assoc();
$urll = $row['url'];
$ownerid = $row['usrid'];
$siteidd = $row['id'];
If there are more than one row, then:
while ($row = $res->fetch_assoc()) {
// process $row
}

Related

Display count of values in PHP array

I can not seem to come over this easy problem.
I have the following code, where I select the column "status" from a table. There's three different values of "status", 0, 1 and 3.
I would like to count how many 0's there are, 1's and 2's, so that I can display them via <?php echo $accepted; ?> etc.
<?php
$sql = "SELECT status FROM applications";
$result = mysqli_query($db, $sql);
$row = mysqli_fetch_array($result, MYSQLI_NUM);
$array = array_count_values($row);
$pending = $array[0];
$accepted = $array[1];
$denied = $array[2];
?>
MYSQL_NUM changed to MYSQLI_NUM accordingly to comment, thank you.
A simple way is get these count using SQL
$sql = "SELECT `status`, COUNT(*) as cnt FROM applications GROUP BY `status`";
you have an array of so you should
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
echo "status = " . $array['status'] . ' = ' . $array['cnt'] .'<br />';
}

mysql query array counter

Apologies if I have the terminology wrong.
I have a for loop in php which operates a mysql query...
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
}
The results that are given when I insert a...
print_r ($job_results);
On screen -> Array()
If I change the query from $user_id_pc[$i] to $user_id_pc[14] for example I receive one set of results.
If I include this code after the query and inside the for loop
echo $i;
echo $user_id_pc[$i] . "<br>";
I receive the number the counter $i is on followed by the data inside the array for that counter position.
I am not sure why the array $job_results is empty from the query using the counter $i but not if I enter the number manually?
Is it a special character I need to escape?
The full code
<?php
print_r ($user_id_pc);
//Select all columns to see if user has a profile
$query = "SELECT * FROM user_profile WHERE user_id = '" . $_SESSION['user_id'] . "'";
//If the user has an empty profile direct them to the home page
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 0)
{
echo '<br><div class="alert alert-warning" role="alert"><h3>Your appear not to be logged on please visit the home page to log on or register. <em>Thank you.</em></h3></div>';
}
//Select data from user and asign them to variables
else
{
$data = mysqli_query($dbc, $query);
if (mysqli_num_rows($data) == 1)
{
$row = mysqli_fetch_array($data);
$cw_job_name = $row['job_description'];
$cw_rate = $row['hourly_rate'];
$job_mileage = $row['mileage'];
$job_postcode = $row['postcode'];
$response_id = $row['user_profile_id'];
}
}
for ($i = 0; $i <count($user_id_pc); $i++)
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
$job_data = mysqli_query($dbc, $query2);
$job_results = array();
while ($row = mysqli_fetch_array($job_data))
{
array_push($job_results, $row);
}
echo $i;
?>
<br>
<?php
}
print ($query2);
print $user_id_pc[$i];
?>
This is primarily a syntax error, the correct syntax should be:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '{$user_id_pc[$i]}'";
Note that this is correct syntax but still wrong!! For two reasons the first is that it's almost always better (faster, more efficient, takes less resources) to do a join or a subquery or a simple IN(array) type query rather than to loop and query multiple times.
The second issue is that passing parameters in this manner leave your vulnerable to sql injection. You should use prepared statements.
The correct way
if(count($user_id_pc)) {
$stmt = mysqli_stmt_prepare(" SELECT job_title, job_info FROM job_description WHERE postcode_ss = ?");
mysqli_stmt_bind_param($stmt, "s", "'" . implode("','",$user_id_pc) . "'");
mysqli_stmt_execute($stmt);
}
Note that the for loop has been replaced by a simple if
You have to check the query variable, instead of:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '$user_id_pc[$i]'"
have you tried this:
$query2 = " SELECT job_title, job_info FROM job_description WHERE postcode_ss = '" . $user_id_pc[$i] . "' ";
And another thing, try something different like this:
while ($row = mysqli_fetch_array($job_data))
{
$job_results[] = array("job_title" => $row["job_title"], "job_info" => $row["job_info");
}
Then try to print the values.
Sorry but I like foreach(), so your working code is:
<?php
// To store the result
$job_results = [];
foreach($user_id_pc as $id ){
// selecting matching rows
$query2 ="SELECT job_title, job_info FROM job_description WHERE postcode_ss = '".$id."'";
$job_data = mysqli_query($dbc, $query2);
// checking if query fetch any result
if(mysqli_num_rows($job_data)){
// fetching the result
while ($row = mysqli_fetch_array($job_data)){
// storing resulting row
$job_results[] = $row;
}
}
}
// to print the result
var_dump($job_results);

php Print_f array error

I am doing a php to push GCM notification, before that I need to grab data from database and send to google GCM server.
Code is below:
mysql_select_db($database_gcm_gcm, $gcm_gcm);
$result = mysql_query("SELECT id, RegID FROM gcm_user ORDER BY id ASC") or die(mysql_error());
$list_arr = array(array());
for($i=0; ($row = mysql_fetch_array($result)); $i++){
$list_arr[$i]=$row;
print_r ($row . '<br>');
}
print_r($result . '<br>');
?>
It suppose to show a result like
Array
(
[0] => 1, RegID 1
[1] => 2, RegID 2
)
How ever it only shows word "Array" and Resource id #4.
Which part I am doing wrong?
Thanks
$list_arr = [];
$resultCount = mysql_num_rows($result);
if ($resultCount > 0)
{
while($row = mysql_fetch_array($result))
{
$list_arr[$i] = $row['id'].','.$row['RegID '];
}
}
You cannot print a query with a for. You have to use do while.
Try this code:
mysql_select_db($database_gcm_gcm, $gcm_gcm);
$result = mysql_query("SELECT id, RegID FROM gcm_user ORDER BY id ASC") or die(mysql_error());
$row = mysql_fetch_array($result);
do{
echo $row['id'] ." " .$row['RegID'] ."<br />";
} while ($row = mysql_fetch_array($result));

MySQL contains variable

I have an array in PHP that is looping through a set of names (and a corresponding quantity). I would like to print the ones found in a MYSQL database table (to which I've succesfully connected). I'm currently using the code:
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
$row = mysql_fetch_array($result);
echo $quan." ".$row['name']. "\n";
}
}
For some reason, this only prints the last quantity and name in the array. Help?
For example, if the array has key-value pairs of {A-4, B-2, C-3}, and table contains {A, B, D} as names ... it'll only print "2 B".
Change the code to the following:
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_array($result)) {
echo $quan." ".$row['name']. "\n";
}
}
}
You have to loop through the result. By the way, stop using mysql_query()! use MySQLi or PDO instead ( and be careful of SQL Injection ; you can use mysqli_real_escape_string() to handle input parameters ). For MySQLi implementation , here it is :
foreach ($arr as $name => $quan) {
$query = "SELECT * FROM table WHERE name='$name'";
$result = mysqli_query($query) or die(mysqli_error());
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
echo $quan." ".$row['name']. PHP_EOL;
}
}
}
And rather "\n", I prefer using PHP_EOL ( as shown above )
And as the comment suggests, the SQL statement can be executed once, as follow:
$flipped_array = array_flip($arr); // flip the array to make "name" as values"
for($i = 0; $i < count($flipped_array); $i++) {
$flipped_array[$i] = '\'' . $flipped_array[$i] . '\''; // add surrounding single quotes
}
$name_list = implode(',', $arr);
$query = "SELECT * FROM table WHERE name IN ($name_list)";
// ... omit the followings
e.g. in $arr contains "peter", "mary", "ken", the Query will be:
SELECT * FROM table WHERE name IN ('peter','mary','ken')
sidenote: but I don't understand your query. You only obtain the name back from the query? You can check number of rows, or you can even group by name, such as:
SELECT name, COUNT(*) AS cnt FROM table GROUP BY name ORDER BY name
to get what you want.
UPDATE (again): based on the comment of OP, here is the solution :
$flipped_array = array_flip($arr); // flip the array to make "name" as values"
for($i = 0; $i < count($flipped_array); $i++) {
$flipped_array[$i] = '\'' . $flipped_array[$i] . '\''; // add surrounding single quotes
}
$name_list = implode(',', $arr);
$query = "SELECT name, COUNT(*) AS cnt FROM table WHERE name IN ($name_list) GROUP BY name HAVING COUNT(*) > 0 ORDER BY name";
$result = mysqli_query($query) or die(mysqli_error());
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)) {
echo $quan." ".$row['name']. ": " . $row['cnt'] . PHP_EOL;
}
}
The above query will show the name appearing in the table only. Names not in table will not be shown. Now full codes ( be cautious of SQL Injection , again )

Get sum of MySQL column in PHP

I have a column in a table that I would like to add up and return the sum. I have a loop, but it's not working.
while ($row = mysql_fetch_assoc($result)){
$sum += $row['Value'];
}
echo $sum;
You can completely handle it in the MySQL query:
SELECT SUM(column_name) FROM table_name;
Using PDO (mysql_query is deprecated)
$stmt = $handler->prepare('SELECT SUM(value) AS value_sum FROM codes');
$stmt->execute();
$row = $stmt->fetch(PDO::FETCH_ASSOC);
$sum = $row['value_sum'];
Or using mysqli:
$result = mysqli_query($conn, 'SELECT SUM(value) AS value_sum FROM codes');
$row = mysqli_fetch_assoc($result);
$sum = $row['value_sum'];
$query = "SELECT * FROM tableName";
$query_run = mysql_query($query);
$qty= 0;
while ($num = mysql_fetch_assoc ($query_run)) {
$qty += $num['ColumnName'];
}
echo $qty;
Try this:
$sql = mysql_query("SELECT SUM(Value) as total FROM Codes");
$row = mysql_fetch_array($sql);
$sum = $row['total'];
Let us use the following image as an example for the data in our MySQL Database:
Now, as the question mentions, we need to find the sum of a particular column in a table. For example, let us add all the values of column "duration_sec" for the date '09-10-2018' and only status 'off'
For this condition, the following would be the sql query and code:
$sql_qry = "SELECT SUM(duration_sec) AS count
FROM tbl_npt
WHERE date='09-10-2018' AND status='off'";
$duration = $connection->query($sql_qry);
$record = $duration->fetch_array();
$total = $record['count'];
echo $total;
MySQL 5.6 (LAMP) . column_value is the column you want to add up. table_name is the table.
Method #1
$qry = "SELECT column_value AS count
FROM table_name ";
$res = $db->query($qry);
$total = 0;
while ($rec = $db->fetchAssoc($res)) {
$total += $rec['count'];
}
echo "Total: " . $total . "\n";
Method #2
$qry = "SELECT SUM(column_value) AS count
FROM table_name ";
$res = $db->query($qry);
$total = 0;
$rec = $db->fetchAssoc($res);
$total = $rec['count'];
echo "Total: " . $total . "\n";
Method #3 -SQLi
$qry = "SELECT SUM(column_value) AS count
FROM table_name ";
$res = $conn->query($sql);
$total = 0;
$rec = row = $res->fetch_assoc();
$total = $rec['count'];
echo "Total: " . $total . "\n";
Method #4: Depreciated (don't use)
$res = mysql_query('SELECT SUM(column_value) AS count FROM table_name');
$row = mysql_fetch_assoc($res);
$sum = $row['count'];
$row['Value'] is probably a string. Try using intval($row['Value']).
Also, make sure you set $sum = 0 before the loop.
Or, better yet, add SUM(Value) AS Val_Sum to your SQL query.
$result=mysql_query("SELECT SUM(column) AS total_value FROM table name WHERE column='value'");
$result=mysql_result($result,0,0);
Get Sum Of particular row value using PHP MYSQL
"SELECT SUM(filed_name) from table_name"
$sql = "SELECT SUM(Value) FROM Codes";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
sum = $row['SUM(price)'];
}
echo sum;

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