mysqli result amounts depending on fetch inside while - php

I am fetching rows from a mysql table (jobs). Inside of that fetch, I am also fetching from another table (accounts) [to receive account api keys all depending on what ID_ASSOC is attacted to the job]: below is the code
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
while($rows = mysqli_fetch_assoc($query)){
$username = $rows['twitter_username'];
$consumer_key = $rows['consumer_key'];
$consumer_secret = $rows['consumer_secret'];
$access_token = $rows['access_token'];
$access_token_secret = $rows['access_token_secret'];
}
echo $job_poster_id ;
echo "<br/>";
echo $twitter_username;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
OUTPUT:
specific-message
4
admin
----------------------------------
When I do this, I only get one row output..and I can't seem to find out why. I want the above out put to repeat as many times as it has rows, and it's only doing one row (with the account fetch in the code). However when I do it without the internal fetch (accounts fetch), it returns multiple rows just as desired. Why is this? (below is sample code WITHOUT the accounts fetch):
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
echo $job_poster_id ;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
OUTPUT:
specific-message
4
----------------------------------
specific-message
1
----------------------------------
specific-message
2
----------------------------------

$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
The problem is that you're using $query for the inner and the outer query.
When the inner query runs, and it steps through the loop, it's iterating to the end of the result set; when the outer while loop runs, mysqli_fetch_assoc($query) is returning false, because you're already at the end of the result set - just not the result set you were expecting.
You can fix this by renaming one of the $query variables.

Related

how to fetch all user data from database?

i want fetch all users data but am getting only one user details, please help me to solve
$conn = mysqli_connect("localhost", "root", "", "bitmining");
$sql6="SELECT username FROM users";
if($result = mysqli_query($conn, $sql6)){
while ($row=mysqli_fetch_array($result)){
//Hashrate Data Fetch
$investedusername = $row['username'];
$sql3="SELECT sum(hashrate_amount) as total FROM buyhashrate WHERE invested_username='$investedusername'";
$result = mysqli_query($conn, $sql3);
$row = mysqli_fetch_assoc($result);
//Total Value of Hashrate
echo $row['total'] . " GH/s";
echo "<br />";
}
$result->close();
}
Your re-using the $result field, change your second reference to something like ...
$result1 = mysqli_query($conn, $sql3);
$row = mysqli_fetch_assoc($result1);
This will stop it reseting the value your using for your main loop in
while ($row=mysqli_fetch_array($result)){
You use the $result variable for both mysqli_query
$result = mysqli_query($conn, $sql3);
While other answers are right about your mistake, I want to give you a better solution.
Try to use a join like this:
$conn = mysqli_connect("localhost", "root", "", "bitmining");
$sql="SELECT SUM(hashrate_amount) AS total FROM users AS t1 LEFT JOIN buyhashrate AS t2 ON (t1.username=t2.invested_username) GROUP BY t1.username";
if($result = mysqli_query($conn, $sql)){
while ($row=mysqli_fetch_array($result)){
//Total Value of Hashrate
echo $row['total'] . " GH/s";
echo "<br />";
}
$result->close();
}
This way you do just one query from database. But using your method you have n+1 queries which n is the number of users. So for one hundred users there is 101 queries.

Getting number of rows form SQL Server table

I have a problem with getting the right value after I counted the rows from a table. I searched on the web but didn't find an answer.
In the database i have a table with all the categories in it they all have an id, and i would like to count using this column.
I have this PHP code, it works but is there an other and better to get over this?
$sql2 = "SELECT COUNT(id) FROM categories";
$stmt2 = sqlsrv_query($conn, $sql2);
$res = sqlsrv_fetch_array($stmt2, SQLSRV_FETCH_ASSOC);
foreach($res as $row)
{
$rows = $row;
}
//if there are categories display them otherwise don't
if ($rows > 0)
{
$sql = "SELECT * FROM categories";
$stmt = sqlsrv_query($conn, $sql);
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
echo "<a href='#' class='cat_links'>" . $row['category_name'] . " - <font size='-1'>" . $row['category_description'] . "</font></a>";
}
}
else
{
echo "<p style='text-align: center'>No categories yet.</p>";
}
I think has to be a better way to convert the $stmt2 variable from a SQL resource to an actual number, or to convert the $res variable from an array to an number. If I try to echo the whole array using foreach, it will only print out the number of rows. This is why I use it to count the rows now.
I can't use the sqlsrv_num_rows function because I then get an error, or no answer.

How to display the content of more then one SQL database row

I am trying to retrieve the data stored in an unknown number of mySQL database rows and display them using HTML. At the moment I can display data from one row.
Each row has a unique id number, I was planning to iterate through by comparing this number to the variable counter. But it would then leave me with the issue of displaying the results in HTML. At the moment I am just echoing variables that contain data from the rows. However what I want to create is a HTML list that increases in length depending on how many rows are in the table.
Here is my current PHP code for retrieving a row from the database is:
$sql = "SELECT * FROM project_tasks WHERE project_name='$proj_name' AND task_id='$counter'";
$query = mysqli_query($db_conx, $sql);
$row = $query->fetch_assoc();
$task_id = $row["task_id"];
$proj_name = $row["project_name"];
$task_name = $row["task_name"];
$task_importance = $row["task_importance"];
$task_description = $row["task_description"];
$task_deadline = $row["task_deadline"];
$task_members = $row["task_members"];
$task_budget = $row["task_budget"];
At the moment I am just displaying some of the results in HTML using this code:
<div id="inner_container">
<?php echo "$task_id $proj_name $task_name $task_deadline"; ?>
</div>
$sql = "SELECT * FROM project_tasks WHERE project_name='$proj_name' AND task_id='$counter'";
$query = mysqli_query($db_conx, $sql);
while($row = $query->mysqli_fetch_assoc()) {
$task_id = $row["task_id"];
$proj_name = $row["project_name"];
$task_name = $row["task_name"];
$task_importance = $row["task_importance"];
$task_description = $row["task_description"];
$task_deadline = $row["task_deadline"];
$task_members = $row["task_members"];
$task_budget = $row["task_budget"];
echo "$task_id $proj_name $task_name $task_deadline";
}
Since you have built an associative array using fetch_assoc all you need to do is loop through that array. The OO example on http://php.net/manual/en/mysqli-result.fetch-assoc.php should get you what you need. A quick example:
$sql = "SELECT * FROM project_tasks WHERE project_name='$proj_name' AND task_id='$counter'";
$query = mysqli_query($db_conx, $sql);
echo '<div id="inner_container">';
while ($row = $query->fetch_assoc()) {
$proj_name = $row["project_name"];
$task_name = $row["task_name"];
$task_deadline = $row["task_deadline"];
echo "$task_id $proj_name $task_name $task_deadline";
};
/* free result set */
$row->free();
echo '</div>;

PHP - how to print each row of an array

I feel like I'm missing something really simple here. Here's my sql query:
$getpages = "SELECT id FROM pages WHERE account = 2 ORDER BY page_order";
$showpages = #mysqli_query ($dbc, $getpages);
$row = mysqli_fetch_array($showpages, MYSQLI_NUM);
I then print the first result:
echo $row[0];
And get a correct value, the id of the first page (by page order):
10
So I submit a form which simply turns that $row[0] into $row[1].
And nothing prints. I don't understand.
You have to do this:
while ($row = mysqli_fetch_array($showpages, MYSQLI_NUM)) {
$id = $row[id];
// do something with the id
echo $id . "<br/>"; // Echo every id
}
This will iterate through all of the results
mysqli_fetch_array is a special function as each time you call it, it outputs the NEXT row.
So:
while ($row = mysqli_fetch_array(stuff)){
$var = $row['sql column'];
// In your case "$var = $row[0];" to get the id for the first row found.
}
is how you use it.
This will keep running the function until mysqli_fetch_array() eventually returns false. These each time it will output a new row. And the $row array is the array of one row in the sql table.
Use :
print_r($row);
in the while loop to see exactly what's being returned for use.
$getpages = "SELECT id FROM pages WHERE account = 2 ORDER BY page_order";
$showpages = #mysqli_query ($dbc, $getpages);
$row = mysqli_fetch_array($showpages, MYSQLI_NUM);
Question how many rows are fetched in this query in this case id, So MYSQL will result to 1 row affected, But since you are using
$row = mysqli_fetch_array($showpages, MYSQLI_NUM);
This code you used will not output anything try
$row[n] as n-1
$result->fetch_assoc() or mysqli_fetch_assoc($result)
catch one line at a time, from a mysqli_result:
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
source: http://www.w3schools.com/php/php_mysql_select.asp

PHP MySQL single column/row display?

I'm trying to get a single result from my database, just one name.
I tried using;
$row = mysql_fetch_array(mysql_query("SELECT * FROM persons WHERE id = '$id'"));
echo $row['name'];
But that din't work, any other way to simply show only one result?
Thanks in advance!
[EDIT:]
(I'm using PHP 5.3)
<?php
include("connection.php");
$id = $_GET['deletid'];
$result = mysql_query("SELECT * FROM persons WHERE id = '$id' LIMIT 1");
if(!$result){
echo mysql_error();
}
if ($row = mysql_fetch_array($result)){
echo $row['name'];
}
echo "<p>id:$id</p>";
?>
If you need just the name and you need just one result you should rewrite your query as follow:
$row = mysql_fetch_array(mysql_query("SELECT name FROM persons WHERE id = '". (int) $id ."' LIMIT 1"));
Now to get the result you should just get it with a
$row['name'];
EDIT
Now that you posted your entire code i got what's wrong: You are deleting that result before getting its name. Basically you delete that user and then you attempt to get its name.
EDIT
<?php
include("connection.php");
if (empty($_GET['deleteid'])) {
exit('"deleteid" is empty');
}
$id = mysql_real_escape_string($_GET['deletid']);
$result = mysql_query("SELECT name FROM persons WHERE id = '". (int) $id ."' LIMIT 1");
if(!$result){
echo mysql_error();
}
$row = mysql_fetch_assoc($result); // for just one result you don't need of any loop
echo $row['name'];
echo "<p>id:". htmlspecialchars($id) ."</p>";
?>
try
$row = mysql_fetch_array(mysql_query("SELECT name FROM persons WHERE id = ". (int) $id));
echo $row['name'];

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