Cannot figure out where to put ORDER BY - php

I know where not to put it, at least.
I inherited this project from my predecessor. I know very little about PHP and SQL Server, but here we are. The first while loop contains the problem. I need to ORDER BY UserType, but obviously it is just ordering one entry every time because of where it is in the while loop. The result is meant to be that the results are shown with one user type first in alphabetical order, then the second user type in alphabetical order.
Google, and searching here, have only been moderately enlightening because most examples don't seem quite this complicated. It may be something simple, but I don't see it. Any help would be appreciated.
$sql = "SELECT UserID FROM vw_AgentService WHERE StateID = ".$state.
"AND OrgID = ".$org.
"AND ServiceID = ".$service; //JOIN
$stmt = sqlsrv_query($conn, $sql);
if ($stmt === false) {
die(print_r(sqlsrv_errors(), true));
}
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
$sqlb = "SELECT * FROM vw_UserInfo WHERE UserId = ".$row["UserID"].
" AND (UserType = 1 OR UserType = 2) ORDER BY UserType DESC, UserLastName"; //Herein lies the problem
$stmtb = sqlsrv_query($conn, $sqlb);
if ($stmtb === false) {
die(print_r(sqlsrv_errors(), true));
}
while ($rowb = sqlsrv_fetch_array($stmtb, SQLSRV_FETCH_ASSOC)) {
$sqlc = "SELECT * FROM vw_UserPhotoPath WHERE UserID = ".$row["UserID"];
$stmtc = sqlsrv_query($conn, $sqlc);
if ($stmtc === false) {
die(print_r(sqlsrv_errors(), true));
}
while ($rowc = sqlsrv_fetch_array($stmtc, SQLSRV_FETCH_ASSOC)) {
if ($rowc['PhotoFilePath'] === null) {
echo "<a href=\"profile.php?id=".$rowb["UserID"].
"\">".
"<li><img src=\"profile/blank-avatar.png\" width=\"100\" />";
} else {
echo "<a href=\"profile.php?id=".$rowb["UserID"].
"\">".
"<li><img src=\"http://demo-bc.cmfirsttech.com:8081/".$rowc['PhotoFilePath'].
"\" width=\"100\" />";
}
}
echo $rowb["UserFirstName"].
" ".$rowb["UserLastName"].
"<br/><span class=\"info\">".$rowb["UserTitle"].
", ".$rowb["UserCompany"]; // Should the ORDER BY actually go here?
//.</span></a></li>"; //rm </br>, add ", "
}
}
I've been told that the ORDER BY usually goes at the end, but I am not sure I understand the logic because at that point it is just echoing the data I already have.

The ORDER BY clause should be added to the end of the SQL statement.
Something like this;
$sqlc = "SELECT * FROM vw_UserPhotoPath WHERE UserID = ".$row["UserID"] . " ORDER BY UserType";
That's assuming UserType is in your source view (vw_UserPhotoPath).
EDIT 1 - Should have read the question properly
You actually need your ORDER BY on the very first SQL call, ie;
$sql = "SELECT UserID FROM vw_AgentService WHERE StateID = ".$state." AND OrgID = ".$org. "AND ServiceID = ".$service . " ORDER BY UserType";
This is because the second query only returns one record. This first query is what's getting the list so where the order by should be placed.

Related

max(id) and limit 10, but use them in different places

I have two tables, posts and sections. I want to get the last 10 posts WHERE section = 1,
but use the 10 results in different places. I make a function:
function sectionposts($section_id){
mysql_set_charset('utf8');
$maxpost1 ="SELECT max(id) from posts WHERE section_id = $section_id ORDER BY ID DESC LIMIT 20";
$maxpost12 =mysql_query($maxpost1);
while ($maxpost_rows = mysql_fetch_array($maxpost12 ,MYSQL_BOTH)){
$maxpost2 = $maxpost_rows[0];
}
$query = "SELECT * FROM posts WHERE id = $maxpost2";
$query2 = mysql_query($query);
return $query2;
}
$query2 = sectionposts(6);
while ($rows = mysql_fetch_array($query2)){
echo $rows['title'] . "<br/>" . "<br/>";
echo $rows['id'] . "<br/>" . "<br/>";
echo $rows['image_section'] . "<br/>";
echo $rows['subject'] . "<br/>";
echo $rows['image_post'] . "<br/>";
}
How can it take these ten results but use them in different places, and keep them arranged from one to ten.
this was the old case and i solve it but i found another problem, that, if the client had deleted a post as id = 800 "so there aren't id = 800 in DB" so when i get the max id minus $NUM from it, and this operation must be equal id = 800, so i have a programing mistake here, how can i take care of something like that.
function getmax_id_with_minus ($minus){
mysql_set_charset('utf8');
$maxid ="SELECT max(id) FROM posts";
$maxid1 =mysql_query($maxid);
while ($maxid_row = mysql_fetch_array($maxid1)){
$maxid_id = $maxid_row['0'];
$maxid_minus = $maxid_id - $minus;
}
$selectedpost1 = "SELECT * FROM posts WHERE id = $maxid_minus";
$query_selectedpost =mysql_query($selectedpost1);
return ($query_selectedpost);
}
<?php
$ss = getmax_id_with_minus (8);
while ($rows = mysql_fetch_assoc($ss)){
$main_post_1 = $rows;
?>
anyway "really" thanks again :) !
A few thoughts regarding posted code:
First and foremost, you should stop using mysql_ functions as they are being deprecated and are vulnerable to SQL injection.
$maxpost1 ="SELECT max(id) from posts WHERE section_id = $section_id ORDER BY ID DESC LIMIT 20";
When you SELECT MAX, MIN, COUNT, AVG ... functions that only return a single row, you do not need an ORDER BY or a LIMIT.
Given that you are only asking for the MAX(id), you can save work by combining your queries like so:
SELECT * FROM posts
WHERE id = (SELECT MAX(id) from posts WHERE section_id = $section_id)
If I'm understanding what you're trying to do (please correct me if I'm wrong), your function would look something like:
function sectionposts($section_id) {
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
$stmt = mysqli_prepare($link, "SELECT title, id, image_section, subject, image_post FROM posts "
. "WHERE section_id = ? ORDER BY id DESC LIMIT 10");
mysqli_stmt_bind_param($stmt, $section_id);
return mysqli_query($link, $stmt)
}
$result = sectionposts(6);
while ($row = mysqli_fetch_assoc($result)) {
echo $rows['title'] . "<br /><br />";
echo $rows['id'] . "<br /><br />";
echo $rows['image_section'] . "<br />";
echo $rows['subject'] . "<br />";
echo $rows['image_post'] . "<br />";
}
Try this instead, to save yourself a lot of pointless code:
$sql = "SELECT * FROM posts WHERE section_id=$section_id HAVING bar=MAX(bar);"
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($result);
echo ...;
echo ...;
The having clause lets you find the max record in a single operation, without the inherent raciness of your two-query version. And unless you allow multiple records with the same IDs to pollute your tables, removing the while() loops also makes things far more legible.
Seems like you want to store them in an array.
$posts = array(); //put this before the while loop.
$posts[] = $row; //put this in the while loop

While loop not printing all answers

This might be really simple, but i cannot figure out the problem with this code:
$sql = mysql_query("select * from Temporary_Stock_Transfer where Emp_ID = '$emp_id' and Company_ID = '$company_id'");
if(mysql_num_rows($sql) == 0) {
echo "<tr><td colspan='3'><i>You currenty have no items</i></td></tr>";
}else {
while($row = mysql_fetch_array($sql)) {
echo mysql_num_rows($sql);
echo 'reached';
$book_id = $row[1];
$sql = mysql_fetch_row(mysql_query("select title from Book where Book_ID = '$book_id'"));
echo "<tr><td>".$sql[0]."</td><td>".$row[2]."</td><td><span class='label label-important'>Remove</span></td></tr>";
}
}
Now based on my database the query is returning 2 results, the echo mysql_num_rows($sql) also gives out 2. However the reached is echoed only once. Does anyone see a potential problem with the code?
P.S: My bad, $sql is being repeated, that was a silly mistake
It will only echo once because of this line :
$sql = mysql_fetch_row(mysql_query("select title from Book where Book_ID = '$book_id'"));
your overwriting the $sql variable.
Why not just run a single query and join the data you require ?
try ($sql2 instead of $sql and $sql[0] to $sql2[0]):
$sql = mysql_query("select * from Temporary_Stock_Transfer where Emp_ID = '$emp_id' and Company_ID = '$company_id'");
if(mysql_num_rows($sql) == 0) {
echo "<tr><td colspan='3'><i>You currenty have no items</i></td></tr>";
}else {
while($row = mysql_fetch_array($sql)) {
echo mysql_num_rows($sql);
echo 'reached';
$book_id = $row[1];
$sql2 = mysql_fetch_row(mysql_query("select title from Book where Book_ID = '$book_id'"));
echo "<tr><td>".$sql2[0]."</td><td>".$row[2]."</td><td><span class='label label-important'>Remove</span></td></tr>";
}
}
I think it'll be because you are ressetting the sql variable
$sql = mysql_fetch_row(mysql_query("select title from Book where Book_ID = '$book_id'"));
when you are still using it in the while loop. :P
It's probably this line:
$sql = mysql_fetch_row(mysql_query("select title from Book where Book_ID = '$book_id'"));
Try using a different variable name.
You are changing the $sql variable inside the loop, therefore the next time you run $row = mysql_fetch_array($sql), the value will be different (probably won't be any based on your code).
Inside the while loop you are re-using the same variable: $sql. This is used to control the terminating condition of the loop.
Using a different variable name, e.g. $sql2 should leave the original variable intact and the loop should proceed as expected.

Loop through array, query each value until certain condition is met

I'm a bit of a newb to PHP and MySQL. I seem to be having an issue with something. How do I loop through an array, querying each value in the array until the query meets a certain condition.. In this case it would be that the number of rows returned from the query is less than five. Here is what I have:
$query1="SELECT UserID FROM Users where RefID='$userid'";
$result1=mysql_query($query1);
while ($row = mysql_fetch_array($result1, MYSQL_NUM) && $sql2querynum < '5')
{
echo ($row[0]);
echo "
";
$sql2 = "SELECT * FROM Users WHERE RefID=$row[0]";
$sql2result = mysql_query($sql2);
$sql2querynum = mysql_numrows($sql2result);
}
Problem is, for every value it echoes out, I get the following warning:
mysql_numrows(): supplied argument is not a valid MySQL result resource
Like I said, I'm a newb, so maybe I'm not even going about doing this the right way.
try this
$query1="SELECT UserID FROM Users where RefID='$userid'";
$result1=mysql_query($query1);
if(mysql_num_rows($result1)<5)
{
while ($row = mysql_fetch_array($result1))
{
echo ($row[0]);
echo "
";
$sql2 = "SELECT * FROM Users WHERE RefID=$row[0]";
$sql2result = mysql_query($sql2);
$sql2querynum = mysql_numrows($sql2result);
}
}
$query1="SELECT UserID FROM Users where RefID='$userid'";
$result1=mysql_query($query1);
while ($row = mysql_fetch_array($result1, MYSQL_NUM) && $sql2querynum < '5')
{
echo ($row[0]);
echo "
";
$sql2 = "SELECT * FROM Users WHERE RefID={$row[0]}";
$sql2result = mysql_query($sql2);
$sql2querynum = mysql_numrows($sql2result);
}
Use { } for variables in " " ... and why you are not using joins ?

loop problem within a statement

i hope someone can help about to scream!
basically I am trying to do a few things with the statement below;
First i want to check if the user id exists in member_categories_position.
If it Does i want then to exclude all entries from the second statement where member_id equals all results from the first statement
the third statement is the else statement that displays if the member_id is not present in the member_categories position.
PROBLEM - the result from the first system loops fine, however when i try and insert into the second statement (!='$memid') is produces no results and has no effect. I think the problem is that $memid is a looped result.
How do i get the second statement to say that any member_id that is in member_categories_position will not show in that statement?
$sql2 = "
SELECT *
FROM member_categories_position a
JOIN member_users b
ON b.id = a.member_id";
$rs2 = mysql_query($sql2);
while ($row = mysql_fetch_array($rs2))
{
$memid = "".$row['member_id']."";
}
if(mysql_num_rows($rs2) != 0)
{
$new= "
SELECT *
FROM member_categories
JOIN member_users
ON member_categories.member_id=member_users.id
JOIN member_config
ON member_categories.member_id=member_config.member_id
WHERE
member_categories.categories='$category'
AND member_categories.member_id !='$field'
GROUP BY member_config.member_id
ORDER BY RAND() limit 0,42";
$rs = mysql_query($new);
while ($row = mysql_fetch_assoc($rs))
{
echo "result excluding member ids from the first statement";
}
echo "<div class=\"clear\"></div>";
}
else
{
$new= "
SELECT *
FROM member_categories
JOIN member_users
ON member_categories.member_id=member_users.id
JOIN member_config
ON member_categories.member_id=member_config.member_id
WHERE
member_categories.categories='$category'
GROUP BY member_config.member_id
ORDER BY RAND() limit 0,42";
$rs = mysql_query($new);
while ($row = mysql_fetch_assoc($rs))
{
echo "Result with all member ids";
}
echo "<div class=\"clear\"></div>";
} } <-- (second is a stray from original post)
$memid is not in scope since it appears to be defined inside the loop. Try defining $memid = ''; at the top of your script.. like this.
$memid = '';
$sql2 = "
SELECT *
That way it will be defined when you use it below..

Pulling data and printing it in an HTML table

From a MySQL table called "submission" containing the fields "loginid, submissionid, title, url, datesubmitted, displayurl", I would like to print an HTML table thats contains all "title" and corresponding "datesubmitted" where "loginid" equals "$profile." The code I am trying to use is below. It isn't working. Any ideas why it isn't working?
Thanks in advance,
John
$profile = $_GET['profile'];
$sqlStr = "SELECT loginid, submissionid, title, url, datesubmitted, displayurl
FROM submission
WHERE loginid = $profile
ORDER BY datesubmitted DESC";
$result = mysql_query($sqlStr);
$arr = array();
echo "<table class=\"samplesrec\">";
while ($row = mysql_fetch_array($result)) {
echo '<tr>';
echo '<td class="sitename1">'.$row["title"].'</td>';
echo '</tr>';
echo '<tr>';
echo '<td class="sitename2">'.$row["datesubmitted"].'</a></td>';
echo '</tr>';
}
echo "</table>";
Your query is probably failing.
Try echoing the return from mysql_error(); after trying the query to see what the issue might be.
You should also protect your input against injection. If loginID is a username, you need to surround a string in a mySQL query with quotes - if loginID is a username. If it's an integer you may be okay.
There are more robust ways to do this but simply:
$profile = mysql_real_escape_string($_GET['profile']);
$sqlStr = "SELECT loginid, submissionid, title, url, datesubmitted, displayurl
FROM submission
WHERE loginid = '$profile'
ORDER BY datesubmitted DESC";
$result = mysql_query($sqlStr);
if($result) {
// Handle output
}
else {
echo 'query failed';
// don't leave this here in production!
echo mysql_error();
}
One problem I can see is you are not checking in the return value of mysql_query()
mysql_query() returns false if it fails to execute the query. So you need to do a check, something like:
$result = mysql_query($sqlStr);
if(! $result) {
//....error occured...prepare $message
die($message);
}
your question regards to debugging, the most important programming art. Noone can find an error for you, you have to do it yourself. With help of little tricks.
change $profile = $_GET['profile']; to $profile = intval($_GET['profile'];)
change $result = mysql_query($sqlStr); to
$result = mysql_query($sqlStr) or trigger_error(mysql_error()." in ".$sqlStr);
andd following 2 lines at the top of your code, run it again and see what it say. if still nothing, you don't have matching records in your table.
ini_set('display_errors',1);
error_reporting(E_ALL);

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