I have a mysql query which is being to find the stock items in a certain location belonging to a certain group. Hence this is going through 4 levels of while loops. No i have given the user the ability to select the locations they want to view the stocks from. This is being achieved using checkboxes which are sent using ajax in an array. The array exploded in PHP using $offices = explode(",", $locations);. However now i want to use the locations selected in my mysql query.
$location are in the form of location1, location2, location3, location4
//selecting all locations using the statement below, however i want to select the locations that where selected by user.
$sql4 = mysql_query("select OfficeID, OfficeTitle from Office where 'wanted locations');
while($row3 = mysql_fetch_array($sql4)) {
$curr_location = $row3[0];
$sql3 = mysql_query("select Quantity from Stock_Management where Book_ID = '$curr_book' and Location_ID = '$curr_location'");
while($row3 = mysql_fetch_array($sql3)) {
echo "<td>".$row3[0]."</td>";
}
}
echo "</tr>";
I want to select the locations based on the selected locations by user, now this can be achievable using a for loop by i don't know how to include that in my sql query!
$locations = mysql_real_escape_string($locations);
$locations = str_replace(",","','",$locations);
$sql = "select OfficeID, OfficeTitle from Office WHERE location in ('$locations')";
$offices = explode(",", $locations);
$loc = implode("','", $offices);
This helps is creating the variable $loc to location1','location2',location3
$sql4 = mysql_query("select OfficeID, OfficeTitle from Office where OfficeTitle IN ('$loc')");
This creates the mysql query to be:
$sql4 = mysql_query("select OfficeID, OfficeTitle from Office where OfficeTitle IN ('location1','location2',location3')");, which solves the purpose for now.
SELECT OfficeID, OfficeTitle FROM Office WHERE OfficeID IN ( $locations ); ??
Also, look up mysql_real_escape_string and 'separation of concerns`
Related
can someone help me on the proper way of using a variable with multiple values in MySQL LIKE clause?
by the way, I have two tables teacher and student.
Teacher table contains. (fname,lname,subject,yr,sec) columns and for the student table (lrn,yr,sec)
what I'm trying to achieve is to count the total number of students of the teacher from all sections.
Here's my query for getting teacher's sections:
<?php
$result= mysqli_query($con, "select *from teacher where lname
= 'Uy' and subject = 'Science & Technology 7'" );
while($rowx = mysqli_fetch_array($result)) {
$section = $rowx['sec'];
}
?>
echoing $section output this: Our Lady of FatimaOur Lady of Guadalupe
Heres my query to count the students.
<?php
$ratequery = mysqli_query($con, "SELECT *,
(SELECT COUNT(*) FROM student WHERE
sec LIKE '%$section%') AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
?>
Output: the count works but only the first value [Our Lady of Fatima] of $section is getting recognized by LIKE clause, so only 1 section is getting counted.
by the way, sorry for my bad English and explanation.
Use Section as a input for second query:
<?php
$final_result = array();
$result= mysqli_query($con, "select *from teacher where lname
= 'Uy' and subject = 'Science & Technology 7'" );
while($rowx = mysqli_fetch_array($result))
{
$section = $rowx['sec'];
$ratequery = mysqli_query($con, "SELECT *, (SELECT COUNT(*) FROM student WHERE
sec =$section) AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
array_push($final_result,array("section"=>$section,"rateresult"=>$rateresult));
}
var_dump($final_result); // Display section wise data
?>
somehow I have managed to achieve my goal it but in other way . I used WHERE IN clause instead of LIKE clause. stored my query results into an array.. then used implode to add seperator.
Thanks for the help mr. #ashnu
<?php
$query="$con, select *from teacher where lname = 'Uy' and subject = 'Science & Technology 7'";
$result = mysqli_query($query) or die;
$sections = array();
while($row = mysqli_fetch_assoc($result))
{
$sections[] = $row['sec'];
}
$string = implode("','",$sections)
?>
<?php
$ratequery = mysqli_query($con, "SELECT *,
(SELECT COUNT(*) FROM student WHERE sec IN ('$string')) AS responseCount FROM student");
$rateresult = mysqli_fetch_array($ratequery);
?>
I am new to PHP coding and just trying to fix some functionality on my site that was left over from the lead developer.
The site, [Vloggi], is a marketplace. So I need to show the name of the job poster in the assignments page . The table I have the jobs in only has the ID, not the name.
So I need a join, but I've tried and it breaks the entire site.
The SQL has 17 tables, I need to display the User Name (usr_name) contained in table 3, the organisation contained in table 7 (usrg_orgname) with the job posting user (vlop_usr_id) details in table 14.
The primary key is users.usr_id, which is linked to users_gor.usrg_usr_id and vlog-ops.vlog_usr_id.
Table 3: users
usr_id, usr_email, usr_password, usr_fbuser, usr_fbtoken, usr_name, usr_loc_name, usr_loc_lat1, usr_loc_lon1, usr_loc_lat2, usr_loc_lon2, usr_status, usr_gor, usr_vgr, usr_token, usr_regtoken,
table 7: users_gor
usrg_usr_id, usrg_creditops, usrg_creditvlog, usrg_creditvlogette, usrg_destination, usrg_orgname, usrg_orgtype, usrg_location, usrg_website, usrg_jobtitle, usrg_phone, usrg_address1, usrg_address2, usrg_state, usrg_postcode, usrg_country
Table 14: vlog-ops
vlop_id, vlop_title, vlop_description, vlop_tags, vlop_deadline, vlop_quantity, vlop_quantityposted, vlop_vser_id, vlop_usr_id,vlop_loc_name, vlop_loc_lat1, vlop_loc_lon1, vlop_loc_lat2, vlop_loc_lon2, vlop_campaign, vlop_rules, vlop_tips, vlop_status
So in main.php i have written the following Sql lookup
in main.php, I have the following SQL lookups:
$sql = "SELECT * FROM users_gor WHERE usrg_usr_id = ".$db->quote($user_info['usr_id'])." LIMIT 1";
$rows = $db->select($sql);
$users_gor = $rows[0];
$sql = "SELECT * FROM users_vgr WHERE usrv_usr_id = ".$db->quote($user_info['usr_id'])." LIMIT 1";
$rows = $db->select($sql);
$users_vgr = $rows[0];
$sql = "SELECT * FROM users WHERE usr_id = ".$db->quote($user_info['usr_id'])." LIMIT 1";
$rows = $db->select($sql);
$users = $rows[0];
$sql = "SELECT * FROM vlog-ops WHERE vlop_usr_id ".$db->quote($user_info['usr_id'])." LIMIT 1";
$rows = $db->select($sql);
$users = $rows[0];
$sql = "SELECT usr_name AS vlop_usr_name FROM users WHERE usr_id = ".$db->quote($user_info['usr_id'])." LIMIT 1";
$rows = $db->select($sql);
$users = $rows[0];
And then in the page template itself, I have written
<?php echo $vlop['vlop_vser_id'] ?>
<?php echo $vlop['vlop_usr_name'] ?>
The first one works, the second doesn’t. What I want eventually is to display the user name and the organisation name in a table.
Whenever I try a JOIN or a NATURAL JOIN or a LEFT JOIN it breaks and the entire site goes blank.
Any help for a newbie would be appreciated with a million thanks.
When you use JOIN you need to specify how you're joining them.
In the query below I'm assuming you're looking for the fields in bold from your question.
$query='SELECT u.usr_name, g.usrg_orgname, v.vlop_usr_id FROM users u
JOIN vlog-ops v on u.usr_id = v.vlop_usr_id
JOIN users_gor g on u.usr_id = g.usrg_usr_id';
I believe I got the name of the fields right but if not just replace them with the correct ones.
Once you have the data fetched, you just loop through the results:
$result = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($result)) {
echo 'User name = ' . $row['u.usr_name'];
echo 'Org name = ' . $row['g.usrg_orgname'];
echo 'Job posting user id = ' . $row['v.vlop_usr_id'];
}
I have a small issue that I can't figure out.
I have to pull data from two different tables, in one loop. I've never done that before, so I have no idea how. I tried two different queries. That looked like this:
$query = "SELECT * FROM colors ";
$color_select = mysqli_query($connection, $query);
$second_query = "SELECT * FROM votes";
$vote_select = mysqli_query($connection, $second_query);
And then put them into a loop:
while($row = mysqli_fetch_assoc($color_select) && $second_row = mysqli_fetch_assoc($vote_select))
{
$color = $row['Colors'];
$votes = $second_row['Votes'];
echo "<tr><td>$color</td><td>$votes</td></tr>";
}
But that didn't work. I didn't expect it to, just wanted to try. :) Maybe someone experienced can help me out. Thanks.
At the end of the day I need a table displayed, that has two columns, one of them contains the color name from one DB table and the other one contains a number of votes.
As requested: table structures.
Table: colors has only one field Colors.
Table: votes has four fields city_id, City, Colors and Votes
*************************EDIT**************************************
So fixed up the query as suggested, but is still shows nothing.
Here is the edited code:
$query = "SELECT * FROM colors,votes WHERE colors.Colors=votes.Colors";
$color_votes_select = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($color_votes_select))
{ $color = $row['Colors'];
$votes = $row['Votes']; }
if table having relation.
try this in single query .
SELECT
`colors`.*,votes.*
FROM
`colors`
INNER JOIN
`votes` ON
`votes`.colorId = `colors`.Id
Most imp *****You should have some relationship between tables
Otherwise workaround
Run query on color, Save it in ArrayA
Run query on vote, Save it in ArrayB
Create New Array ArrayC
$arrayC = array();
Loop array A or C if they both contact same row count
array_push($ArrayC, key and value of color, key and value of votes);
Final loop ArrayC to print tr and td
First Relate These two tables, write color_id in votes table.
$query = "SELECT * FROM colors,votes where colors.id=votes.color_id";
$color_select = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($color_select))
{
$color = $row['Colors'];
$votes = $row['Votes'];
}
Try this:
$query = "SELECT colors FROM colors";
$color_select = mysqli_query($connection, $query) or die (mysqli_error());
$second_query = "SELECT votes FROM votes"; //you only need column votes right?
$vote_select = mysqli_query($connection, $second_query) or die (mysqli_error());;
while( $row = mysqli_fetch_assoc($color_select) && $second_row = mysqli_fetch_assoc($vote_select)){
$color[] = $row['colors'];
$votes[] = $second_row['votes'];
echo "<tr><td>$color</td><td>$votes</td></tr>";
}
Short explanation:
It will fetch the select and store into an array (because what you were doing is selecting multiple rows into one single variable) and then just display with the echo.
I'm working in PHP and MySQL to create and list a membership directory. I have three tables - company, contact and branch. Companies have Contact people, and some but not all companies have Branches, which also have Contact people. I am using LEFT JOIN in my first query to connect the contact people to their respective company, and loading the results into an array, which works using the following code:
// Retrieve all the data from the "company" table
$query = "SELECT * FROM company LEFT JOIN contact ON company.company_id = contact.company_id WHERE comp_county = 'BERNALILLO' ORDER BY company.comp_name, contact.cont_rank";
$result = mysql_query($query)
or die(mysql_error());
// Build the $company_array
$company_array = array();
while($row = mysql_fetch_assoc($result)) {
$company_array[] = $row;
}
Now I am trying to figure out how to run a second query, which needs to select from my branch table all branches whose company_id match the company_id stored in my above array: $company_array. I then want to store the results of the second query into a second array called $branch_array. I have tried this:
// Retrieve all the matching data from the "branch" table
foreach($company_array as $row) {
$query2 = "SELECT * FROM branch LEFT JOIN contact ON branch.branch_id = contact.branch_id WHERE branch.company_id = '".$row['company_id']."' ORDER BY branch.br_name, contact.cont_rank";
$result2 = mysql_query($query2)
or die(mysql_error());
}
// Build the $branch_array
$branch_array = array();
while($row2 = mysql_fetch_assoc($result2)) {
$branch_array[] = $row2;
}
But this does not seem to work... Can anyone give me an example of how to do this? The query needs to run so that it checks each different company_id in my $company_array for a match in the branch table - hopefully the question makes sense. Thanks.
i think your while should be inside your foreach statement so that your $branch_array can be filled with results of all company_id values, not only the last one :
foreach($company_array as $row) {
$query2 = "SELECT * FROM branch LEFT JOIN contact ON branch.branch_id = contact.branch_id WHERE branch.company_id = '".$row['company_id']."' ORDER BY branch.br_name, contact.cont_rank";
$result2 = mysql_query($query2)
or die(mysql_error());
// Build the $branch_array
$branch_array = array();
while($row2 = mysql_fetch_assoc($result2)) {
$branch_array[] = $row2;
}
}
I have a MySQL database containing a user's country and whether they are an individual or an organisation. The field names are 'country' and 'type'.
Using PHP, I'd like to 'count' the number of countries, the number of individuals and the number of organisations in the database and then display the numbers in the following example format:
<p>So far, <strong>500</strong> individuals and <strong>210</strong> organisations from <strong>40</strong> countries have registered their support.</p>
I am currently listing the total number of records using the below code if this helps:
<?php
$link = mysql_connect("localhost", "username", "password");
mysql_select_db("database_name", $link);
$result = mysql_query("SELECT * FROM table_name", $link);
$num_rows = mysql_num_rows($result);
echo " $num_rows\n ";
?>
My PHP / MySQL skills are very limited so I'm really struggling with this one.
Many thanks in advance!
Ben
To get the number of countries:
SELECT COUNT(DISTINCT country) AS NumCountries FROM tableName
To get the number of individuals, or the number of organisations:
SELECT COUNT(*) AS NumIndividuals FROM tableName WHERE type = 'individual'
SELECT COUNT(*) AS NumOrganisations FROM tableName WHERE type = 'organisation'
What you are looking for is a count based on a grouping. Try something like this:
$sql = "SELECT type, count(*) as cnt FROM users GROUP BY type";
$result = mysql_query($sql);
$counts = array();
while ($row = mysql_fetch_assoc($result)) {
$counts[$row['type']] = $row['cnt'];
}
This will give you an array like
Array (
'individual' => 500,
'organization' => 210
)
For counting the countries, use the first statement as posted by Hammerite.
EDIT: added a verbose example for counting the countries
$sql = "SELECT COUNT(DISTINCT country) AS NumCountries FROM users";
$result = mysql_query($sql);
$number_of_countries = 0;
if ($row = mysql_fetch_assoc($result)) {
$number_of_countries = $row['NumCountries'];
}
This altogether you can then print out:
printf('<p>So far, <strong>%d</strong> individuals and <strong>%d</strong> '.
'organisations from <strong>%d</strong> countries have registered '.
'their support.</p>', $counts['individual'], $counts['organization'],
$number_of_countries);
The answer is to retrieve the answer by using the COUNT(*) function in SQL:
SELECT COUNT(*) AS individual_count FROM user WHERE type = 'individual';
SELECT COUNT(*) AS organization_count FROM user WHERE type = 'organization';
SELECT COUNT(*) AS country_count FROM user GROUP BY country;
The last will group your query set by the country name, and will result in one row for each country. Using COUNT on this result set will give the count of distinct coutries.
You can then fetch this value by using mysql_fetch_assoc on your $result from mysql_query, and the answer will be contained in 'invididual_count', 'organization_count' and 'country_count' for each query.
Thank you for all of your help with this (especially Cassy).
I think it's worthwhile displaying the full code in case anybody else comes across a similar requirement in the future:
<?php
$link = mysql_connect("localhost", "username", "password");
mysql_select_db("database_name", $link);
$sql = "SELECT type, COUNT(*) as cnt FROM table_name GROUP BY type";
$result = mysql_query($sql);
$counts = array();
while ($row = mysql_fetch_assoc($result)) {
$counts[$row['type']] = $row['cnt'];
}
$sql = "SELECT COUNT(DISTINCT country) AS NumCountries FROM table_name";
$result = mysql_query($sql);
$number_of_countries = 0;
if ($row = mysql_fetch_assoc($result)) {
$number_of_countries = $row['NumCountries'];
}
printf('<p><strong>So far, <em class="count">%d</em> individuals and <em class="count">%d</em> organisations from <em class="count">%d</em> countries have registered their support.', $counts['Individual'], $counts['Organisation'], $number_of_countries); ?>
If you're just looking for the number of rows returned try this:
$result = mysql_db_query($db, $query);
$num_rows = mysql_num_rows($result);
Another option would be to execute a separate query with the mysql count function and use the result from that.