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one variable row inside another variable row
I have script something like -
$sql = "SELECT * FROM `users`"
$q = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($q);
$sql1 = "SELECT * FROM `other_table`";
$q1 = mysql_query($sql1) or die(mysql_error());
$row1 = mysql_fetch_array($q1);
$item = $row1[$row['username']];
How can I set one variable row inside another, since it don't work. Basically, I need to select username, and then select column with user username from other table, in which is written user points.
I was thinking about adding -
$sql = "SELECT * FROM `users`"
$q = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($q);
$sql1 = "SELECT `".$row['username']."` FROM `other_table` WHERE `uid` = 1";
$q1 = mysql_query($sql1) or die(mysql_error());
$row1 = mysql_fetch_array($q1);
$item = $row1[xxxxxxxxxx]; // DONT KNOW HOW TO DEFINE IT, so it takes out found variable (there is only one).
I have question already, but not with full info, and I don't know how to delete it :(!
Here is my table -
Users
'id'--'name'
'1'--'Bil'
'2'--'Conor'
'3'--'Ilian'
Other_table (which holds points for users)
'id'--'Bil'--'Conor'--'Ilian'
'1'--'2'--'3'--'55'
Don't ask, why I don't hold the points in the same table, since if I could, I would do that ;)!
It sounds like you need to redesign the tables to start with, I would suggest a table structure like this:
Table Users
ID | Name
Table Points
UserID | Points
This way you can add a foreign key constraint between the tables and then do a simple join query like:
SELECT U.Name, P.Points FROM Users AS U INNER JOIN Points AS P ON U.ID = P.UserID
Then your php code would look like:
$sql = "SELECT U.Name, P.Points FROM Users AS U INNER JOIN Points AS P ON U.ID = P.UserID";
$q = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($q);
Here are the create table queries for you:
CREATE TABLE Users (ID INT NOT NULL AUTO_INCREMENT,
Name VARCHAR(50) NOT NULL, PRIMARY KEY (ID))
CREATE TABLE Points (UserID INT NOT NULL, Points INT,
FOREIGN KEY(UserID) REFERENCES Users(ID) ON UPDATE CASCADE ON DELETE CASCADE)
EDIT:
This should allow you to get the value (NOTE: untested as I don't have a php/mysql instance to mess with currently), but for a more robust solution you really should look into redesigning your table schema.
$userName = $row['username'];
$sql1 = "SELECT [". $userName ."] FROM other_table WHERE [uid] = 1";
$q1 = mysql_query($sql1) or die(mysql_error());
$row1 = mysql_fetch_array($q1);
$item = $row1[$userName];
SELECT users.id, users.name, other_table.points FROM users LEFT JOIN other_table ON other_table.name=users.name
Related
I have a simple blog where I'm practicing some php and mysql. I'm trying to display the username of the post author (you know where it says posted by Author).
I have two tables blog_members and blog_posts which are related 1 to many so I got a memberID field into the blog_posts. I'm trying to get the username of the member who's the author of the post.
I was thinking to do a join or something but I can't figure this out.
Here's what I was trying to do but it's not working because I'm sure I'm not using it properly.
$query1 = "SELECT username from blog_members JOIN blog_posts ON memberID = memberID ";
$result1 = mysqli_query($link, $query1);
$row1 = mysqli_fetch_array($result1);
PS: I got it working one way by using SESSION to get the userID but that works only if the user is logged is which is not the case, I want to display the name in any case.
Thanks!
Use inner join this way
And with a proper sanitize use $your_user_id for match
$query1 = "SELECT username
from blog_members
INNER JOIN blog_posts ON blog_members.memberID = blog_posts.memberID
WHERE blog_posts.memberID = '" .$your_user_id . "';";
JOIN syntax is wrong in your query.
Use following query:
$query1 = "SELECT username from blog_members JOIN blog_posts ON blog_members.memberID = blog_posts.memberID ";
$result1 = mysqli_query($link, $query1);
$row1 = mysqli_fetch_array($result1);
Try something like this, usng INNER JOIN :
$query1 = "SELECT blog_members.username FROM blog_members INNER JOIN blog_posts ON blog_members.memberID = blog_posts.memberID ";
reference : http://www.w3schools.com/sql/sql_join.asp
Here is a solution using a simple WHERE condition (same performance Explicit vs implicit SQL joins) :
SELECT a.username FROM blog_members a, blog_posts b WHERE a.memberID = b.memberID
But if you need more information about MySQL Join : https://www.sitepoint.com/understanding-sql-joins-mysql-database/
Hope this helps !
Basically, I am seeking to know if there is a better way to accomplish this specific task.
Basically, what happens is I query the db for a list of "project needs" -- These are each uniquer and only appear once.
Then, I search another table to find out how many members have the required "skills - which are directly correlated to the project needs".
I accomplished exactly what I was trying to do by running a second query and then inserting them into an array like this:
function countEachSkill(){
$return = array();
$query = "SELECT DISTINCT SKILL_ID, SKILL_NAME FROM PROJECT_NEEDS";
$result = mysql_query($query) or die(mysql_error());
$num_rows = mysql_num_rows($result);
while($row = mysql_fetch_assoc($result)){
$query = "SELECT COUNT(*) as COUNT FROM MEMBER_SKILLS WHERE SKILL_ID = '".$row['NEED_ID']."'";
$cResult = mysql_query($query);
$cRow = mysql_fetch_assoc($cResult);
$return[$row['SKILL_ID']]['Count'] = $cRow['COUNT'];
$return[$row['SKILL_ID']]['Name'] = $row['SKILL_NAME'];
}
arsort($return);
return $return;
}
But I feel like there has to be a better way (perhaps using some kind of join?) that would return this in a result set to avoid using the array.
Thanks in advance.
PS. I know mysql_ is depreciated. It is not my choice on which to use.
SELECT P.SKILL_ID, P.SKILL_NAME, COUNT(M.SKILL_ID) as COUNT FROM PROJECT_NEEDS P INNER JOIN MEMBER_SKILLS M
ON P.SKILL_ID=M.SKILL_ID
GROUP BY P.SKILL_ID, P.SKILL_NAME
I've adjusted Nriddens answer to accomodate for the select distinct, Im under the belief that his adjustment would be ok given SKILL_ID is a primary key
function countEachSkill(){
$return = array();
$query = "
SELECT
COUNT(*) AS COUNT,
PROJECT_NEEDS.SKILL_NAME,
PROJECT_NEEDS.SKILL_ID
FROM
(SELECT DISTINCT
SKILL_ID, SKILL_NAME
FROM
PROJECT_NEEDS) AS PROJECT_NEEDS
INNER JOIN
MEMBER_SKILLS
ON
MEMBER_SKILLS.SKILL_ID = PROJECT_NEEDS.SKILL_ID
GROUP BY PROJECT_NEEDS.SKILL_ID";
$result = mysql_query($query) or die(mysql_error());
$num_rows = mysql_num_rows($result);
while($row = mysql_fetch_assoc($result)){
$return[$row['SKILL_ID']]['Count'] = $row['COUNT'];
$return[$row['SKILL_ID']]['Name'] = $row['SKILL_NAME'];
}
arsort($return);
return $return;
I am subquerying on the select distinct because I dont believe you have a dedicated skills table with an auto inc primary key, if that was there I wouldn't be using a subquery.
Can you test this query
select project_needs.*,count(members_skills.*) as count from project_needs
inner join members_skills
on members_skills.skill_id=project_needs.skill_id Group by project_needs.skill_name, project_needs.skill_id
I have two tables:
Table 1
contacts
id
email
name
lastname
postcode
straat
huisnr
woonplaats
klantnummer
bsn
land
debiteurnummer
Table 2
contacts_group
id
mail
group_name
How can I select and order this 2 tables in one query. I've tried union and left join, but did not work.
$result = mysqli_query($database->connection,
"SELECT *
FROM contacts
WHERE owner = '$session->username'
ORDER BY name ASC ,bedrijfsnaam ASC")
or die(mysqli_error());
while($roww = mysqli_fetch_array($result)){
echo $roww['email'];
echo $roww['name'];
}
Table contacts_group:
$result = mysqli_query($database->connection,
"SELECT *
FROM contacts_group
WHERE owner = '$session->username'
ORDER BY group_name ASC")
or die(mysqli_error());
while($roww = mysqli_fetch_array($result)){
echo $roww['mail'];
echo $roww['group_name'];
}
Escape your variables. You should not put the php variable unescaped in sql query. You can suffer from sql injections.
If you want to join two tables by foregin key you can do:
SELECT * FROM contacts, JOIN contacts_group ON contacts_group.id = contacts.group_id
WHERE contacts.owner = '$session->username' ORDER BY contacts.name
But you are missing group_id in contacts table or some foregin key to connect two tables.
Before do it read this topic, after try to use mysql join construction
<?php
include('includes/config.php');
$topi = $_GET['id']; //id of url
mysql_select_db("ban", $con);
$query = "SELECT * FROM `basic` WHERE id = '$topi' LIMIT 0, 30";
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
$aa = $row['item'];
$cc = $row['moreinfo'];
$dd = $row['contactinfo'];
$ff = $row['id'];
In this script, I get information from the table basic, but I want to retrieve data from another table named users. How can I retrieve data from two tables at once?
users table consists of following columns:
email
username
ID
You need to JOIN the two tables on a common value, called a foreign key. Once you've posted the structure of the users table as requested in the comments, I can provide a more complete example.
EDIT: See example. This calls explicit column names instead of SELECT *.
$query = "SELECT
basic.id,
basic.item,
basic.moreinfo,
basic.contactinfo,
users.email,
users.username
FROM basic JOIN users ON basic.id = users.id
WHERE id = '$topi'
LIMIT 0 , 30";
You would use a JOIN onto the other table.
$query = "SELECT *
FROM basic b
JOIN users u ON b.user_id = u.user_id
WHERE id = '$topi'
LIMIT 0, 30";
Something like that, but based on your fields.
Please Note: the ON clause specifies what you will be looking for a match on.
Hey guys need some more help
I have 3 tables USERS, PROFILEINTERESTS and INTERESTS
profile interests has the two foreign keys which link users and interests, they are just done by ID.
I have this so far
$statement = "SELECT
InterestID
FROM
`ProfileInterests`
WHERE
userID = '$profile'";
Now I want it so that it selects from Interests where what it gets from that query is the result.
So say that gives out 3 numbers
1
3
4
I want it to search the Interests table where ID is = to those...I just don't know how to physically write it in PHP...
Please help.
Using a JOIN:
Best option if you need values from the PROFILEINTERESTS table.
SELECT DISTINCT i.*
FROM INTERESTS i
JOIN PROFILEINTERESTS pi ON pi.interests_id = i.interests_id
WHERE pi.userid = $profileid
Using EXISTS:
SELECT i.*
FROM INTERESTS i
WHERE EXISTS (SELECT NULL
FROM PROFILEINTERESTS pi
WHERE pi.interests_id = i.interests_id
AND pi.userid = $profileid)
Using IN:
SELECT i.*
FROM INTERESTS i
WHERE i.interests_id IN (SELECT pi.interests_id
FROM PROFILEINTERESTS pi
WHERE pi.userid = $profileid)
You are on the right track, lets say you execute the query above using this PHP code:
$statement = mysql_query("SELECT InterestID FROM `ProfileInterests`
WHERE userID = '$profile'");
Then you can use a PHP loop to dynamically generate an SQL statement that will pull the desired IDs from a second table. So, for example, continuing the code above:
$SQL = "";
while ($statementLoop = mysql_fetch_assoc($statement)) {
//Note the extra space on the end of the query
$SQL .= "`id` = '{$statementLoop['InterestID']}' OR ";
}
//Trim the " OR " off the end of the query
$SQL = rtrim($SQL, " OR ");
//Now run the dynamic SQL, using the query generated above
$query = mysql_query("SELECT * FROM `table2` WHERE {$SQL}")
I haven't tested the code, but it should work. So, this code will generate SQL like this:
SELECT * FROM `table2` WHERE `id` = '1' OR `id` = '3' OR `id` = '4'
Hope that helps,
spryno724
Most likely you want to join the tables
select
i.Name
from
ProfileInterests p
inner join
interests i
on
p.interestid = i.interestid
where
p.userid = 1