I read an article and its comments from mysql database with two separate queries as
$result = mysql_query("SELECT * FROM articles WHERE article_id='$id'");
$row = mysql_fetch_array($result);
$title=$row['title'];
........
AND
$result = mysql_query("SELECT * FROM comments WHERE article_id='$id'");
while($row = mysql_fetch_array($result)) {
$comment_title=$row['title'];
.........
}
Is the best way to read this set of data from database? OR
Is it possible to catch the data through one query or one transaction?
NOTE: My issue is that first query for article is only for one row; but the second one needs a loop to process (and display in html) several comments.
$result = mysql_query("SELECT articles.title as article_title, comments.title as comment_title FROM articles LEFT JOIN comments ON articles.article_id = comments.article_id WHERE articles.article_id = '$id'");
$row = mysql_fetch_array($result);
$article_title=$row['article_title'];
$comment_title=$row['comment_title '];
You can do it in this way -
SELECT a.*, c.* FROM articles a
LEFT JOIN comments c
ON a.article_id = c.article_id
WHERE a.article_id='$id';
Is the best way to read this set of data from database?
Sure, it is.
Is it possible to catch the data through one query
Yes, it's possible but there is no point in doing that.
Why do you ask?
Related
$read = "SELECT * FROM elmtree
WHERE id ='$getid' AND
INNER JOIN elmtree_users ON elmtree.userid = elmtree_users.id";
The above query will not pull the information from the database to publish to the website.
Im trying to pull the item from the database with the $getid but also join the item id with the userid who uploaded it. Then using a while loop to print out the item to screen.
Any help would be greatly appreciated.
The SQL syntax you show isn't correct. A join clause describes which tables will be available for the rest of the query; all tables you mention need to be part of the ‘FROM’ clause, so that is where the ‘JOIN’ also belongs.
SELECT *
FROM elmtree
INNER JOIN elmtree_users ON elmtree.userid = elmtree_users.id
WHERE id ='$getid'
Your SQL query was not correctly written. The AND is not used in joining tables and the WHERE statement should be after the JOIN.
Try the following:
$read = "SELECT * FROM elmtree INNER JOIN elmtree_users ON(elmtree.userid = elmtree_users.id) WHERE elmtree.id ='$getid'";
UPDATE
Could you try the following in your code (replacing $this->database with your database variable) and post the result:
if ($result = $conn->query($read)) {
while ($row = $result->fetch_assoc()) {
...
}
} else {
throw new Exception("Database Error [{$this->database->errno}] {$this->database->error}");
}
$read = "SELECT * FROM elmtree INNER JOIN elmtree_users ON(elmtree.userid = elmtree_users.id) WHERE elmtree.itemid ='$getid'
Big thanks to Omari Celestine to finding the answer to the problem!
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 $return;
I am subquerying on the select distinct because I dont believe you have a dedicated skills table with an auto inc primary key, if that was there I wouldn't be using a subquery.
Can you test this query
select project_needs.*,count(members_skills.*) as count from project_needs
inner join members_skills
on members_skills.skill_id=project_needs.skill_id Group by project_needs.skill_name, project_needs.skill_id
Hi guys in the code below you can see what my JSON returns.
{"lifehacks":[{
"id":"2",
"URLtoImage":"http:\/\/images.visitcanberra.com.au\/images\/canberra_hero_image.jpg",
"title":"dit is nog een test",
"author":"1232123",
"score":"2",
"steps":"fdaddaadadafdaaddadaaddaadaaaaaaaaaaa","category":"Category_2"}]}
What the JSON returns is fine. The only problem is it is only displaying lifehacks if it has one like or more. So what should I change about my Query so it would display lifehacks without likes aswell.
//Select the Database
mysql_select_db("admin_nakeitez",$db);
//Replace * in the query with the column names.
$result = mysql_query("select idLifehack, urlToImage, title, Lifehack.Users_fbId, idLifehack, steps, Categorie, count(Lifehack_idLifehack) as likes from Lifehack, Likes where idLifehack = Lifehack_idLifehack AND idLifehack > " . $_GET["id"]. " group by idLifehack;", $db);
//Create an array
$json_response = array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$row_array['id'] = $row['idLifehack'];
$row_array['URLtoImage'] = $row['urlToImage'];
$row_array['title'] = $row['title'];
$row_array['author'] = $row['Users_fbId'];
$row_array['score'] = $row['likes'];
$row_array['steps'] = $row['steps'];
$row_array['category'] = $row['Categorie'];
//push the values in the array
array_push($json_response,$row_array);
}
echo "{\"lifehacks\":";
echo json_encode($json_response);
echo "}";
//Close the database connection
fclose($db);
I hope my problem is clear like this. Thank you in advance I can't figure it out myself.
You need a LEFT JOIN here. Your query has an INNER JOIN.
select
idLifehack,
urlToImage,
title,
Lifehack.Users_fbId,
idLifehack,
steps,
Categorie,
count(Lifehack_idLifehack) as likes
from Lifehack
left join Likes on idLifehack = Lifehack_idLifehack
where idLifehack > whatever
group by idLifehack;
There's an excellent explanation of the different join types here.
A couple additional points...
Use prepared statements in your PHP. Your code is wide-open to SQL Injection, which has ruined careers and led to millions of innocent people having their personal information stolen. There are plenty of web sites showing how to do this so I won't go into it here, though I'll say my favorite is bobby-tables.
Avoid the implicit join anti-pattern in your queries. This is an implicit join:
FROM a, b
WHERE a.id = b.id
Use explicit joins instead; they separate your join logic from your filtering (WHERE) logic:
FROM a
INNER JOIN b ON a.id = b.id
I first search all questions info. from "question" table including title, content, user etc.
the Code:
$sql = "select * FROM question where id>0 ORDER BY id ASC";
$result1 = mysql_query($sql);
$res=Array();
And then I want to search the user's point from "user" table. So I must search point for each user in each row from the result1
The Code:
while($rows=mysql_fetch_assoc($result1))
{
$res[]=$rows;
$user = $rows['user'];
$sql2 = "select point from user where name='$user'";
$result2 = mysql_query($sql2);
}
My problem is how to combine all the users' point(result2) with the questions info.(result1) together so that I can return a json for each row.
Use left join, as my understanding this work for you
$sql = "SELECT q.*, u.point AS point FROM question AS q LEFT JOIN user AS u ON q.user = u.name WHERE q.id > 0 ORDER BY q.id ASC";
$result = mysql_query($sql);
It's better go with the joins here i am giving you the query.i hope it may helps you
select * from question q,user u where q.id>0 ORDER BY id ASC
try something like this:using left join
select question.*,user.point FROM question left join user on user.name= question.name where id>0 ORDER BY id ASC
I am doing a foreach loop and selecting each row that equals that section name. However, I need to also show all results that don't equal any of the section names. Below is the part of my code that is selecting rows where it equals the section name.
$result2 = mysqli_query($con, "SELECT * FROM sections ORDER BY `order`");
$sectionnames = array();
while($row = mysqli_fetch_array($result2)) {
$sectionnames[] = $row['sectionname'];
}
foreach ($sectionnames as $sectionname) {
$result = mysqli_query($con,"SELECT * FROM faq WHERE section = '$sectionname' ORDER BY `order`");
Do you mean something like this:
SELECT * FROM `faq` WHERE `section` NOT IN (SELECT `sectionname` FROM `sections`)
which would select all faq items that do not have a section that shares a name with any sectionname in the sections table? If not then perhaps what juergen d posted is what you need. It's a little unclear exactly what your end goal is, so feel free to clarify and I will update.
I think it should be,
$result = mysqli_query ($con, "SELECT * FROM faq WHERE section NOT IN
(SELECT sectionname FROM sections)");
You can use a left join to get the data in 1 query
SELECT *
FROM sections s
left join faq f on f.section = s.sectionname
ORDER BY s.`order`, f.`order`
SELECT section FROM Faq
LEFT JOIN Sections
ON Faq.section = Sections.sectionname
WHERE Sections.sectionname IS NULL