Like on title... how to display row data from inner join table if name of the row are the same?? but data is diffrent?
$sql = "SELECT fv.name,fvcount.name,fvcount.datew,fvcount.u_uid
FROM fv
INNER JOIN fvcount ON fv.u_uid = fvcount.u_uid ";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
echo $row['u_uid'];
}
Result will be printed, but on both table 1 of the rows name is like: name
So if i put
$row['name'];
i will have output of inner join table
How to get output from main table and inner joint table?
I can't change name of the row...
Any clue?
The typical solution is to use column aliases. You can do something like:
SELECT fv.name AS fv_name, fvcount.name as fvcount_name, ...
And then use:
$row['fv_name']
Or:
$row['fvcount_name']
Related
I'm working on a system, and this module is supposed to echo the contents of the database.
It worked perfectly until I added some JOIN statements to it.
I've checked and tested the SQL code, and it works perfectly. What's not working is that part where I echo the content of the JOINed table.
My code looks like this:
$query = "SELECT reg_students.*, courses.*
FROM reg_students
JOIN courses ON reg_students.course_id = courses.course_id
WHERE reg_students.user_id = '".$user_id."'";
$result = mysqli_query($conn, $query);
if (mysqli_fetch_array($result) > 0) {
while ($row = mysqli_fetch_array($result)) {
echo $row["course_name"];
echo $row["course_id"];
The course_name and course_id neither echo nor give any error messages.
UPDATE: I actually need to increase the query complexity by JOINing more tables and changing the selected columns. I need to JOIN these tables:
tutors which has columns: tutor_id, t_fname, t_othernames, email, phone number
faculty which has columns: faculty_id, faculty_name, faculty_code
courses which has columns: course_id, course_code, course_name, tutor_id, faculty_id
I want to JOIN these tables to the reg_students table in my original query so that I can filter by $user_id and I want to display: course_name, t_fname, t_othernames, email, faculty_name
I can't imagine that the user_info table is of any benefit to JOIN in, so I'm removing it as a reasonable guess. I am also assuming that your desired columns are all coming from the courses table, so I am nominating the table name with the column names in the SELECT.
For reader clarity, I like to use INNER JOIN instead of JOIN. (they are the same beast)
Casting $user_id as an integer is just a best practices that I am throwing in, just in case that variable is being fed by user-supplied/untrusted input.
You count the number of rows in the result set with mysqli_num_rows().
If you only want to access the result set data using the associative keys, generate a result set with mysqli_fetch_assoc().
When writing a query with JOINs it is often helpful to declare aliases for each table. This largely reduces code bloat and reader-strain.
Untested Code:
$query = "SELECT c.course_name, t.t_fname, t.t_othernames, t.email, f.faculty_name
FROM reg_students r
INNER JOIN courses c ON r.course_id = c.course_id
INNER JOIN faculty f ON c.faculty_id = f.faculty_id
INNER JOIN tutors t ON c.tutor_id = t.tutor_id
WHERE r.user_id = " . (int)$user_id;
if (!$result = mysqli_query($conn, $query)) {
echo "Syntax Error";
} elseif (!mysqli_num_rows($result)) {
echo "No Qualifying Rows";
} else {
while ($row = mysqli_fetch_assoc($result)) {
echo "{$row["course_name"]}<br>";
echo "{$row["t_fname"]}<br>";
echo "{$row["t_othernames"]}<br>";
echo "{$row["email"]}<br>";
echo "{$row["faculty_name"]}<br><br>";
}
}
i have a question. my english isn't well. so i hope i explain well...
i have two tables, tbl_home and tbl_office, the question is
how do i make a select statement from 2 tables which have identical value from column 'case_no' where it is referenced in both table..
$a=$_POST['home_id']
the code above is where i get the home_id from,
while the statement below is how i try to select both tables based on value in column 'case_no' of both table. but it is based on variable $a which i retrieved from form
<?php
$sql2 = "SELECT * FROM tbl_office WHERE case_no IN (SELECT * FROM tbl_home WHERE home_id = '$
$result2=$conn->query($sql2);
while($row = $result2->fetch_assoc()){
$a=$row['case_no'];
$bc=$row['colour'];
echo " $a <br/> ";
echo " $bc2 <br/>";
?>
is the select statement above correct??
soo, i just want anybody to take a look a this specific statement and how to make it right
$sql2 = "SELECT * FROM tbl_office WHERE case_no IN (SELECT * FROM tbl_home WHERE home_id = '$a'";
You need inner join to use:
" SELECT t_office.home_id,t_office.case_no,t_office.name FROM tbl_office
t_office INNER JOIN tbl_home t_home ON t_office.case_no = t_home.case_no;
where t_office.case_no ='$a'";
u can use "inner join" for example:
"SELECT t.home_id,t.case_no,t.name FROM tbl_office
t INNER JOIN tbl_home h ON h.case_no = h.case_no"
**select tbl_home.name,tbl_office.case_no,tbl_office.color from tbl_office
INNER JOIN tbl_home on tbl_office.case_no = tbl_home.case_no
where tbl_office.case_no ='$a';**
I hope this will be working fine until $a(case_no) value is existed in tbl_home or else it doesn't give any rows
I have two database tables like this and want to fetch to my website like this(see the screenshot)
But I can only fetch one table. I don't know how to use group by with JOIN
here is my code
$sql = "SELECT photographer,GROUP_CONCAT(free_image)
FROM free_images_table
GROUP BY photographer";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result))
{
$free_image = explode(',', $row['GROUP_CONCAT(free_image)']);
echo "<tr>";
echo "<td>".$row['photographer_id']."</td>"; ?>
<td>
<?php
for($i=0; $i < count($free_image); $i++ )
{
echo $free_image[$i];
}
?></td>
echo "</tr>";
}
The special table may not have photographer (my website require only freeimage, the special image is optional.
This could be done using a LEFT JOIN (added column aliases for simplicity in php) -
SELECT free_images_table.photographer,
GROUP_CONCAT(DISTINCT free_images_table.free_image) as free_images,
GROUP_CONCAT(DISTINCT special_images_table.special_image) as special_images
FROM free_images_table
LEFT JOIN special_images_table
ON special_images_table.photographer = free_images_table.photographer
GROUP BY photographer
LEFT JOIN is used when you have a record in the 1st table, but not always a matching record in the 2nd table
Then in php, you would create your special image cells the same as your free image cells
$free_image = explode(',', $row['free_images']);
$special_image = explode(',', $row['special_images']);
...
SELECT table_1.free_image, table_2.special_image FROM table1 INNER JOIN table_2 ON table_1.photographer = table_2.photographer
In this case can use INNER JOIN.
Add table name before the field and INNER JOIN with same photographer.
I have a database with several tables. I'm able to query IDs from a single table. What I'd like to do is Use those IDs to query another tables IDs, then use these new IDs to query fields from the final table. Here is what I am currently doing:
Here is how I acquire the first set of IDs:
$returnedPost = mysqli_query($con, "SELECT Region_ID FROM Region WHERE RegionName='" . $queryVar . "'");
function resultToArray($result) {
$rows = array();
while ($row = $result->fetch_assoc()) {
$rows[] = $row;
}
return $rows;
}
$rows = resultToArray($returnedPost);
//$rows[x]['Region_ID'];//returns Region_ID 1...n
I'd like to use the IDs in $rows to be able to query a new set of IDs from other tables as follows:
$newTbl = mysqli_query($con, "SELECT Location_ID FROM Location WHERE Region_ID=" . $rows[$x]['Region_ID']);
$rows2 = resultToArray($newTbl);
$finalTbl = mysqli_query($con, "SELECT Field1, Field2 FROM Posts WHERE Location_ID=" . $rows2[$x]['Location_ID']);
Can someone please tell me how I can accomplish this? Thanks.
you can use INNER JOIN in one query to get at this data, maybe something like this
SELECT P.Field1,P.Field2
FROM Region R
INNER JOIN Location L ON R.Region_ID = L.Region_ID
INNER JOIN Posts P ON L.Location_ID = P.Location_ID
WHERE R.RegionName = Your_Region_QueryVar
My code right now is this
$extract = mysqli_query($dbc, "SELECT * FROM items");
$numrows = mysqli_num_rows($extract);
while($row = mysqli_fetch_assoc($extract)) {
$id = $row['i_id'];
$iname = $row['i_name'];
echo "<tr><td><a href='capture.php?item_id=$id'>$iname</a></td><td>Incomplete</td></tr>";
}
What i want to do is run a check to see if the item has already been completed.
I have a separate table that contains the information, for instance say that Item 1 has been completed then I would like to be able to change my echo to something like:
echo "<tr><td>$iname</td><td>Complete!</td></tr>";
How would I accomplish this? Would I need to use some form of a join statement?
I want to be able to display all the items in the table but not duplicate them i initially thought that an if statement on the echo to see if item complete then do this else do that
Here are the two tables
Items item_collection
------ ---------------
i_id i_id
i_name complete
caption
image
c_id
g_id
You can use join condition like this (assuming complete is a varchar field)
SELECT a.i_id, a.i_name,
CASE WHEN i_status = '1' THEN 'Complete!' ELSE 'Incomplete' END AS complete_status
FROM items a
LEFT OUTER JOIN item_collection b ON a.i_id = b.i_id
select
case
when ic.complete = 1 then 'Complete'
else 'Incomplete'
end as item_status
from items i
left join item_collection ic on i.i_id = ic.i_id
where i.i_name = 'your_item_name'
Assuming that ic.complete = 1 when item is complete.
Something along the lines of SELECT * FROM table1 WHERE table1.id not in (SELECT Id FROM table2)