{Connecting 2 tables together (with teacher ID and teacher name)
I have created the join and I have tested in SQL. Seems good. I am trying to print it on the screen.
$classandteacher = "SELECT person_name FROM people RIGHT OUTER JOIN classes ON classes.instructor_id=people.instructor_id ASC";
$result = mysqli_query($dbc, $classandteacher){
while($row = mysqli_query($dbc, $result));
$teacher = $row["person_name"];
echo ("Teacher: " . $teacher . "<br>");}
It's because the While Loop isn't correctly formatted.
Remove the ; and put it into {}
You also need to use mysqli_fetch_assoc($result) to get the Array from the query
$classandteacher = "SELECT person_name FROM people RIGHT OUTER JOIN classes ON classes.instructor_id=people.instructor_id ASC";
$result = mysqli_query($dbc, $classandteacher);
if(!$result) { //If your MYSQL query is throwing an error
error_log(mysqli_error());
}
while($row = mysqli_fetch_assoc($result))
{
$teacher = $row["person_name"];
echo "Teacher: " . $teacher . "<br>";
}
Or try using
mysqli_fetch_array instead
$classandteacher =
"SELECT
person_name FROM
people RIGHT OUTER JOIN classes ON
classes.instructor_id
=
peopleinstructor_id ASC";
$result =
mysqli_query($dbc,$ classandteacher);
if(!$result) { //If
your MYSQL query is
throwing an error
error_log(mysqli_er
ror() );}
while($row =
mysqli_fetch_array(
$result))
{
$teacher = $row
["person_name"];
echo "Teacher: " . $
teacher . "<br>"; }
Related
I have a SQL database of university students, with the following details:
Table_name: register
Column_names: position, tertinst
The data in the database will look something like this:
Coach..........UCT
Athlete........Tukkies
Official.......University of JHB
Athlete........Tukkies
Athlete........Tshwane Tech
Manager........UCT
I need to count the amount of students who are athletes(position), per university(tertinst) and the output has to be something like this:
UCT.....................735
University of Jhb.......668
Tukkies.................886
this is my attempt:
$positionx = 'Athletes';
include_once 'core/includes/db_connect.php';
$sql = "SELECT tertinst, COUNT(position) FROM register WHERE position = '$positionx' GROUP BY tertinst ";
$result = $conn->query($sql);
while ($row = mysql_fetch_array($result)) {
echo $row['COUNT(tertinst)'] . '......' . $row['COUNT(position)'];
}
$conn->close();
I get no result when the code is executed and I have searched for a different solution for hours, without success.
I made a few mistakes, and corrected the syntax in the sql count, as well as the echo. Here is the solution to my problem:
<?php
include_once 'core/includes/db_connect.php';
$sql = "SELECT tertinst, COUNT(*) total FROM register WHERE position = 'Athletes' GROUP BY tertinst ";
$result = $conn->query($sql);
while ($row = $result->fetch_assoc()) {
$tertinst = $row["tertinst"];
echo $tertinst . '......' . $row['total'] . '<br>';
}
$conn->close();
I'm trying to use foreach loop, for showing my DB but I always get error. What I want is to print each row and column. My code is like this :
$sql = "SELECT a.*, b.klasifikasi FROM kl_stre as b
INNER JOIN data_latih as a
ON a.id_stres = b.id_stres
ORDER BY a.id_dl";
$result = mysqli_query($conn, $sql);
foreach ($result as $dt_train => $row_dt_train):
foreach ($row_dt_train as $attr => $attr_dt_train):
echo $result[$row_dt_train][$attr_dt_trian]; // this line is the problem
endforeach;
endforeach;
the error I get is
Warning: Illegal offset type in C:\xampp\htdocs\knn\array_db.php on line 43
Would you mind explaining what is wrong with this code and how to solve this problem ?
you need to first check your query
if (!$result) {
die('query is not correct'.mysqli_error($conn));//$con is your database connection paste it after $result = mysqli_query($conn, $sql);
}
after that fetch record and simply remove foreach and try to use while() loop
while ($row=mysqli_fetch_array($result)) {
echo $row['your database field name'];
}
your code will be
$sql = "SELECT a.*, b.klasifikasi FROM kl_stre as b
INNER JOIN data_latih as a
ON a.id_stres = b.id_stres
ORDER BY a.id_dl";
$result = mysqli_query($conn, $sql);
if (!$result) {
die('query is not correct'.mysqli_error($conn));
}
while ($row=mysqli_fetch_array($result)) {
echo $row['your database field name'];
}
I am using a code something like below to get data from the second table by matching the id of first table. Code is working well, but I know it slow down the performance, I am a new bee. Please help me to do the same by an easy and correct way.
<?php
$result1 = mysql_query("SELECT * FROM table1 ") or die(mysql_error());
while($row1 = mysql_fetch_array( $result1 ))
{
$tab1_id = $row1['tab1_id'];
echo $row['tab1_col1'] . "-";
$result2 = mysql_query("SELECT * FROM table2 WHERE tab2_col1='$tab1_id' ") or die(mysql_error());
while( $row2 = mysql_fetch_array( $result2 ))
{
echo $row2['tab2_col2'] . "-";
echo $row2['tab2_col3'] . "</br>";
}
}
?>
You can join the two tables and process the result in a single loop. You will need some extra logic to check if the id of table1 changes, because you'll only want to output this value when there's a different id:
<?php
// Join the tables and make sure to order by the id of table1.
$result1 = mysql_query("
SELECT
*
FROM
table1 t1
LEFT JOIN table2 t2 ON t2.col1 = t1.id
ORDER BY
t1.id") or die(mysql_error());
// A variable to remember the previous id on each iteration.
$previous_tab1_id = null;
while($row = mysql_fetch_array( $result1 ))
{
$tab1_id = $row['tab1_id'];
// Only output the 'header' if there is a different id for table1.
if ($tab1_id !== $previous_tab1_id)
{
$previous_tab1_id = $tab1_id;
echo $row['tab1_col1'] . "-";
}
// Only output details if there are details. There will still be a record
// for table1 if there are no details in table2, because of the LEFT JOIN
// If you don't want that, you can use INNER JOIN instead, and you won't need
// the 'if' below.
if ($row['tab2_col1'] !== null) {
echo $row['tab2_col2'] . "-";
echo $row['tab2_col3'] . "</br>";
}
}
Instead of having 2 while loops, what you can do is join the 2 tables and then iterate over the result.
If you're not sure what join is look here: https://dev.mysql.com/doc/refman/5.1/de/join.html
Also here is a fairly simple query written using join: Join Query Example
You can use this. One relation with two tables:
<?php
$result1 = mysql_query("SELECT tab2_col2, tab2_col3 FROM table1, table2 where tab2_col1 = tab1_id ") or die(mysql_error());
while($row1 = mysql_fetch_array( $result1 ),)
{
echo $row2['tab2_col2'] . "-";
echo $row2['tab2_col3'] . "</br>";
}
?>
Like Sushant said, it would be better to use one JOIN or simpler something like that:
SELECT * FROM table1, table2 WHERE `table1`.`id` = `table2`.`id
I am fetching rows from a mysql table (jobs). Inside of that fetch, I am also fetching from another table (accounts) [to receive account api keys all depending on what ID_ASSOC is attacted to the job]: below is the code
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
while($rows = mysqli_fetch_assoc($query)){
$username = $rows['twitter_username'];
$consumer_key = $rows['consumer_key'];
$consumer_secret = $rows['consumer_secret'];
$access_token = $rows['access_token'];
$access_token_secret = $rows['access_token_secret'];
}
echo $job_poster_id ;
echo "<br/>";
echo $twitter_username;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
OUTPUT:
specific-message
4
admin
----------------------------------
When I do this, I only get one row output..and I can't seem to find out why. I want the above out put to repeat as many times as it has rows, and it's only doing one row (with the account fetch in the code). However when I do it without the internal fetch (accounts fetch), it returns multiple rows just as desired. Why is this? (below is sample code WITHOUT the accounts fetch):
$sql = "SELECT * FROM jobs";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
echo $job_poster_id ;
echo "<br/>";
echo "----------------------------------";
echo "<br/>";
}
OUTPUT:
specific-message
4
----------------------------------
specific-message
1
----------------------------------
specific-message
2
----------------------------------
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_assoc($query)){
echo $row['action'];
echo "<br/>";
$job_poster_id = $row['id_assoc'];
$sql = "SELECT * FROM accounts WHERE id_assoc='$job_poster_id'";
$query = mysqli_query($db_conx, $sql);
The problem is that you're using $query for the inner and the outer query.
When the inner query runs, and it steps through the loop, it's iterating to the end of the result set; when the outer while loop runs, mysqli_fetch_assoc($query) is returning false, because you're already at the end of the result set - just not the result set you were expecting.
You can fix this by renaming one of the $query variables.
I have some PHP/MySQL code that pulls data from multiple different tables (using Inner Joins). It looks something like this:
$query = "SELECT * FROM table1 INNER JOIN table2 USING (key)";
$data = mysqli_query($dbc, $query);
while ($row = mysqli_fetch_array($data)) {
echo $row[1];
}
So the code is simple enough, but what I want to do is echo the table each row is in inside of that while loop, since it could be in one of 2 tables.
I saw there was some old mysql functions like mysql_field_table and mysql_tablename that would do the trick, but they all seem to be deprecated.
Would appreciate any advice on how to accomplish this.
You could select the data with a special identifier for each table instead of using *
select table1.row1 as t1r1, table2.row1 as t2r1,..... from .....
And inside php you could look for the strings t1 and t2 and do stuff accordingly.
What you can do is echo more than one field from your result table (which hopefully now contains information from both the tables.
$query = "SELECT * FROM table1 INNER JOIN table2 USING (key)";
$data = mysqli_query($dbc, $query);
while ($row = mysqli_fetch_array($data))
{
echo $row[1] . " " . $row[2];
}
...and then $row[3] etc...
Or access the column name/field by the column name or alias provided by AS.
$query = "SELECT * FROM table1 INNER JOIN table2 USING (key)";
$data = mysqli_query($dbc, $query);
while ($row = mysqli_fetch_assoc($data))
{
echo $row["c_name1"] . " " . $row["c_name2"];
}
Where "c_name1" and "c_name2" are your column names.
Use an AS keyword to rename all columns.
$select_clause = '';
$tables = array('tblA', 'tblB');
foreach($tables as $tbl) {
mysql_query('SHOW COLUMNS FROM ' . $tbl);
while ($row = mysql_fetch_assoc())
$select_clause .= '`'.$tbl.'`.`'.$row['Field'].'` AS `'.$tbl
.'_'.$row['Field'].'`,';
}
$select_clause = substr($select_clause, 0, -1);
mysql_query('SELECT '.$select_clause.' FROM /*...*/ ');