I'm looking to loop the following script, as to save lines of code, and manual hardcoding. The script pulls specific values from my database and assigns them to variables that are used across my website.
$w1001a = $conn->query("SELECT columnB FROM table WHERE columnA='w1001a' limit 1")->fetch_object()->content;
$w1001b = $conn->query("SELECT columnB FROM table WHERE columnA='w1001b' limit 1")->fetch_object()->content;
....
$w1025b = $conn->query("SELECT columnB FROM table WHERE columnA='w1001b' limit 1")->fetch_object()->content;
Thanks in advance!
Don't loop the query, use one query to get all the info.
The below is just dummy code:
$stmt = $conn->query("SELECT columnA, columnB FROM table WHERE columnA IN ('w1001a', 'w1001b', ...)");
$result = array();
while ($obj = $stmt->fetch_object()) {
$result[$obj->columnA] = $obj->columnB;
}
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);
?>
Right now I am selecting some data from Mysql which I want to echo out, but I don't know how..
Code
<?php
// Database connection
require_once($_SERVER["DOCUMENT_ROOT"] . "/includes/config.php");
require_once($_SERVER["DOCUMENT_ROOT"] . "/includes/opendb.php");
// News 1
$searchroutenews1 = "SELECT newsid FROM NewsHomepage WHERE id = '1'";
$handlenews1 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews1);
$news1 = mysqli_fetch_row($handlenews1);
$searchroutenewsimg1 = "SELECT newsimg FROM NewsHomepage WHERE id = '1'";
$handlenewsimg1 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg1);
$NewsImg1 = mysqli_fetch_row($handlenewsimg1);
// News 2
$searchroutenews2 = "SELECT newsid FROM NewsHomepage WHERE id = '2'";
$handlenews2 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews2);
$news2 = mysqli_fetch_row($handlenews2);
$searchroutenewsimg2 = "SELECT newsimg FROM NewsHomepage WHERE id = '2'";
$handlenewsimg2 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg2);
$NewsImg2 = mysqli_fetch_row($handlenewsimg2);
// News 3
$searchroutenews3 = "SELECT newsid FROM NewsHomepage WHERE id = '3'";
$handlenews3 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenews3);
$news3 = mysqli_fetch_row($handlenews3);
$searchroutenewsimg3 = "SELECT newsimg FROM NewsHomepage WHERE id = '3'";
$handlenewsimg3 = mysqli_query($GLOBALS["___mysqli_ston"], $searchroutenewsimg3);
$NewsImg3 = mysqli_fetch_row($handlenewsimg3);
?>
After this I require_once this in an other file, and then I echo the variables $news1, $news2, $news3, $NewsImg1, $NewsImg2 and $NewsImg3. But if I echo this variables out now it says: array.
You can fetch all this information via a single query, instead of the 6 you currently are running. Then it's a matter of putting mysqli_fetch_*() as the argument of a while, as you'll then fetch all the rows, until that function returns null - at which point you've fetched all the rows returned by the query.
$result = mysqli_query($GLOBALS["___mysqli_ston"], "SELECT newsid, newsimg FROM NewsHomepage WHERE id IN (1, 2, 3)");
while ($row = mysqli_fetch_assoc($result)) {
echo $row['newsid']." ".$row['newsimg'];
}
Change however you need it to be displayed inside the while loop, and use the two variables as they are inside.
Alternatively, you can use WHERE id BETWEEN 1 AND 3 instead, but using IN (1, 2, 3) can more easily be changed to the exact ids you need.
http://php.net/mysqli-result.fetch-assoc
First you should read http://php.net/manual/en/mysqli-result.fetch-row.php
you can find there mysqli_result::fetch_row -- mysqli_fetch_row — Get a result row as an enumerated array mysqli_fetch_row always return array so now you can't echo array thats why it gives you array.
You can try foreach loop or for loop or while loop to display your data. there are also various methods to get array value.
Below is an example you can use.
while ($news1 = mysqli_fetch_row($handlenews1)) {
echo $news1[0];
}
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 trying to display a field from my MySQL database. It's in the table tblproducts in the row with the id is set to 1. And under the column qty.
This is the code I'm using:
<?php
mysql_connect("localhost","username","password");
mysql_select_db("database_name");
$available = "SELECT qty FROM tblproducts WHERE id = 1";
$result = mysql_query($available);
echo $result;
?>
However, I keep getting this message: Resource id #2
I've done a bit of research and seen where other people are having similar problems but most of them are trying to display their data in an HTML table whereas I just need the data from 'qty' to display. And of course I'm definitely not a MySQL guru.
Can anyone help me out with this please?
Try changing this:
$result = mysql_query($available);
To this:
$result = mysql_result(mysql_query($available), 0);
Let's start from the start. (I'll assume you have the connection set)
Form the query
$query = "SELECT `qty`
FROM `tblproducts`
WHERE `id` = 1";
Execute the query
$run = mysql_query($query);
Now, put the result in an assoc array
$r = mysql_fetch_array($run);
See the contents of the array
echo $r['qty'];
It's also advised that you move up from mysql to either mysqli, or PDO. PDO is preferred as you're not bound to the MySQL database model.
Try this:
Here you need to generate associative array and then get the resulting row.
$query = "SELECT `qty` FROM `tblproducts` WHERE `id` = 1";
$run = mysql_query($query);
$r = mysql_fetch_array($run);
echo $r['qty'];
-
Thanks
I'm struggling whit this...
Basically, I have a script that will run every day (cron job).
This script has to retrieve data from a table and it will post the data in another table (same database).
Here is my example script:
<?php
$c = mysql_connect("localhost", "user", "password");
$db = mysql_select_db("mydb", $c);
$query_sel = "SELECT id, rating, rating_count FROM mytableone";
$result_sel = mysql_query($query_sel) or die(mysql_error());
$ids = array();
while($id = mysql_fetch_array($result_sel))
$ids[] = $row;
foreach($ids as $id){
$lid = $row['id'];
$etvalue = $row['user_rating'];
$etcount = $row['rating_count'];
mysql_query("INSERT INTO mytabletwo VALUES ('$lid','$etvalue','$etcount',CURRENT_DATE())");
}
?>
My main idea is to insert data (as mysql query in foreach statement) "for each id from mytableone".
Where is my fault?
Thanks
You can do this with one SQL query
INSERT INTO mytabletwo SELECT id,rating,rating_count,CURRENT_DATE() FROM mytableone
As to the fault you are using $row instead of $id[...]
The easiest way to do this is:
INSERT INTO mytabletwo(idColumnName, ratingColumnName, rating_countColumnName, dateColumnName) SELECT id, rating, rating_count, CURRENT_DATE() FROM mytableone;
(That is one statement.)