SQL SELECT Query use for search - php

I have three variables:
$school
$class
$subject
And a search String $searchquery
I want to search my database for the rows that match the Search Query... Currently I am using this:
$searchquery = str_replace(" ","|",$searchquery);
$sql = "
SELECT * FROM filedetails
WHERE school RLIKE '$searchquery'
OR class RLIKE '$searchquery'
OR subject RLIKE '$searchquery'";
Now, how this works, Suppose I enter Mathematics Class X. It converts the string into Mathematics|Class|X which is fine and search the Database and prints the results...
But there is one error:
It is printing all the rows which have either Mathematics or Class or X in them, so my output includes all the Classes of all Schools of Mathematics and All Subjects of All Schools of Class X...
Whereas I want it to show me All Schools having Mathematics and Class X...
P.S. - I have tried using AND instead of OR, but now I get empty result because no School Matches my Query String...

One way of doing is first make different 2-D Arrays of all the data searches and then take their intersection, so this will give the common row which matches all the keywords...
$sql = "SELECT * FROM filedetails";
$query = mysql_query($sql);
$array1 = array();
while ($row = mysql_fetch_array($query))
{
$array1[] = $row;
}
$sql = "SELECT * FROM filedetails WHERE school RLIKE '$searchquery'";
$query = mysql_query($sql);
$array2 = array();
while ($row = mysql_fetch_array($query))
{
$array2[] = $row;
}
$sql = "SELECT * FROM filedetails WHERE subject RLIKE '$searchquery'";
$query = mysql_query($sql);
$array3 = array();
while ($row = mysql_fetch_array($query))
{
$array3[] = $row;
}
$sql = "SELECT * FROM filedetails WHERE class RLIKE '$searchquery'";
$query = mysql_query($sql);
$array4 = array();
while ($row = mysql_fetch_array($query))
{
$array4[] = $row;
}
Now, what this code is doing that it is searching individually for every field and making separate result for each Query of each field, now We will be taking Intersection of the arrays so as to get the common row having the keywords
But there is one problem, suppose the $searchquery contains Details about School and Class, but no information about Subject, then the $array3 will be null, as the query will result NULL, then we need to consider handling null arrays which have been solved in Another Question Here...
Hope this helps... :)

Related

Using variable with multiple values in MySQL LIKE clause

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);
?>

php while loop + where

list table
id
name
illness
recipe
directions
$illness = 'Highblood';
$list = mysql_query("SELECT * FROM list ",$con);
while($row=mysql_fetch_array($list, MYSQL_ASSOC)){
$test[] = $row['recipe'];
$test2[] = $row['directions'];
}
I want to only get the rows
$row['recipe'];
$row['directions'];
if the value of illness column is equals to highblood.
how do I do this?
I suggest to make some search on google to learn MySQL if you can't do basic query like this.
mysql_query("SELECT * FROM list WHERE illness = '".$illness."'",$con);
Select only required rows.
$list = mysql_query("SELECT recipe,directions FROM list WHERE illness='Highblood'",$con);
while($row=mysql_fetch_array($list, MYSQL_ASSOC)){
$test[] = $row['recipe'];
$test2[] = $row['directions'];
}

Accessing two different tables for one loop

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.

How To Use Key Value in Stored Array in New Select Query

I'm working in PHP and MySQL to create and list a membership directory. I have three tables - company, contact and branch. Companies have Contact people, and some but not all companies have Branches, which also have Contact people. I am using LEFT JOIN in my first query to connect the contact people to their respective company, and loading the results into an array, which works using the following code:
// Retrieve all the data from the "company" table
$query = "SELECT * FROM company LEFT JOIN contact ON company.company_id = contact.company_id WHERE comp_county = 'BERNALILLO' ORDER BY company.comp_name, contact.cont_rank";
$result = mysql_query($query)
or die(mysql_error());
// Build the $company_array
$company_array = array();
while($row = mysql_fetch_assoc($result)) {
$company_array[] = $row;
}
Now I am trying to figure out how to run a second query, which needs to select from my branch table all branches whose company_id match the company_id stored in my above array: $company_array. I then want to store the results of the second query into a second array called $branch_array. I have tried this:
// Retrieve all the matching data from the "branch" table
foreach($company_array as $row) {
$query2 = "SELECT * FROM branch LEFT JOIN contact ON branch.branch_id = contact.branch_id WHERE branch.company_id = '".$row['company_id']."' ORDER BY branch.br_name, contact.cont_rank";
$result2 = mysql_query($query2)
or die(mysql_error());
}
// Build the $branch_array
$branch_array = array();
while($row2 = mysql_fetch_assoc($result2)) {
$branch_array[] = $row2;
}
But this does not seem to work... Can anyone give me an example of how to do this? The query needs to run so that it checks each different company_id in my $company_array for a match in the branch table - hopefully the question makes sense. Thanks.
i think your while should be inside your foreach statement so that your $branch_array can be filled with results of all company_id values, not only the last one :
foreach($company_array as $row) {
$query2 = "SELECT * FROM branch LEFT JOIN contact ON branch.branch_id = contact.branch_id WHERE branch.company_id = '".$row['company_id']."' ORDER BY branch.br_name, contact.cont_rank";
$result2 = mysql_query($query2)
or die(mysql_error());
// Build the $branch_array
$branch_array = array();
while($row2 = mysql_fetch_assoc($result2)) {
$branch_array[] = $row2;
}
}

Mysqli fetch array nth row

I have the following code that fetches a single row:
$query = "SELECT *
FROM translations
WHERE iddoc = '$id'
AND submitted = 1;";
$result = mysqli_query($query);
$row = mysqli_fetch_array($result);
I know this is useful in a while loop, where you can just loop through the results.
But I need to be able to grab specific rows that meet this condition. Something following this pseudo code:
$query = "SELECT *
FROM translations
WHERE iddoc = '$id'
AND submitted = 1;";
$result = mysqli_query($query);
$arrayofrows = mysqli_fetch_arrayofrows($result);
$arrayofrows[0] //(to get the first row)
$arrayofrows[1] //(2nd row)
and so on...
How can I do this?
You can try something like this
$arrayofrows = array();
while($row = mysqli_fetch_array($result))
{
$arrayofrows = $row;
}
You can now have
$arrayofrows[0] //(to get the first row)
$arrayofrows[1] //(2nd row)
I think the function you are looking for is mysqli_fetch_all as shown below. This avoids the need to create an extra looping structure or a function to do something already provided.
$query = "SELECT *
FROM translations
WHERE iddoc = '$id'
AND submitted = 1;";
$result = mysqli_query($query);
$arrayofrows = mysqli_fetch_all($result);
$arrayofrows[0] //(to get the first row)
$arrayofrows[1] //(2nd row)
It depends on whether you require the entire result set back or not but I think the LIMIT could be used like:
$query = "SELECT *
FROM translations
WHERE iddoc = '$id'
AND submitted = 1 LIMIT 200,200;";
Otherwise as others say you will need to convert to an array (which is what fetch_all does) and then get the element from that array.

Categories