Php search in database to show data, not working - php

i'm trying to display some data from my database, and i dosen't work when i try to display some of the data, specified by a search bar. Here is the code. I hope someone can fix my problem.
Ive tried for a few hours, but nothing seems to work.
I removed most of the html code, to even be able to post.
<?php
include_once 'includes/dbh.inc.php';
?>
<!DOCTYPE html>
<html lang="en">
<body>
<form>
<input type="text" name="search" placeholder="Search for graphicscard"/>
<input type="submit" value="Search" />
<?php
?>
</form>
<table width="600" border="1" cellpaddin="1" cellspacing="1">
<tr>
<th>Graphicscard</th>
<th>Coin</th>
<th>Hashrate</th>
<th>Bios-modded</th>
<th>Core-clock</th>
<th>Memory-clock</th>
<th>Power draw</th>
<th>Additional info</th>
</tr>
<?
if(isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^`0-9a-z]#i","", $searchq);
$query = mysqli_query("SELECT * FROM 'cards' WHERE 'name' LIKE '$search%' OR coin LIKE '$search%'") or die ("No results");
$count = mysqli_num_rows($queryQ);
while ($row = mysqli_fetch_array($query)) {
echo "<tr>";
echo "<td>".$row['name']."</td>";
echo "<td>".$row['coin'] . "</td>";
echo "<td>".$row['hashrate'] . "</td>";
echo "<td>".$row['bios_mod'] . "</td>";
echo "<td>".$row['core_clock'] . "</td>";
echo "<td>".$row['memory_clock'] . "</td>";
echo "<td>".$row['power_draw'] . "</td>";
echo "<td>".$row['additional_info'] . "</td>";
echo "</tr>";
}
}
?>

The mysqli_query() used as a function needs the link id returned by mysqli_connect() o mysqli_init()
PHP mysqli_query() Function
So you need a $connection var:
$connection=mysqli_connect("localhost","my_user","my_password","my_db");
and the use it on the query:
mysqli_query($connection, "SELECT * FROM 'cards' WHERE 'name' LIKE '$search%' OR coin LIKE '$search%'") or die ("No results");

You are using $search in the query, but the correct variable should be $searchq .

Do you get correct query result and how many rows in it? As I can see were is a mistake in your query:
SELECT * FROM 'cards' WHERE 'name' LIKE...
Name of table and fields should be in another quoats - `
SELECT * FROM cards WHERE name LIKE...
Or this is typing issue on posted message?
On this row: $count = mysqli_num_rows($queryQ); - you variable of query result is - $query not $queryQ

Here is an alternative way:
select * from cards where name like 'ABC%'
union
select * from cards where coin like 'XYZ%'
union
select * from cards where gender like 'PQR%';

Change your code on this:
if(isset($_POST['search'])) {
$searchq = $_POST['search'];
$searchq = preg_replace("#[^`0-9a-z]#i","", $searchq);
$query = mysql_query("SELECT * FROM cards WHERE name LIKE '$searchq%' OR coin LIKE '$searchq%'") or die ("No results");
$count = mysql_num_rows($query);
while ($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>".$row['name']."</td>";
echo "<td>".$row['coin'] . "</td>";
echo "<td>".$row['hashrate'] . "</td>";
echo "<td>".$row['bios_mod'] . "</td>";
echo "<td>".$row['core_clock'] . "</td>";
echo "<td>".$row['memory_clock'] . "</td>";
echo "<td>".$row['power_draw'] . "</td>";
echo "<td>".$row['additional_info'] . "</td>";
echo "</tr>";
}
}
If you use mysqli please see this referenc on how to use it
in your tag form add method of sending data POST: <form method="POST">

Related

I want a simple php table inside a while loop, with add or subtract

I am new to php and need some help please.
I'd like to create a simple php while loop.
Inside the loop I want to be able to add or subtract values.
My code seems to be doing what I want, except that i do not know what to put in the $count variable.
The aim is to have one table, and inside that table either add or subtract values of 5.
Here is my code.
<?php
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "<tr>";
echo "<td>" . $row["id"] . "</td>";
echo "<td>" . $row["userName"] . "</td>";
echo "<td>" . $row["email"] . "</td>";
echo "<td>" . $row["balance"] . "</td>";
$count = $row["id"];
?>
<td>
<form name="form" method="post" action="">
<input type="submit" name="add" value="+">
<input type="submit" name="subtract" value="-">
<?php
if (isset($_POST['add'])) {
$sql2 = "UPDATE users SET balance = balance + 5 where id = $count ";
$result2 = $conn->query($sql2);
}
if (isset($_POST['subtract'])) {
$sql2 = "UPDATE users SET balance = balance - 5 where id = $count";
$result2 = $conn->query($sql2);
}
echo $row["id"];
?>
</form>
</td>
</tr>
<?php
}
} else {
echo "0 results";
}
?>
You must send the row id back to the php script, not just the action 'add' or 'subtract'. You can identify the row by the id you want to update. The code will look something like this:
if (isset($_POST['id'], $_POST['add'])) {
$stmt = $conn->prepare('UPDATE users SET balance = balance + 5 WHERE id = ?');
if (!$stmt) {
die('Prepare failed: '.$conn->error);
}
$stmt->bindParam('d', $_POST['id']);
$result = $stmt->execute();
if (!$result) {
die('Execute failed: '.$stmt->error);
}
}
if (isset($_POST['id'], $_POST['substract'])) {
// ...
}
You add this code before the while loop and remove the existing query stuff from inside the while loop.

How to add a button to my PHP form that deletes rows from my MYSQL database [duplicate]

This question already has answers here:
How to add a delete button to a PHP form that will delete a row from a MySQL table
(5 answers)
Closed 1 year ago.
I am new to php coding.
I am adding each row delete button, but it should not working.
This is my html code:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "emp",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
$sql = "SELECT * FROM venu ";
$result = mysql_query($sql) or die(mysql_error());
?>
<table border="2" style= " margin: 0 auto;" id="myTable">
<thead>
<tr>
<th>name</th>
<th>id</th>
<th>rollnumber</th>
<th>address</th>
<th>phonenumber</th>
</tr>
</thead>
<tbody>
<?php
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['rollnumber'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['phonenumber'] . "</td>";
echo "<td><form action='delete.php' method='POST'><input type='hidden' value='".$row["address"]."'/><input type='submit' name='submit-btn' value='delete' /></form></td></tr>";
echo "</tr>";
}
?>
</tbody>
</table>
</body>
</html>
This is my delete code:
<?php
$connection = mysql_connect('localhost', 'root','');
if (!$connection)
{
die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db( "emp",$connection);
if (!$select_db)
{
die("Database Selection Failed" . mysql_error());
}
error_reporting(0);
session_start();
$name = $_POST['name'];
$id = $_POST['id'];
$rollnumber = $_POST['rollnumber'];
$address = $_POST['address'];
$phonenumber = $_POST['phonenumber'];
if($name!='' and $id!='')
{
$sql = mysql_query("DELETE FROM 'venu' WHERE name='balaji'AND id='93'AND rollnumber='93'AND address='bangalore'AND phonenumber='1234567890'");
echo "<br/><br/><span>deleted successfully...!!</span>";
}
else{
echo "<p>ERROR</p>";
}
mysql_close($connection);
?>
I am trying to delete each row using a button, but it is not working.
In your html view page some change echo "<td><a href='delete.php?did=".$row['id']."'>Delete</a></td>"; like bellow:
<?php
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['rollnumber'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['phonenumber'] . "</td>";
echo "<td><a href='delete.php?did=".$row['id']."'>Delete</a></td>";
echo "</tr>";
}
?>
PHP delete code :
<?php
if(isset($_GET['did'])) {
$delete_id = mysql_real_escape_string($_GET['did']);
$sql = mysql_query("DELETE FROM venu WHERE id = '".$delete_id."'");
if($sql) {
echo "<br/><br/><span>deleted successfully...!!</span>";
} else {
echo "ERROR";
}
}
?>
Note : Please avoid mysql_* because mysql_* has beed removed from
PHP 7. Please use mysqli or PDO.
More details about of PDO connection http://php.net/manual/en/pdo.connections.php
And more details about of mysqli http://php.net/manual/en/mysqli.query.php
First you need to change your button like
echo "<td>Delete</td>";
this will send the ID of the row which you want to delete to the delete.php
Secondly you need to change a bit your delete.php currently is wide open for SQL injections. Try using MySQLi or PDO instead
if(isset($_GET['id'])) {
$id = $_GET['id'];
$stmt = $mysqli->prepare("DELETE FROM venu WHERE id = ?");
$stmt->bind_param('i', $id);
$stmt->execute();
$stmt->close();
}
Of course if you need to add more parameters in delete query you should pass them also with the button..
EDIT: Simple example for update record
You can put second button on the table like
echo "<td>Update</td>";
Then when you click on it you will have the ID of the record which you want to update. Then in update.php
if(isset($_GET['id'])) {
$id = $_GET['id'];
$stmt = $mysqli->prepare("UPDATE venu SET name = ?, rollnumber = ?, address = ? WHERE id = ?");
$stmt->bind_param('sisi', $name, $rollnumber, $address, $id);
$stmt->execute();
$stmt->close();
}
Here ( in update.php ) you can have form which you can fill with new data and pass to variables $name, $rollnumber, $address then post it to update part.
Something to start with: PHP MySqli Basic usage (select, insert & update)
change up your query to use the dynamic value entered by the user, right now it is hard coded in there.
session_start();
require_once 'conn.php';
class myClass extends dbconn {
public function myClassFunction(){
try {
$id = $_GET['id'];
if(isset($_GET['id'])) {
$sql = "DELETE FROM tablename WHERE id = ?";
$stmt = $this->connect()->query($sql);
$stmt->bind_param('i', $id);
header("location: ../filepath/index.php");
}
} catch (PDOException $e) {
echo $sql . "<br>" . $e->getMessage();
}
}
}
This line is wrong, you need to set the WHERE clause to the data you get from the hidden input value
$sql = mysql_query("DELETE FROM 'venu' WHERE name='balaji'AND id='93'AND rollnumber='93'AND address='bangalore'AND phonenumber='1234567890'");
Should be:
$sql = mysql_query("DELETE FROM 'venu' WHERE address='"._POST['address']."'");
And in the little form you are using, change:
<input type='hidden' value='".$row["address"]."'/>
to:
<input type='hidden' name='address' value='".$row["address"]."'/>

Produce dynamic link from array

I have a searchpage, which works fine. Search results are displayed in a table with tr hover and a window.location to take the user to a different page.
What I am trying to achieve is a dynamic link for the window.location based on data from the array. All the data in the db belongs to 4 different categories, testkat, and I would like to direct the user to the right page depending on the value from testkat, and then using the 'testid' to fill in the data.
I have been trying numerous ways to achieve my goal, and searched both SE, Google etc, but no luck. I'm pretty new to PHP so using the right search term might have something to do with it.
From my point of view I'm thinking that I have to store the value from testkat in a variable, lets say $link. And from there make an IF statement, something like:
if ($results('testkat') == 'something') {
$link = "something.php?id='$testid'";
}
if ($results('testkat') == 'something_else') {
$link = "something_else.php?id='$testid'";
}
And from there put $link in the window.location
Here's my code:
<?php
$conn = mysql_connect("localhost", "root", "") or die("Couldn't do it: ".mysql_error());
mysql_select_db("db") or die(mysql_error());
mysql_set_charset('utf8',$conn);
$query = $_POST['query'];
$query = htmlspecialchars($query);
$query = mysql_real_escape_string($query);
$raw_results = mysql_query("SELECT * FROM db WHERE (`id` LIKE '%".$query."%') OR (`name` LIKE '%".$query."%') OR (`age` LIKE '%".$query."%')") or die(mysql_error());
$count = mysql_num_rows($raw_results);
if(isset($_POST['query'])) {
echo "<br>";
echo "Your search for <span class=\"bold\">" .$query. "</span> returned " . $count . " hits";
echo "<br>";
if(mysql_num_rows($raw_results) > 0){
echo "<br>";
echo "<table class=\"tbl-text\">";
echo "<tr class=\"tablelist\"><th>Heading 1</th><th>Heading 2</th><th>#</th><th>Heading 3</th><th>Heading 4</th><th>Heading 5</th>";
while($results = mysql_fetch_array($raw_results)){
echo "<tr onclick=\"window.location='#'\" style=\"cursor:pointer\" class=\"tr-hover\">";
echo "<td class=\"bordered\">" .$results['testid'] . "</td>";
echo "<td class=\"bordered\">" .$results['testkat'] . "</td>";
echo "<td class=\"bordered\">" .$results['intnr'] . "</td>";
echo "<td class=\"bordered\">" .$results['pro'] . "</td>";
}
echo "</table>";
}
else{
}
}
?>
Update:
Forgot to tell about the error. When doing it the way I think it should be done, I get an error message in the IF statement saying: Fatal error: Function name must be a string.
Referring to this one:
if ($results('testkat') == 'something') {
$link = "something.php?id='$testid'";
}
I know about MySQLi and PDO, working on it.
Eager to learn, so any hints and tricks are greatly appreciated :)
Chris
That method looks fine. You don't need the single quotations around $testid though
$link = "something_else.php?id=$testid";
As you've mentioned you should stop using mysql, get learning :)
Managed to fix it, and posting if someone else are having the same problem.
First, rewrote the whole thing to MySQLi.
Then I put an IF statement after the WHILE LOOP like this:
Connecting to db ->
if(isset($_POST['query'])) {
$query = $_POST['query'];
$query = htmlspecialchars($query);
$sql = $db->query("SELECT * FROM db WHERE (`?` LIKE '%".$query."%') OR (`?` LIKE '%".$query."%') OR (`?` LIKE '%".$query."%')");
$result = mysqli_query($db, sql);
$hits = $sql->num_rows;
echo "<br>";
echo "Your search for <span class=\"bold\">" .$query. "</span> returned " . $hits . " results";
echo "<br>";
if($sql->num_rows > 0){
echo "<br>";
echo "<table class=\"tbl-text\">";
while ($row = mysqli_fetch_array($sql)) {
if ($row['category'] == 'cat01'){
$link = 'cat01.php?id=' . $row['testid'] . '';
}
if ($row['category'] == 'cat02'){
$link = 'cat02.php?id=' . $row['testid'] . '';
}
if ($row['category'] == 'cat03'){
$link = 'cat03.php?id=' . $row['testid'] . '';
}
if ($row['category'] == 'cat04'){
$link = 'cat04.php?id=' . $row['testid'] . '';
}
echo "<tr onclick=\"window.location='$link'\" style=\"cursor:pointer\" class=\"tr-hover\">";
echo "<td class=\"bordered\">" .$row['testid'] . "</td>";
>>> more echo
}
There are probably more efficient ways to do this, but at least I got the results I was after, and the script is also more secure now using MySQLi

How to use MySQL query results as options in HTML forms

<?php
$con = new mysqli('localhost', 'root' ,'', 'world');
$query = 'SELECT * FROM city ORDER BY Name';
if ($result = mysqli_query($con, $query)) {
echo "<table>";
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr>"
echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['CountryCode'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
}
mysqli_close($con);
?>
This simple code will display every entries in this database. Now I would like to add a selective display option by choosing the CountryCode.
$query2 = 'SELECT DISTINCT CountryCode FROM city ORDER BY CountryCode';
How do I use the result I got from the query above and make it become radio buttons to choose what to display?
Similar to what you are doing already: Something like
while ($row = mysqli_fetch_assoc($result)) {
echo "<input type='radio' name='whatever' value='".$row['Name']."'>". $row['Name'];
}
Ofcourse the field names can be whatever you like them to be, as long as you select them in the query

Printing two separate SQL table data on the dynamically populated table - PHP

Can someone please help me on the following code? Simply put, I'm trying to get two separate SQL tables' data, one on the horizontal side (brands) and the other (distributors) on the vertical side of a dynamically populated table.
my issue is, if you go through the code I cant get the text boxes populated under each respective brand name I get to display from the database. The text boxes are appearing only for the first brand name column.
My second issue if how do I assign a unique ID or a name to a dynamically populated text box here?
<?php
$q=$_GET["q"];
include ("../connection/index.php");
$sql="SELECT * FROM distributors WHERE rsm='".$q."'";
$sqlq="SELECT * FROM brands";
$result = mysqli_query($db,$sql) or die ("SQL Error_er1");
$resultq = mysqli_query($db,$sqlq) or die ("SQL Error_er2");
echo "<table border='1'>
<tr>
<th>Distributor</th>";
"<tr>";
while($rowq = mysqli_fetch_array($resultq))
{
echo "<td>" . $rowq['bname'] . "</td>";
}
"</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['dname'] . "</td>";
echo "<td><input type='text' name='txt1'></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($db);
?>
You're creating loop without relating to eachother, and the same goes for the execution of the querys...
If you want to solve it you will have to nestle the loops together, something like:
<?php
$q=$_GET["q"];
include ("../connection/index.php");
$sql="SELECT * FROM distributors WHERE rsm='".$q."'";
$result = mysqli_query($db,$sql) or die ("SQL Error_er1");
echo "<table border='1'>
<tr>
<th>Distributor</th>";
"<tr>";
//Go through all distributors
while($rowq = mysqli_fetch_array($result)) {
echo "<td>" . $rowq['bname'] . "</td>";
//Show all brands for current distributor
$sqlq="SELECT * FROM brands where distributor_id = " . $rowq['rsm'];
$resultBrands = mysqli_query($db,$sql) or die ("SQL Error Brands");
while($row = mysqli_fetch_array($resultBrands))
{
$id = $row['rsm'];
echo "<tr>";
echo "<td>" . $row['dname'] . "</td>";
echo "<td><input type='text' name='textBox[]'></td>";
echo "</tr>";
}
//End show all brands for current distributor
}
//End Go through all distributors
A better solution though, would be something like (of course $q has to validated before input inside of query and also binded with bind_param()).
<?php
$q=$_GET["q"];
include ("../connection/index.php");
$sql = " SELECT * FROM distributors d";
$sql .=" LEFT JOIN brands b ON (d.brand_id = b.brand_id)";
$sql .=" WHERE d.rsm=$q";
$result = mysqli_query($db,$sql) or die ("SQL Error");
echo "<table border='1'>";
while($rowq = mysqli_fetch_array($result))
{
$id = rowq['rsm'];
echo "<tr>";
echo "<td>" . $rowq['dname'] . "</td>";
echo "<td>" . $rowq['bname'] . "</td>";
echo "<td><input type='text' name='textBox[]'></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($db);
?>
Take notice of the name='textBox[]'. From PHP you can access the variable with $_POST['textBox'] (or $_GET['textBox'] and PHP will return an array).

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