PHP new strings update in DB with unshift? - php

I have a problem to adding more strings in my database.
The idea is: SELECT information, then added array together, after these UPDATE to database.
These are in one code, but UPDATE not working with summed arrays only separately.
With echo I see the array_unshift is working well, the data is good, but not updating.
Need I change something on the server? Maybe version?
(I don't get mysqli_error!)
//CHECKBOX KIOLVASÁSA DB-BŐL!
$sql = ("SELECT id, checkbox FROM osszesito WHERE id = '$id'");
//$result = mysqli_query($conn, $sql);
//if (mysqli_num_rows($result) > 0) {
if ($result = mysqli_query($conn, $sql)) {
while($row = mysqli_fetch_assoc($result)) {
//EREDETI SOR LISTÁZÁSA
$original_array = array( $row["checkbox"] );
$x ="";
echo 'Eredeti sor: ';
foreach ($original_array as $x)
{
echo "$x "."<br><br>";
}
//EREDETI SOR KIEGÉSZÍTÉSE AZ ÚJ ADATTAL
array_unshift($original_array, $chb);
$last ="";
echo "Új sor: "."<br>";
foreach ($original_array as $last)
{
echo $last."<br>";
}
//ÚJ SOR FRISSÍTÉSE A DB-BEN!
//$sqla = "UPDATE osszesito SET checkbox = '$chb' WHERE id = '$id' ";
$sqla = "UPDATE osszesito SET checkbox = '$last' WHERE id = '$id' ";
if (mysqli_query($conn, $sqla)) {
echo "ÚJ SOR ELMENTVE!";
//header("Location: /megrendelesek/index.php");
} else {
echo "Hiba a beírás során: " . mysqli_error($conn);
}
}
///////////////////////////////////////////////
//LEZÁRÁS
} else {
echo "Jelenleg nincs megrendelés az adatbázisban!";
}
mysqli_close($conn);

Related

updating my database values through php

Hi I am trying to do a Registration that the users will put their name password and their answers to some questions and then an admin will manually answer to it if it's accepted.I did the system that loads their name password and answers in the database,and I also ran the things that will show the answers to the admin,but I can't figure a way to change a value just for one user not for all of them,I will leave you my codes and everything over here.
Here is my admin.viewapplications.php code
(Here,it shows everything fine,but I can't figure a way that the button to act just for one id not for all)
<?php
//include(__DIR__ . "/signup.php");
include("../resources/config.php");
//$name = $_POST['Name'];
//$mg = $_POST['MG'];
//$pg = $_POST['PG'];
//$rk = $_POST['RK'];
$sql = "SELECT id, name, tutorial, MG, PG, RK FROM rp_users WHERE tutorial = 2";
//$tutorial = "SELECT tutorial FROM rp_users";
$result = mysql_query($sql);
//$result2 = mysql_query($tutorial);
//$value = mysql_fetch_object($result2)
/*if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}*/
//if($value > 1)
//
while($row = mysql_fetch_array($result))
{
//$tutorial = row["tutorial"];
//f($tutorial == 2)
//}
$id = $row["id"];
$name = $row["name"];
$mg = $row["MG"];
$pg = $row["PG"];
$rk = $row["RK"];
echo "ID: " . $id."<br> <br>";
echo "Nume: " . $name."<br> <br>";
echo "MG: " . $mg."<br> <br>";
echo "PG: " . $pg."<br> <br>";
echo "RK: " . $rk."<br> <br>";
echo '<form action="./?p=applicationaccept" method="POST">';
echo '<input type="submit" name="accept" value="Accepta">';
echo '</form><br>';
echo '<form action="./?p=applicationdeny" method="POST">';
echo '<input type="submit" name="deny" value="Respinge">';
echo '</form><br> <br> <br>';
}
//}
//
?>
And here is my applicationaccept.php
<?php
include("../admin/admin.viewapplications.php");
include("../resources/config.php");
$iduser = $id;
$sql = "UPDATE rp_users SET tutorial=0";
$result = mysql_query($sql);
if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}
/*while($row = mysql_fetch_array($result))
{
}*/
?>
I think what you want to do is a simple UPDATE to your MySQL database..
but make sure you format the PHP code you're using otherwise it'll give you an ERROR!
Also you have to use 'mysqli' now in PHP!
<?php
$someID = '1';
$sql = "UPDATE `rp_users` SET `tutorial`= '0' WHERE `id` = $someID";
$result = mysqli_query($link, $sql);
if($result)
{
echo "Success";
}
else
{
echo ("Error");
}
?>
BTW I forgot to mntion the '$link' is the connection to your database!
As of my understanding of your question if your form action is applicationaccept.php and you are trying to update for one user in applicationaccept.php file, try this:
<?php
include("../admin/admin.viewapplications.php");
include("../resources/config.php");
$iduser = $_POST["id"]; // pass id as parameter in form
$sql = "UPDATE rp_users SET tutorial=0";// change this line to following line
$sql = "UPDATE rp_users SET tutorial=0 where id=$iduser";
$result = mysql_query($sql);
if($result)
{
echo "Succes";
}
else
{
die(mysql_error());
}
?>
Be aware your code is vulnerable

How to access while array outside the loop in PHP

I want to access while array $row2 outside the loop so that I can compare in if else condition, because $row2 array contains many number. Is there any solution make $row array accessible outside the loop or any other method?
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while($row = mysqli_fetch_array($query)) {
$row2 $row['id'];
}
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
?>
If you want to loop over two dependent things you need to nest them or else $row2 will always contain the last value in the while loop:
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($query)) {
$row2 = $row['id']; // I added an = sign here.
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
}
If this isn't what you want please clarify your question.
Pass your variable to a separate function that performs whatever task you want.
<?php
while($row = mysqli_fetch_array($query))
{
MyFunction($row2);
}
function MyFunction($passed_row2_value){
// perform whatever task you want here.
}
?>
This shows with some format the ids retrieved by the query and fills the blanks without format.
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
$row2 = [];
while($row = mysqli_fetch_array($query)) {
$row2[$row['id']] = true;
}
for($i=1;$i<=10;$i++) {
if ($row2[$i]) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
} else {
echo $i.'<br>';
}
}
?>

Repeating SQL insert statement in loop till id has certain value?

<?php
include ("db.php");
$id2 = "SELECT MAX(id) FROM osalejad;";
$id = $conn->query($id2);
if (id<= 8){
while($id<= 7) {
$min = 1;
$max = 20;
$suvanumber = rand($min, $max);
$bar = (string) $suvanumber;
$prii= "vaba2" . $bar;
$sql="INSERT INTO osalejad (name) VALUES ('$prii')";
$result = $conn->query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
}
}
else {
echo '8 osalejat on juba kirjas';
}
?>
I have a question, i want to insert data into MYSQl table, word with a random number, but problem is, it never work as it should - if i have 3 records in name column, then till, it adds 8 more records, instead of 5 records. Is something outside of scope atm?
I will use the DB in this code.
<?php
include("db.php");
$sql = "SELECT * FROM `osalejad`;";
$result = $conn->query($sql);
$var = array();
if ($result->num_rows > 0) {
// output data of each row
while ($row = mysqli_fetch_array($result)) {
$var[] = $row["name"];
}
} else {
/// echo "0 results";
}
//shuffle the array, and encode into json.
shuffle($var);
$brack=json_encode($var);
$testbrack= json_encode(array_chunk($var,2));
$final = '{';
$final .= '"teams":';
$final .= $testbrack;
$final .= ',"results":[[[[0,0],[null,null]],[[null,null],[null,null]]]]}';
//updating last tournament bracket.
$sql1 = "UPDATE lan_brackets SET json='$final' ORDER BY tid DESC LIMIT 1;";
$roundOne=$conn->query($sql1);
?>

Update Multiple Rows (PHP + MySQL)

I am working on a lead management system - and as the database for it grows the need for more bulk functions appears - and unfortunately I am getting stuck with one of them. The database stores many different leads - with each lead being assigned to a specific closer; thus the database stores for each lead the lead id, name, closer name, and other info. The main lead list shows a checkbox next to each lead which submits the lead id into an array:
<input type=\"checkbox\" name=\"multipleassign[]\" value=\"$id\" />
Now this all goes to the following page:
<?php
include_once"config.php";
$id = $_POST['multipleassign'];
$id_sql = implode(",", $id);
$list = "'". implode("', '", $id) ."'";
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$closer = mysql_result($result,$i,"business_name");
$businessname = mysql_result($result,$i,"closer");
echo "$closer - $businessname";
echo"<br>";
++$i; } } else { echo "The database is empty"; };
echo "<select name=\"closer\" id=\"closer\">";
$query2 = "SELECT * FROM members ";
$result2 = mysql_query($query2);
$num2 = mysql_num_rows ($result2);
if ($num2 > 0 ) {
$i2=0;
while ($i2 < $num2) {
$username = mysql_result($result2,$i2,"username");
$fullname = mysql_result($result2,$i2,"name");
echo "<option value=\"$fullname\">$fullname</option>";
++$i2; } } else { echo "The database is empty"; }
echo "</select>";
?>
I want to be able to use the form on this page to select a closer from the database - and then assign that closer to each of the leads that have been selected. Here is where I have no idea how to continue.
Actually - i got it. I don't know why I didn't think of it sooner. First off I passed the original $list variable over to the new page - and then:
<?php
include_once"config.php";
$ids = $_POST['list'];
$closer = $_POST['closer'];
$query = "UPDATE `promises` SET `closer` = '$closer' WHERE id IN ($ids) ";
mysql_query($query) or die ('Error updating closers' . mysql_error());
echo "A new closer ($closer) was assigned to the following accounts:";
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$businessname = mysql_result($result,$i,"business_name");
echo "<li>$businessname";
++$i; } } else { echo "The database is empty"; };
?>
The updated page before this:
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$closer = mysql_result($result,$i,"business_name");
$businessname = mysql_result($result,$i,"closer");
echo "$closer - $businessname";
echo"<br>";
++$i; } } else { echo "The database is empty"; };
echo "<form name=\"form1\" method=\"post\" action=\"multiple_assign2.php\">";
echo "<input type=\"hidden\" name=\"list\" value=\"$list\" />";
echo "<select name=\"closer\" id=\"closer\">";
$query2 = "SELECT * FROM members ";
$result2 = mysql_query($query2);
$num2 = mysql_num_rows ($result2);
if ($num2 > 0 ) {
$i2=0;
while ($i2 < $num2) {
$username = mysql_result($result2,$i2,"username");
$fullname = mysql_result($result2,$i2,"name");
echo "<option value=\"$fullname\">$fullname</option>";
++$i2; } } else { echo "The database is empty"; }
echo "</select>";
echo "<input name=\"submit\" type=\"submit\" id=\"submit\" value=\"Reassign Selected Leads\">";
?>
After you select the leads and submit the form , your script should show them in a list with hidden inputs (with name=leads[] and value=the_lead's_id) and next to each lead there will be a dropdown box () which will be populated with all the closers.
After choosing and sending the second form your script will "run" all-over the leads' ids array and update each and every one of them.
Got the idea or you want some code?

To delete queries! I've tried with the posts already available. It didn't work.

My first page to delete queries selected by user query.php which is working absolutely fine:
<form method=post action="delete.php">
List of queries<br/>
<?php
$ebits = ini_get('error_reporting');
error_reporting($ebits ^ E_NOTICE);
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("testdb") or die(mysql_error());
echo "<br />";
$query = "select * from queries ";
$result = mysql_query($query) or die(mysql_error());
$count=mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
print "<input type='checkbox' name='Query[]' value=\"".$row['queryId']."\"> ";
echo " ". $row['name']." ". $row["address"]." ". $row["contactNo"]."
". $row["query"];
echo "<br />";
}
?>
<input type="submit" value="Delete" name="Delete">
<br/>
</form>
I've tried with following codes for second page delete.php but nothing seems to work.
Code1:
<?php
if($_POST['Delete'])
{
if(count($_POST['checkbox']) > 0) {
foreach($_POST['checkbox'] as $checkbox)
{
$del_id=$checkbox;
$sql = "DELETE * FROM queries WHERE `queryId`= '$del_id'";
$result = mysql_query($sql);
mysql_error();
}
echo "Selected Rows deleted";
} else {
$NEW="Nothing to Delete";
}
}
?>
Code2:
<?php
if(($_POST['Delete']))
{
$count=array();
$count=$_POST['checkbox'];
for($i=0;$i<count($count);$i++){
$del_id = $checkbox[$i];
$sql = "DELETE FROM queries WHERE queryId='$del_id' ";
$result = mysql_query($sql);
}
$NEW="Selected records Deleted";
}
var_dump($_POST['checkbox']);
var_dump($count);
?>
Your checkbox names are "Query", but you're accessing it as $_POST['checkbox']. This should be $_POST['Query'] instead.
EDIT checking from your updated code:
if($_POST['Delete']) {
if(count($_POST['Query']) > 0) {
foreach($_POST['Query'] as $checkbox) {
$del_id=$checkbox;
$sql = "DELETE * FROM queries WHERE queryId= '$del_id'";
$result = mysql_query($sql);
mysql_error();
}
echo "Selected Rows deleted";
}
else {
$NEW="Nothing to Delete";
}
}
Instead of this:
$del_id=$checkbox;
do this:
// if queryId is numeric
$del_id=intval($checkbox);
This makes sure that the value you're working with is numeric, instead of potential malicious input from your user. I'm going under the assumption that queryId is numeric. If it's not, then you need to do this:
// if queryId is not numeric:
$del_id = mysql_real_escape_string($checkbox);
Your DELETE syntax is incorrect:
$sql = "DELETE * FROM queries WHERE queryId= '$del_id'";
You want just DELETE FROM. Also if the value for queryId is numeric, you don't need the quotes around it:
$sql = "DELETE FROM `queries` WHERE `queryId` = $del_id";
Finally, your MySQL error call doesn't do anything useful as is:
mysql_error();
Here's how you should do this, along with the rest of the code:
if($_POST['Delete']) {
if(count($_POST['Query']) > 0) {
foreach($_POST['Query'] as $checkbox) {
$del_id= intval($checkbox);
$sql = "DELETE FROM `queries` WHERE `queryId` = $del_id";
$result = mysql_query($sql);
if(!$result) {
echo "There was an error in the query: " . mysql_error();
}
}
echo "Selected Rows deleted";
}
else {
$NEW="Nothing to Delete";
}
}

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