This PHP code is for cheking for names who user is following.They are his friends or if they are not his $friends must show <td> <input type="submit" id="downloadbut" value="Follow"></td> if there are his $friends must show echo '<td> <span id="liketextvote">Following</span></td>'; if the user is following them. But my problem is I can't get result for $friends. The problem is
Notice: Array to string conversion in
C:\xampp\htdocs\Edu\1111111111111\userinfo.php on line 109 Array
. But inside while loop result is: megiantongeorgekatimery - this is example from MySQL database. I don't have idei how can I get this results. Someone can help?
$userfile = $_GET['username'];
$username = $user_data['username'];
if ($username != $userfile) {
$addfriends = mysql_query("SELECT `follow` FROM `friends` WHERE `username` = '$username' ORDER BY id DESC");
$friends = array();
while ($query_row = mysql_fetch_array($addfriends)) {
$val = $query_row['follow'];
echo $val; // hier I can get result example five name megiantongeorgekatimery
$friends[] = $val;
}
echo $friends; // hier I can get result example five name megiantongeorgekatimery
if (!in_array($userfile, $friends)) {
echo '<form action="addfriends.php?username='.$userfile.'" method="post">
<td> <input type="submit" id="downloadbut" value="Follow"></td>
</form>';
} else {
echo '<td> <span id="liketextvote">Following</span></td>';
}
}
You are trying to echo an array when you do echo $friends.
If that is just a debug statement, use print_r instead
print_r($friends);
You start out by using $friends as an array.
If you want to print the data out from the array, either loop through it using foreach or join in all together before printing.
foreach ($friends as $friend) {print "$friend<br>";}
OR
print join("<br>",$friends);
Related
The code below is to display the quiz(questions and answers)
When submitting, I am getting error:
"Array ( ['1'] => 1 ) 1
Notice: Undefined offset: 1 in C:\xampp\htdocs\HR\functions\functions_applicants.php on line 152
2".
<form method="post" action="index.php?results">
<?php
for($i=1; $i<27; $i++) {
$query = query("SELECT * FROM assess_questions WHERE question_id =$i");
confirm($query);
while($row = fetch_array($query)) {
?>
<div>
<h4><?php echo $row['question']; ?></h4>
<input type="radio" name="quizcheck['<?php echo $row['question_id']; ?>']"
value=1><?php echo $row['A']; ?><br>
<input type="radio" name="quizcheck['<?php echo $row['question_id']; ?>']"
value=2><?php echo $row['B']; ?><br>
<input type="radio" name="quizcheck['<?php echo $row['question_id']; ?>']"
value=3><?php echo $row['C']; ?><br>
<input type="radio" name="quizcheck['<?php echo $row['question_id']; ?>']"
value=4><?php echo $row['D']; ?><hr>
</div>
<?php
}
}
?>
<input class="btn btn-info" type="submit" name="submit_answers"
value="Next">
</form>
THIS IS THE FUNCTION TO CHECK FOR THE ANSWER. THIS IS WHERE IM GETTING THE ERROR FROM. ITS THE $i that's causing the error.
if(isset($_POST['submit_answers'])) {
$result = 0;
$i = 1;
$average = 0;
$item = ($_POST['quizcheck']);
print_r($item) ;
$query = query("SELECT * FROM assess_questions");
confirm($query);
while($row = fetch_array($query)) {
print_r($row['answer_id']);
$checked = ($row['answer_id']) == $item[$i];
if($checked) {
$result++;
}
$i++;
}
}
The clue is in the contents of $item which you have done a print_r on and got the result:
Array(['1'] => 1)
You're getting this result because in your html your radio buttons are labelled quizcheck['n'] where n is the question id. So presumably in this case you have pressed the first radio button in the first question. You should change the line which gives the radio buttons a name to
<input type="radio" name="quizcheck[<?php echo $row['question_id']; ?>]" value=1><?php echo $row['A']; ?><br>
(i.e. remove the single quotes around <?php echo $row['question_id']; ?>). This will make $item look like:
Array ( [1] => 1 )
so the test
($row['answer_id']) == $item[$i];
will work. Note the parentheses around $row['answer_id'] are unnecessary.
The other issue you are going to run into is that your form obviously doesn't require the user to submit an answer to every question. This means that in your while loop you need to check whether the user has submitted an answer to be checked against the result. If you are not going to make it compulsory to answer all questions, you will need to put an array_key_exists check around the result checking code:
if (array_key_exists($i, $item)) {
$checked = $row['answer_id'] == $item[$i];
if ($checked) {
$result++;
}
}
you begin with $i = 1;
in which case you'll avoid $item[0]; (first position) and it will mismatch $checked = ($row['answer_id']).
start with $i=0; as if there's only one answer $i[1] will not exists but $i[0];
****EDIT****
first check your query result for not being void:
example, having this db connection function/method:
<?php
function connection(){
try{
//host, user, passwd, DB name)
$mysqli = new mysqli($host, $user, $passwd, $dbName);
return $mysqli;
}catch (Exception $mysqlin){
echo "Error establishing connection with ddbb ".$mysqlin->getMessage();
}
}
?>
And modifying your code:
if(isset($_POST['submit_answers'])) {
$result = 0;
//indexes must start at 0 unless you ensure that you don't need 0 value and your algorithm will not keep trying to get a value out of bounds / array offset
$i = 0;
$average = 0;
//i assume that $_POST['quizcheck'] is an array, otherwise the $item[$i] will be an out of bounds, which match with your issue (array offset). But i ensured some parts of your structure below for you to learn:
$item = ($_POST['quizcheck']);
print_r($item) ;
//you'll must prepare this statement after, for security
$sql = query("SELECT * FROM assess_questions");
//we get the result of this query on $result, if possible
if($result = connection()->query($sql)){
//we get the first row inside $rs (usually called Record Set) as associative (it means that you'll call the values as the column name on DB unless you use AS "Alias" on your query.
$rs = mysqli_fetch_assoc($result);
//if the record set is not void:
while($rs!="") {
//assuming that your DB col is called answer_id
print_r($rs['answer_id']);
//checked will get the value of matching those two arrays, so make sure you control nullables:
if($checked = ($rs['answer_id']) == $item[$i]){
//only if $checked value could be set
if($checked) {
$result++;
}
}
$i++;
}
//repeat this to get the next value, when it has no more values it will be void so it will escape from while loop.
$rs = mysqli_fetch_assoc($result);
}
}
Never assume that you'll get always a value.
Never assume that users will put numbers in some input only because you told them to do it.
Never use dates without checking.
Never state an algorithm with not-controlled vars/function outputs.
Control all data all over across your code and comment it, it will help you to avoid issues and modify your code months after you coded it.
Hope it helps you.
I recommend you to get a hosting to test/code in a controlled environment.
My form have multiple checkboxes in it (each with the code):
<input type="checkbox" name="id[]" value="<? echo $row['id'] ?>">
when user select id = 2 , 3 and 9 and submit the value. In my update php i am using following code to get selected value
echo "Check box test<pre>" ;
print_r($id);
echo "</pre>";
if(!empty($_POST['id'])) {
foreach($_POST['id'] as $check) {
echo $check."\n";
}
}
// update data in mysql database
$sql="UPDATE table SET display = '2' WHERE id IN ($check)";
i am always getting last selected id updated like 9
but i am not getting result as i wanted like
$sql="UPDATE table SET display = '2' WHERE id IN (2,3,9)";
. please Help what to do. i am very new to php.
You are doing nothing with $check, Make id like this
$ids="";
if(!empty($_POST['id'])) {
foreach($_POST['id'] as $id) {
$ids[] = $id;
}
}
$check = implode(",", $ids);
Because you are iterating the loop and $check will have latest value for the iteration.
foreach($_POST['id'] as $check) {
echo $check."\n";
}
Why can't you try instead ?
$check[] = $_POST['id'];
I think you need to get a better understanding of echo and the diference between server-side and client-side.
What you are looking for is implode
if(!empty($_POST['id'])){
$check = '(' . implode(',', $_POST['id']) . ')';
}
I have two php page.
In the first I have looping checkbox array :
<td><input type="checkbox" name="cek[]" value=" <?php echo "$kodeinventarisit" ?>"></td>`
Then i submit form from page one to page two :
<?php
include 'koneksi.php';
$cek = $_POST['cek'];
$jumlah_dipilih = count($cek);
for($x=0;$x<$jumlah_dipilih;$x++){
$jojo = $cek[$x];
$coba = "select * from msstok where kodeinventarisit = '$jojo' ";
$cobaquery = mysql_query($coba);
$hasil = mysql_fetch_array($cobaquery);
$jenis = $hasil['jenis'];
?>
<input name="kode" type="text" id="license" value="<?php echo htmlentities($jenis) ; ?>" readonly="readonly" />
<?php
echo "$jojo";
}
?>
The problem is in the sql query return nothing, I try echo "$jojo" and it's print the value but in the text field is empty..
Does anyone have suggestions on how to fix this?
Thank You Very Much
1
What you are doing is bad.
Load your data before your loop and loop every result to print them.
2
Protect your sql request from injection.
Connect
$db = new mysqli("","root","","");
Prepare your request
$sql = "select * from msstok where kodeinventarisit = ? ";
$res = $db->prepare($sql);
$res->bind_param("sssd",$jojo);
Get results
$res->execute();
Documentation : http://php.net/manual/fr/book.mysql.php
If you want to pass the array you need to check if arrive in you second page.
<pre>
print_r($_POST['cek']);
</pre>
Now, if arrive here, you can read the values like this:
<?php
// If is array(), then you can go to loop
if(is_array($_POST['cek']))
{
// Run the loop
foreach($_POST['cek'] as $value)
{
// Show values per line
echo $value. "<br/>";
}
}
?>
You can read only 1 value of your array
<?php echo $_POST['cek'][0]; ?>
<?php echo $_POST['cek'][1]; ?>
<?php echo $_POST['cek'][2]; ?>
Conclusion
You can't pass array to SQL in query. If you want to use it, this is the only way with implode.
$coba = "SELECT * FROM msstok WHERE kodeinventarisit IN (".implode(',', $jojo).")";
$records = mysql_query($coba, $connection);
while ($row = mysql_fetch_array($records)) {
echo "Name: " . $rows['name'] . "<br />"; // replace the name for column you want
}
I have this table with checkboxes, my idea is to be able to delete the rows where the checkboxes have been checked.
With the help of Charaf jra, I was able to POST the uniqID of the row so I could DELETE it using mysql query on my delete.php page.
My problem now is in order to pass the uniqID I had to add it on the table, which doesnt look good. I mean, I dont want the ID number showing on the table. I have been reading on how to hide it, but none of the explanations I have read apply to my case.
Here is my code updated:
if ($arch = $pdo->prepare("SELECT name, age, uniqID FROM table WHERE id = ?")) {
$arch ->execute(array($id));
$data = $arch->fetchAll();
echo '<div class="coolTable" ><form method="post" action="delete.php"><table><tr><td>Name</td><td>Age</td><td>Check</td></tr>';
foreach ($data as $row){
echo '<tr>';
foreach ($row as $col){
$col=nl2br($col);
echo '<td>'.$col.'</td>';
}
echo '<td><input type="checkbox" name="checkbox[]" value="'.$col.'" id="checkbox"></td>'; //this captures the column ID so I can pass it through the `POST`
echo '</tr>';
}
echo '</table><input type="submit" value="Delete Selected"/></form></div>';
}
This works perfect. The only big problem is that I dont want the uniqID to be shown. Can anyone tell me how to hide it from the table and still be able to pass it through the POST?
if ($arch = $pdo->prepare("SELECT name, age, uniqID FROM table WHERE id = ?")) {
$arch ->execute(array($id));
$data = $arch->fetchAll();
echo '<div class="coolTable" ><form method="post" action="delete.php"><table><tr><td>Name</td><td>Age</td><td>Check</td></tr>';
foreach ($data as $row){
echo '<tr>';
echo '<td>'.nl2br($row['name']).'</td>';
echo '<td>'.nl2br($row['age']).'</td>';
echo '<td><input type="checkbox" name="checkbox[]" value="'.nl2br($row['uniqID']).'" id="checkbox"></td>'; //this captures the column ID so I can pass it through the `POST`
echo '</tr>';
}
echo '</table><input type="submit" value="Delete Selected"/></form></div>';
}
PHP arrays can be used as dictionaries (if you've seen python or ruby). PDO returns an array of pairs key-value, where your key is the name your column and the value, the value of your column. So you can access those values via
iteration (foreach)
iteration (index - direct access if you know the position of your element inside the array)
key (direct access without knowing the position of your element inside the array)
Assuming you want to loop the columns in each row, as opposed to manually echoing them as alkis suggests, get the results as an associative array and then unset $row['uniqID'];
if ($arch = $pdo->prepare("SELECT name, age, uniqID FROM table WHERE id = ?")) {
$arch ->execute(array($id));
$data = $arch->fetch(PDO::FETCH_ASSOC);
echo '<div class="coolTable" ><form method="post" action="delete.php"><table><tr><td>Name</td><td>Age</td><td>Check</td></tr>';
foreach ($data as $row){
if ( isset($row['uniqID']) ) {
unset($row['uniqID']);
}
echo '<tr>';
foreach ($row as $col){
$col = nl2br($col);
echo '<td>'.$col.'</td>';
}
echo '<td><input type="checkbox" name="checkbox[]" value="'.$col.'" id="checkbox"></td>'; //this captures the column ID so I can pass it through the `POST`
echo '</tr>';
}
echo '</table><input type="submit" value="Delete Selected"/></form></div>';
}
I have a form that is dynamically created based off multiple mysql tables. This form sends to an external page for processing.
this means that my $_POST data will always be different. I need to extract the post array, strip it down and create a query.
here's the print_r of the Posted array:
Array ( [userid] => 1 [modid1] => on [fid1] => on [fid3] => on [fid5] => on [fid7] => on [fid8] => on [modid3] => on )
as you can see I have three parts to this userid, modid, and fid. the catch is, the only way I could pass the id's I need is to name the fields that. So each modid and fid are rows in the db. the number after that is the id that needs updating, and of course "on" is from the check box.
so end result would be something like:
to give a better idea here's how I would write the query normally
for modid1:
UPDATE table SET var = var WHERE modid = 1
for fid1
UPDATE table SET var = var WHERE fid = 1
heres the code that generated this array:
<form id="ajaxsubmit" method="post" action="modules/users/updaterights.php">
<?php
$modsql = mysql_query("SELECT * FROM modules")or die("Mod failed " .mysql_error());
while($row = mysql_fetch_array($modsql))
{
echo '<div class="rights">';
echo "<ul>";
$userid = safe($_POST['user']);
$id = $row['id'];
$sql = mysql_query("SELECT * FROM modpermissions WHERE userid = '$userid' AND modid = '$id'")or die("Mod died " .mysql_error());
$sql2 = mysql_fetch_array($sql);
$modper = $sql2['modpermission'];
if($modper == 1){
echo '<li><input type="checkbox" name="modid'.$row["id"].'" checked> <b>'.$row["name"].'</b></li>';
}
if($modper == 0){
echo '<li><input type="checkbox" name="modid'.$row["id"].'"> <b>'.$row["name"].'</b></li>';
}
if($row['features'] == 1)
{
echo "<ul>";
$sql = mysql_query("SELECT * FROM features WHERE modid = '$id'")or die("Features loop failed " .mysql_error());
while($row2 = mysql_fetch_array($sql))
{
$userid2 = safe($_POST['user']);
$id2 = $row2['id'];
$sql3 = mysql_query("SELECT * FROM fpermissions WHERE userid = '$userid2' AND fid = '$id2'")or die("features died " .mysql_error());
$sql4 = mysql_fetch_array($sql3);
$fper = $sql4['fpermission'];
if($fper == 1){
echo '<li><input type="checkbox" name="fid'.$row2["id"].'" checked> '.$row2['feature'].'</li>';
}
if($fper == 0){
echo '<li><input type="checkbox" name="fid'.$row2["id"].'"> '.$row2['feature'].'</li>';
}
}
echo "</ul>";
}
echo "</ul>";
echo '</div>';
}
?>
<p><input type="submit" id="submit" value="Submit" class="button"> <input type="reset" class="reset" value="Reset Form"> </p>
</form>
its a mess I know, im learning. If someone can understand my question and point me in the right direction to accomplish what Im attempting I would be grateful.
First thing to do is to store the old value as well as having the check box (using a hidden field).
I would also suggest as a minimum using a fixed character as a delimeter in your field names so you can explode the field name to easy get the part that is the id.
Also consider using joins rather than looping around one result, and for each one doing another query.
Your output script would look something like this:-
<form id="ajaxsubmit" method="post" action="modules/users/updaterights.php">
<?php
$userid = safe($_POST['user']);
$modsql = mysql_query("SELECT modules.id, modules.features, modules.name, modpermissions.modpermission
FROM modules
LEFT OUTER JOIN modpermissions
ON modules.id = modpermissions.modid
AND modpermissions.userid = '$userid'")or die("Mod failed " .mysql_error());
$PrevModuleId = 0;
while($row = mysql_fetch_array($modsql))
{
if ($PrevModuleId != $row['id'])
{
if ($PrevModuleId != 0)
{
echo "</ul>";
echo '</div>';
}
echo '<div class="rights">';
echo "<ul>";
$PrevModuleId = $row['id'];
}
echo '<li><input type="checkbox" name="modid_'.$row["id"].'" '.(($row['modpermission'] == 1) ? "checked='checked'" : "").'><input type="hidden" name="modid_old_'.$row["id"].'" value="'.$row['modpermission'].'"> <b>'.$row["name"].'</b></li>';
if($row['features'] == 1)
{
echo "<ul>";
$sql = mysql_query("SELECT features.id, features.feature, fpermissions.fpermission
FROM features
INNER JOIN fpermissions
ON features.id = fpermissions.fid
AND fpermissions.userid = $userid
WHERE modid = '$id'")or die("Features loop failed " .mysql_error());
while($row2 = mysql_fetch_array($sql))
{
echo '<li><input type="checkbox" name="fid_'.$row2["id"].'" '.(($row2['fpermission'] == 1) ? "checked='checked'" : "").'><input type="hidden" name="fid_old_'.$row2["id"].'" value="'.$row2['fpermission'].'"> '.$row2['feature'].'</li>';
}
echo "</ul>";
}
}
if ($PrevModuleId != 0)
{
echo "</ul>";
echo '</div>';
}
?>
<p><input type="submit" id="submit" value="Submit" class="button"> <input type="reset" class="reset" value="Reset Form"> </p>
</form>
You can then loop through each entry on the $_POST array, explode the key based on the _ character, check when the values have changed and if needs be do an update Or possibly you can use an INSERT instead, but using ON DUPLICATE KEY update type syntax (this way you can update many rows with different values easily).
Note you also need to put the userid value somewhere in your form (probably as another hidden field) so you have the value to process with the updates.