I have been following a lesson on how to make an admin page. I got all the information out of my database to a table on the page. I have an update button and when I change the information and press the button I receive this error: Warning: undefined array key "WebID" in ..\Update.php on line 3
From my search online everyone is trying to change the code so that if array key does not exist: return null. I tried that and the error does not appear no more, but the table does not change.
Any thoughts?
This is the code:
<?php
require_once("DB/DB.php");
$SearchQueryParameter = $_GET["WebID"];
if (isset($_POST["Update"])) {
$Ename = $_POST["Ename"];
$Eid = $_POST["Eid"];
$Erank = $_POST["Erank"];
$Eemail = $_POST["Eemail"];
$Edate = $_POST["Edate"];
$Epassword = $_POST["Epassword"];
$Specialisms = $_POST["Specialisms"];
global $ConnectingDB;
$sql ="UPDATE emp_data SET Ename='$Ename', Eid='$Eid', Erank='$Erank', Eemail='$Eemail', Edate='$Edate', Epassword='$Epassword',
Specialisms='$Specialisms' WHERE WebID='$SearchQueryParameter'";
$Execute = $ConnectingDB->query($sql);
if ($Execute) {
echo '<script>window.open("adminpage.php?WebID=Recored Updated","_self")</script>';
}
}
?>
<?php
<?php
global $ConnectingDB;
$sql = "SELECT * FROM emp_data WHERE WebID='$SearchQueryParameter'";
$stmt = $ConnectingDB->query($sql);
while ($DataRows = $stmt->fetch()) {
$WebID = $DataRows["WebID"];
$Ename = $DataRows["Ename"];
$Eid = $DataRows["Eid"];
$Erank = $DataRows["Erank"];
$Eemail = $DataRows["Eemail"];
$Edate = $DataRows["Edate"];
$Epassword = $DataRows["Epassword"];
$Specialisms = $DataRows["Specialisms"];
}
?>
Html file used to update:
<form id="UpdateForm" method="post" action="Update.php?WebID<?php echo $SearchQueryParameter; ?>">
<div class="form-group">
<button type="submit" name="Update" class="form-control-submit-button">Update</button>
</div>
you have to write the form action like this.. you missed the = sign
action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>"
<form id="UpdateForm" method="post" action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>">
You missed the = sign, in the url
I want to delete a row from a form select when i click on submit input with an sql query,(i think i am wrong on something, but i don't understand what) as you can see below for my example :
My list and the blue case i want to delete on submit
My actual code, and the $supp i want to do when the user click on submit
`
<form method="POST">
<select>
<?php
// Drop Down
$res = null;
$sql2 = "SELECT `sinistre_type` FROM `form_sinistre`";
$query2 = $db->prepare($sql2);
$query2->execute();
// INIT > PREP > EXEC > SUPP
$supp = "DELETE FROM `form_sinistre` WHERE `sinistre_type` = '$res'";
$query3 = $db->prepare($supp);
$sendbddsupp = $query3->execute();
echo "<option disabled selected>..Choix Possible..</option>\n";
while ($res = $query2->fetch(PDO::FETCH_NUM)) {
echo "<option name='res'>" . $res[0] . "</option>\n";
}
?>
</select>
<input type="submit" value="Supprimer">
</form>
`
Some $_POST config
`
<?php
session_start(); //debut de SESSION
include("config.php"); //Appel de la bdd
// ... INIT VARIABLES ...
$sinistre_type = "";
$sinistre_desc_dmg = "";
$list = "";
if (empty($_POST)) { // SANS COOKIES / POST
} else { // AVEC COOKIES / POST
$sinistre_type = $_POST['nom'];
$sinistre_desc_dmg = $_POST['vent'];
$res = $_POST['res'];
$sql = "INSERT INTO `form_sinistre` (sinistre_type, sinistre_desc_dmg) VALUES (:sinistre_type, :sinistre_desc_dmg)";
$query = $db->prepare($sql);
$query->execute(array(':sinistre_type' => $sinistre_type, ':sinistre_desc_dmg' => $sinistre_desc_dmg));
}
var_dump(isset($_POST['res']));
?>
`
(EDIT : my list is linked with my db and working that why i want to send sql query)
Thanks by advance for your help, if you need more information let me know :)
I have 7 file input fields namely:
<input type="file" accept="image/*" name="imgreq1">
<input type="file" accept="image/*" name="imgreq2">
<input type="file" accept="image/*" name="imgreq3">
<input type="file" accept="image/*" name="imgreq4">
<input type="file" accept="image/*" name="imgreq5">
<input type="file" accept="image/*" name="imgreq6">
<input type="file" accept="image/*" name="imgreq7">
How can I check if the other input fields are empty if I for example uploaded a file in the first and the second input fields. Then submit the form leaving the other fields empty?
Update:
<?php
if(isset($_POST['sumit'])){
$count = count($_FILES);
for($i = 1; $i <= $count; ++$i){
if(is_uploaded_file($_FILES['imgreq'.$i]['tmp_name']) || !file_exists($_FILES['imgreq'.$i]['tmp_name'])){
echo "$i";
// your code
}else{
//to retrieve user_id to stored in the request table in the database
$query = "SELECT * FROM dummyclients_tbl WHERE user_id = '".$_SESSION['user']."'";
if (!$result = mysql_query($query)) {
exit(mysql_error());
}
if(mysql_num_rows($result) > 0){
while($row = mysql_fetch_assoc($result)){
$sid= ''.$row['user_id'].'';
$coll=''.$row['college'].'';
$stat="Pending";
//$query="SELECT document_name FROM document_tbl WHERE document_id = '$passed_id'";
//$dn=mysql_query($query);
//$getname=mysql_fetch_assoc($dn);
//var_dump($getname);
//to analyze the contents of the image selected in filebrowser1
$image1=addslashes($_FILES['imgreq1']['tmp_name']);
$name1=addslashes($_FILES['imgreq1']['name']);
$image1=file_get_contents($image1);
$image1=base64_encode($image1);
//to analyze the contents of the image selected in filebrowser2
$image2=addslashes($_FILES['imgreq2']['tmp_name']);
$name2=addslashes($_FILES['imgreq2']['name']);
$image2=file_get_contents($image2);
$image2=base64_encode($image2);
//to analyze the contents of the image selected in filebrowser3
$image3=addslashes($_FILES['imgreq3']['tmp_name']);
$name3=addslashes($_FILES['imgreq3']['name']);
$image3=file_get_contents($image3);
$image3=base64_encode($image3);
//to analyze the contents of the image selected in filebrowser4
$image4=addslashes($_FILES['imgreq4']['tmp_name']);
$name4=addslashes($_FILES['imgreq4']['name']);
$image4=file_get_contents($image4);
$image4=base64_encode($image4);
//to analyze the contents of the image selected in filebrowser5
$image5=addslashes($_FILES['imgreq5']['tmp_name']);
$name5=addslashes($_FILES['imgreq5']['name']);
$image5=file_get_contents($image5);
$image5=base64_encode($image5);
//to analyze the contents of the image selected in filebrowser6
$image6=addslashes($_FILES['imgreq6']['tmp_name']);
$name6=addslashes($_FILES['imgreq6']['name']);
$image6=file_get_contents($image6);
$image6=base64_encode($image6);
//to analyze the contents of the image selected in filebrowser7
$image7=addslashes($_FILES['imgreq7']['tmp_name']);
$name7=addslashes($_FILES['imgreq7']['name']);
$image7=file_get_contents($image7);
$image7=base64_encode($image7);
//function nga defined sa dalum para i insert ang uploaded files sa databasess
saveimage($sid,$passed_id,$image1,$image2,$image3,$image4,$image5,$image6,$image7,$stat,$coll);
}
}
}
}
}
function saveimage($sid,$passed_id,$image1,$image2,$image3,$image4,$image5,$image6,$image7,$stat,$coll){
$con=mysql_connect("localhost","root","");
mysql_select_db("dummy",$con);
$qry="INSERT INTO request_tbl (user_id,document_id,imgreq1,imgreq2,imgreq3,imgreq4,imgreq5,imgreq6,imgreq7,request_status,college) VALUES ('$sid','$passed_id','$image1','$image2','$image3','$image4','$image5','$image6','$image7','$stat','$coll')";
$result=mysql_query($qry,$con);
if($result){
?>
<script>alert('Requirements Successfully Submitted!');</script>
<?php
}else{
?>
<script>alert('Error while submitting form!');</script>
<?php
}
}
?>
From OP's comment,
But how can you do it using for each loop this feature of php is a deep one ...
Use a simple for loop, in conjunction with is_uploaded_file() function, to check whether user has uploaded a file via HTTP POST or not, like this:
$count = count($_FILES);
for($i = 1; $i <= $count; ++$i){
if(is_uploaded_file($_FILES['imgreq'.$i]['tmp_name'])){
// user has uploaded a file
}
}
Update:
Based on the below discussion with OP, the complete solution would be like this:
<?php
if(isset($_POST['sumit'])){
$count = count($_FILES);
$query = "SELECT * FROM dummyclients_tbl WHERE user_id = '".$_SESSION['user']."'";
if (!$result = mysql_query($query)) {
exit(mysql_error());
}
if(mysql_num_rows($result)){
$row = mysql_fetch_assoc($result);
$sid = $row['user_id'];
$coll =$row['college'];
$query = "INSERT INTO request_tbl (user_id,document_id,imgreq1,imgreq2,imgreq3,imgreq4,imgreq5,imgreq6,imgreq7,request_status,college) VALUES ('$sid','$passed_id'";
for($i = 1; $i <= $count; ++$i){
if(is_uploaded_file($_FILES['imgreq'.$i]['tmp_name']) && $_FILES['imgreq'.$i]['size']){
$query .= ",'" . base64_encode(file_get_contents(addslashes($_FILES['imgreq'.$i]['tmp_name']))) . "'";
}else{
$query .= ",NULL";
}
}
$query .= ",'$stat','$coll')";
saveimage($query);
}
}
function saveimage($qry){
$con = new mysqli("localhost", "root", "", "dummy");
$result=mysqli_query($con, $qry);
if($result){
?>
<script>alert('Requirements Successfully Submitted!');</script>
<?php
}else{
?>
<script>alert('Error while submitting form!');</script>
<?php
}
}
?>
As a sidenote, learn about prepared statement as it prevents your query from any kind of SQL injection attacks. Here's a good read on how you can prevent SQL injection in PHP.
I have created a database with a table and that table with one column of varchar datatype.
Now I want to delete the rows without using
DELETE FROM songs_list WHERE songs_name = 'Shakira Waka Waka'
This is so because I cannot write the name of the checkbox with spaces. For example:
<?php for ($printedsno=1; $printedsno <$noio+1 ; $printedsno++){?>
<?php while($row = mysqli_fetch_assoc($query7) ):?>
<input type="checkbox" name="<?php echo "cb".$row['songs_name'];?>"/>
<?php echo $printedsno?> <?php echo $row['songs_name']?></br>
<?php
break;
endwhile;
}
?>
Here We can not write the checkbox name as <?php echo "cb".$row['songs_name'];?>
That is why I want to use a psudo-Column. But For the Psudo-Column We can not be able to use a where clause beacause this column does'nt exist.
I want to delete the row from table with only column.
<?php
include 'database.php';
$noofsongs = "select * from songs_list";
$query8 = mysqli_query($con, $noofsongs);
$noio = mysqli_num_rows($query8);
$flag = 0;
$del = 0;
$row = mysqli_fetch_assoc($query8);
$song_names = $row['songs_name'];
if(isset($_POST['deletethis'])){
for($j=1; $j<=$noio ;$j++){
while($row = mysqli_fetch_assoc($query8) ){
if (isset($_POST['cb'.$song_names])){
$deletesongsquery = "DELETE FROM `songs_list` WHERE `songs_name` = $song_names" ;
$query4 = mysqli_query($con, $deletesongsquery);
if($query4){ $del++;}
$flag = $flag + 1;
}
}
}
if($del != $flag){echo "Can Not Be Deleted All Selected Songs.";}
else{echo "Deleted $del Selected Songs.";}
}
?>
Can anyone help me in this??
i have a php page called page1.php with this form
<form id="myForm" action="page2.php" method="post">
<label for="name">a label:</label><input type="submit" name="SubmitCar" value="Done" id="fbutton" /> <br />
<br />
<select name="selectCar">
<?php
session_start();
$user = "cardatabase";
$password = "";
$host = "";
$database = "my_cardatabase";
$connessione = mysql_connect($host, $user, $password) or die( mysql_error() . " <br/> could not connect to server");
mysql_select_db($database, $connessione) or die( mysql_error() . " <br/> could not connect datbase");
$id = $_SESSION['myid'];
$query = "select IDCar
from Car";
$result = mysql_query($query, $connessione) or die(mysql_error());
if( mysql_num_rows($result) > 0 ) {
$array = array();
$i = 0;
while ($row = mysql_fetch_array($result)) {
$array[$i] = $row['IDCar'];
++$i;
}
for ($i = 0; $i < count($array); ++$i) {
echo "<option value='$array[$i]'>$array[$i]</option>";
}
}
mysql_close();
?>
</select>
</form>
Simply fill the select box from DB. Now here's the problem. When i reach page2.php i need the value of the select box and i tried this
page2.php
<?php
$value = $_POST['selectCar'];
?>
But it's not working, so i tried to use sessions in this way
page1.php
</form>
<?php
if(isset($_POST['SubmitCar'])){
$_SESSION['idAuto'] = $_POST['selectCar'];
}
?>
out of the form, but still not working. What can i do to get this value in page2.php??
Try using this code to build your <option> , and use mysqli when working with the database
$result = mysqli_query($connessione, $query) or die(mysql_error());
if( mysqli_num_rows($result) > 0 ) {
$i=0;
while ($row = mysqli_fetch_row$result)) {
echo "<option value='".$row[$i]."'>".$row[$i]."</option>";
$i++;
}
}
Also, it's not indicated to have mysql tables with names in uppercase, or fields with names in uppercase.
Page2.php
You forgot to echo it.
<?php
$value = $_POST['selectCar'];
echo $value;
?>
I found a simple workaround for my problem. It might not be perfect but it works.
i deleted the action="page2.php" in the form of page1
in page1.php add this code outside the form
This will do the trick
if(isset($_POST['SubmitCar'])){
$_SESSION['idCar'] = $_POST['selectCar'];
header('Location:page2.php');
}