Undefined index: client_id" on line 1001 [duplicate] - php

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 2 years ago.
any help you offer on this will be useful. Am working on my school project and this error code keeps popping up on the application "Undefined index: client_id" on line 1001"
Here is the code
public function onNewclientaddress(){
$addShipmentForm = Settings::get('addShipmentForm',true);
$data = post();
= \Spot\Shipment\Models\Address::where('user_id', $data['client_id'])->update(['default' => 0]);
if ( $addShipmentForm == "add_form_normal"){
$subitem = new \Spot\Shipment\Models\Address;
$subitem->name = htmlspecialchars($data['street_addr']);
$subitem->user_id = htmlspecialchars($data['client_id']);
$subitem->street = htmlspecialchars($data['street_addr']);
$subitem->city = htmlspecialchars($data['city_id']);
$subitem->zipcode = htmlspecialchars($data['postal_code']);
$subitem->country = htmlspecialchars($data['country_id']);
$subitem->default = 1;
$subitem->created_at = \Carbon\Carbon::now();
$subitem->updated_at = \Carbon\Carbon::now();
$subitem->save();
}
else{
line 1001 is the first line of code in the post. please pardon my English

This error indicates that $data['client_id'] is not being set. You will need to ensure that whatever form is providing this data, is passing client_id correctly.
You can see what is currently in $data with a line like:
die(var_dump($data));
This will output the data on the page.
Ensure your client_id field in your form has a name attribute:
name="client_id"

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ErrorException (E_NOTICE) Undefined variable: actualLabels [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 3 years ago.
Help, i got an error "ErrorException (E_NOTICE) Undefined variable: actualLabels" in my code
$title = "Data Confusion Matrix";
$testing_data = DataTesting::count();
$klasifikasi = Klasifikasi::with('sentimen')->get();
foreach($klasifikasi as $kelas){
$predictedLabels[] = $kelas->sentimen->kategori;
$testing = DataTesting::where('id_testing',$kelas->id_testing)->first();
$twitter = TwitterStream::with('sentimen')->where('id_crawling',$testing->id_crawling)->first();
$actualLabels[] = $twitter->sentimen->kategori;
}
$getPrecision = new ControllerConfusionMatrix($actualLabels, $predictedLabels);
$accuracy = ControllerConfusionMatrix::score($actualLabels, $predictedLabels);
$recall = $getPrecision->getRecall();
$precision = $getPrecision->getPrecision();
Add this line to beginning of your code : $actualLabels = [];
You are getting error because when $klasifikasi is empty, then the statement inside loop is not executed. So $actualLabels variable is not created. In this case you get errro of (E_NOTICE) Undefined variable: actualLabels.
Hope you understand.
define Array()
$predictedLabels = array();
$actualLabels = array();
foreach($klasifikasi as $kelas){
$predictedLabels[] = $kelas->sentimen->kategori;
$testing = DataTesting::where('id_testing',$kelas->id_testing)->first();
$twitter = TwitterStream::with('sentimen')->where('id_crawling',$testing->id_crawling)->first();
$actualLabels[] = $twitter->sentimen->kategori;
}

The LIKE parameter in PDO and php not working [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Using LIKE in bindParam for a MySQL PDO Query [duplicate]
(2 answers)
Closed 3 years ago.
Why isn't this working? I've viewed the past posts on the subject and nothing worked for me.
$sql_strongsdef_ses = "SELECT id, strongs, etym_strongs FROM ".$tablename." WHERE etym_strongs LIKE ?";
$params = array("'%".$getSearchEtymStrongs."%'");
$result_strongsdef_ses = $conn->prepare($sql_strongsdef_ses);
$result_strongsdef_ses->execute($params);
echo "<span style=\"color: red;\">".$sql_strongsdef_ses."</span><br /><br />\n";
//while ($row_ses = $result_strongsdef_ses->fetch()){
while($row_ses = $result_strongsdef_ses->fetch(PDO::FETCH_ASSOC)){
$id_ses = $row_ses['id'];
$strongs_ses = $row_ses['strongs'];
$etymstrongs_ses = $row_ses['etym_strongs'];
}
var_dump($strongs_ses);
I get the error:
Notice: Undefined variable: strongs_ses in \edit.php on line 52

Very strange situation with Undefined index [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
php code:
1. if (isset($data['city_id']))
2. {
3. $city_id = "city_id='". $data['city_id']. "', ";
4. }
And I get:
Notice: Undefined index: city_id on line 3
How can this be?
Just ran your code sample and it works perfectly, I do not get 'undefined index' error - taking us to the big apple
<?php
$data['city_id']='New York';
if (isset($data['city_id']))
{
$city_id = "city_id='". $data['city_id']. "', ";
echo $city_id;
}
?>
output: city_id='New York',
Surely, without the $data['city_index']='New York'; I just get a blank screen, as the if condition is not met - no errors.

Error when writing code php [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
if i write code php same:
<?php
$title = strip_tags($_POST['title']);
?>
unknown error show!
Notice: Undefined index: title in C:\xampp\htdocs\file.php on line 3
$_POST has value, after submitting form, so before that anybody can't use $_POST ..
<?php
if(isset($_POST['title'])){
//Here in condition if(array_key_exists ( 'title' , $_POST )) can also be checked...
//OR if(!empty($_POST)) OR if(!empty($_POST['title'])) can also be put..
$title = strip_tags($_POST['title']);
}
?>
If both form and action in the same page, the first load will show error as there is no data posted. So first checj whether a POST has been made and then assign. Try this
$title = "";
if(isset($_POST['title'])){
$title = strip_tags($_POST['title']);
}

Notice: Undefined offset: 1 in E:\xampp\htdocs\WPTC-PMS5-5-14\edit_column.php on line 12 [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
when the below code works I get the above mentioned error , what should i do ? please help me ...
<?php
//session_start();
include("dbconnect_database.php");
$tname=$_GET['tn'];
$cname=$_GET['cn'];
$des=mysql_query("desc `$tname` `$cname`");
$row=mysql_fetch_array($des);
list($type, $b) = explode('[(]',$row[1]);
list($size) = explode('[)]',$b);
?>
Try changing this:
list($type, $b) = explode('[(]',$row[1]);
to this:
list($type, $b) = explode('[(]',$row[0]);
UPDATE
The error is telling you that 1 is not a valid index for $row, so that is the problem. Just before that line, try var_dump($row). This will tell what the valid indexes for $row are, and you should be able to use that to fix your code.
there is no ' around table and column name
$des=mysql_query("desc $tname $cname");
sidenote: use mysqli_query instead of mysql because use of mysql is been deprecated

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