Multiple Notice: Undefined variable errors when I call a function

Question

I'm trying to write a PHP function Insertdata($table,$field,$data) that will insert some data. Here is what I've got so far:

insertdata.php

<?php
  include('connection.php');

  function Insertdata($table,$field,$data){

    $field_values= implode(',',$field);
    $data_values=implode(',',$data);

    $sql="INSERT into". " ".$table." ".$field_values. "VALUES(".$data_values.")";
    $result=$conn->query($sql);
  }
?>

The call of this function happens in another page:

sample.php

<?php
  include('../model/insertdata.php');
  $table="country";
  $field_values=array("country_name","status");
  $data_values=array("usa",'1');
  $sample=Insertdata($table,$field,$data);
  if($result)
  {
    echo "inserted";
  }
  else
  {
    echo "not inseterd";
  }
?>

Whenever I call the function above the following error gets thrown and no data is inserted:

Notice: Undefined variable: field in /var/www/html/online_test/admin/view/sample.php on line 6
Notice: Undefined variable: data in /var/www/html/online_test/admin/view/sample.php on line 6
Warning: implode(): Invalid arguments passed in /var/www/html/online_test/admin/model/insertdata.php on line 6
Warning: implode(): Invalid arguments passed in /var/www/html/online_test/admin/model/insertdata.php on line 7
INSERT into country VALUES()
Notice: Undefined variable: conn in /var/www/html/online_test/admin/model/insertdata.php on line 10
Fatal error: Call to a member function query() on a non-object in /var/www/html/online_test/admin/model/insertdata.php on line 10

I'm new to PHP and in the process of learning. I need some advice how to get this working.

It appears that `$field` & `$data` are never defined. It does look like you defined `$table`, `$data_values`, & `$field_values`. Typo? — RoyaleWCheese1, Jan 28 '16 at 05:55
9 people answered and not a single soul mentioned this function is outright dangerous, allowing any kind of SQL injection — Your Common Sense, May 11 '20 at 06:01

score 1 · Accepted Answer · answered Jan 28 '16 at 05:55

1

In sample.php change

$sample=Insertdata($table,$field,$data);

to

$sample=Insertdata($table,$field_values,$data_values);

that will fix most of those errors as $field and $data aren't defined in sample.php. The final error has to do with $conn which is presumably defined in connection.php

answered Jan 28 '16 at 05:55

Richard Theobald

781
3
9

thank you i have changed now its throwing only one error Call to a member function query() on a non-object in /var/www/html/online_test/admin/model/insertdata.php on line 10 – Arun Kumaresh Jan 28 '16 at 06:01
1

That error is telling you that $conn isn't an object. The problem is most likely a scope issue. Right inside of the function definition at the very top add `global $conn;`. If that doesn't fix it, then it is an error within `connection.php` – Richard Theobald Jan 28 '16 at 06:04
1

Also, your `if($result)` will need to be `if($sample)` or it will always echo `not inserted` as `$result` is never defined (scope issue). – Richard Theobald Jan 28 '16 at 06:06

score 0 · Answer 2 · answered Jan 28 '16 at 06:02

There is plenty wrong with this code, first:

$table="country";
$field_values=array("country_name","status");
$data_values=array("usa",'1');
$sample=Insertdata($table,$field,$data);

$field and $data don't exist. Make sure you are passing the correct variable names.

Also, $conn is not known within Insertdata() context, you have to either pass this variable as parameter too or make it a global.

Also, the SQL syntax is all wrong. INSERT into country country_name,statusVALUES(usa,1) is not valid SQL. Make sure you are formatting your query properly, and also that you are filtering against SQL Injection.

For example:

$fields = implode('`, `', $field);
$values = implode("', '", array_map(function ($val) use ($conn) { return $conn->real_escape_string($val); }, $data));
$sql = "INSERT INTO `{$table}` (`{$fields}`) VALUES ('{$values}')";

score -1 · Answer 3 · answered Jan 28 '16 at 06:02

-1

You just mis-configured your variables when calling the function.

$table="country";
$field_values=array("country_name","status");
$data_values=array("usa",'1');
$sample=Insertdata($table,$field_values,$data_values);

answered Jan 28 '16 at 06:02

John Delmark Aguilar

1
1

devpro · Answer 4 · 2016-01-28T06:21:44.630

There are so many issues in your code, first of all here is the modified code:

Modified Function:

<?
function Insertdata($table,$field,$data)
{
    $field_values= implode(',',$field);
    $data_values=implode("','",$data);

    $sql= "INSERT INTO $table (".$field_values.") 
    VALUES ('".$data_values."') ";
    $result=$conn->query($sql);
}

Modified Function Calling:

$table = "country";
$field = array("country_name","status");
$data = array("usa",'1');
$result = Insertdata($table,$field,$data);
if($result)
{
    echo "inserted";
}
else
{
    echo "not inseterd";
}
?>

Issues:

Insertdata() function you are using VALUES as a simple string, you need to use quotes between string value otherwise it will display like this VALUES (usa,1) this should be VALUES ('usa','1') or VALUES ('usa',1).
You are using undefined variable in function parameters $field, $data, both of them not defined, this should be $field_values, $data_values.
Third, you are using undefined variable $result in success, failure condition this should be $sample not $result.
You are getting Fatal error because you are not using $conn in function param, limited scope

Aakash Bhardwaj · Answer 5 · 2018-09-14T10:05:46.370

In all of these answers you got a similar error that is 1) Notice: Undefined variable: conn 2) Warning: mysqli_query() expects parameter 1 to be mysqli, null These two errors come to avoid them try this:

function:

<?php 
function Insertdata($table,$field,$data,$conn)
{
    $field_values= implode(",",$field);
    $data_values=implode(",",$data); 

    $sql= "INSERT INTO $table (".$field_values.") VALUES (".$data_values.")";
    $result=$conn->query($sql);
}
if($result){
    echo "inserted";
} else {
    echo "not inseterd";
}
?>

function calling:

<?php 
$table = "country";
$field = array("country_name","status");
$data = array("'usa'","'1'");
$result = Insertdata($table,$field,$data,$conn);

?>

Use this code to avoid those two errors.

ManojKiran A · Answer 6 · 2018-11-08T20:00:30.820

    function InsertRecordToTable($tableName='',$fieldNames='')
    {
        $fields  = implode(',', array_keys($fieldNames));
        $values  = implode(',', mysqli_escape_string(array_values($fieldNames)));
        $insertQuery = "INSERT INTO ".$tableName.'('.$fields.') Values '.'('.$values.')';
    }

Try This Function Example with Argument Passed

$field['FirstName'] = 'Manoj';
$field['LastName'] = 'Kiran';
$field['Age'] = '21';
Insertrecord('testtable',$field);

It is working fine to insert record

score -2 · Answer 7 · edited May 11 '20 at 08:28

-2

Try This:

<?php
   $conn = mysqli_connect('localhost','root','','kailash');


   $table = "country";

   $field = array("country_name","status");
   $data = array("usa",'1');

   $field_values= implode(',',$field); 
   $data_values = implode("','",$data);

   $sql = "INSERT INTO $table($field_values)VALUES('$data_values')";
   $result = $conn->query($sql);
?>

edited May 11 '20 at 08:28

h3t1

1,126
2
18
29

answered May 11 '20 at 02:56

Kailash

1
1

Multiple Notice: Undefined variable errors when I call a function

7 Answers7