i am trying to update quary of "email" by associtive array loop through but when i try this there are take only one last value of array and upadat.

here is code...

<?php

include_once('db_connect.php');

$update_array=array();
                $update_array=array(
                                            array('Email'=>'[email protected]'),
                                            array('Email'=>'[email protected]'),
                                            array('Email'=>'[email protected]'),
                                            array('Email'=>'[email protected]'),
                                            array('Email'=>'[email protected]'));
print_r($update_array)."<br>";
echo "<pre>";

            
           foreach ($update_array as  $value ) {        

            
                $update_Email=$value['Email'];

                $update="UPDATE form SET email='$update_Email' where id";

                    $result= mysqli_query($conn, $update);
          
            }               
                            
if ($result) {
  echo "New record Updated successfully";
} 
else {

  echo "Error:not created ". mysqli_error($conn);
}

mysqli_close($conn);

?>

image of data update,

enter image description here

i tried nasted foreach to featch array value and than passing into upadate but error apper: array ofset on strin in.

help to update all email's data in once and all.

CodePudding user response：

you need to define the id in your where clause, id needs to be of the row you want to update.

<?php

include_once('db_connect.php');

$update_array=array();
$update_array=array(
    array('Email'=>'[email protected]', 'id'=>'1'),
);

foreach ($update_array as $value ) {        
    $update_Email=$value['Email'];
    $update_id = $value['id'];
    
    $update="UPDATE form SET email='$update_Email' where id = $update_id";
    $result= mysqli_query($conn, $update);
}               

if ($result) {
    echo "New record Updated successfully";
} else {
    echo "Error:not created ". mysqli_error($conn);
}
mysqli_close($conn);

?>    

i would also highly recommend you reading about prepared statements to make sure you wont be a target of sql injection.

<?php

include_once('db_connect.php');

$update_array=array();
$update_array=array(
    array('Email'=>'[email protected]', 'id'=>'1'),
);

foreach ($update_array as $value ) {        
    
    $stmt = $conn->prepare("UPDATE form SET email = ? WHERE id = ?");
    $stmt->bind_param("si", $update_Email, $update_id);

    $update_Email=$value['Email'];
    $update_id = $value['id'];

    $stmt->execute();
}           
$stmt->close();
$conn->close();
?>

php.net // Prepared statements