如何使用数组访问循环外的whileloop记录

Question

Hello i selected all records from users and returned them using a php whileloop and array, using implode i could get all the records outside the loop, what i wish to do actually is to access individual record and assign them to a variable individually to be used later in the same page.

this is what i came up with already, it works, but i don't know how to assign the recorded to individual variables, since all the records are displayed using implode

       `<?php
$data = array();
            $query = "SELECT * FROM users ";
            $queryResult = mysqli_query($conn,$query);
            while($_row = mysqli_fetch_array($queryResult)){
              $data[] = $_row['first_name'];
            $data[] = $_row['email'];
            $data[]  = $_row['amount'];
               
                ?>
                <?php }?>
                <?php
                echo "<br />";
               $request = implode("<br />", $data);
              
               echo $request;?>`

please i need assistance on how to achieve or a better way to access whileloop records outside the loop this thanks in advance

This is what i intended doing with that result of the loop in the next query

`<?php

$profit = 10;
$query = "UPDATE storage SET ";
$query .="store_profit = '{$profit}' ";
$query .= "WHERE email = '{$email_from_loop_above}' ";?>`

for clarity purposes.. this script will be executed by a cronjob every 1 minute, initially this is what is did..

            `
<?php
$query = "SELECT * FROM users WHERE email = $_SESSION['email']  ";
            $queryResult = mysqli_query($conn,$query);
            while($_row = mysqli_fetch_array($queryResult)){
              $first_name = $_row['first_name'];
           $email_from_loop_above = $_row['email'];
            $amount  = $_row['amount'];
               
                ?>
                <?php }?>`

Then for the update query this is what i did

    `<?php
//The profit below was gotten from a percent from the amount i return from the loop above.
    $profit = 10;
    $query = "UPDATE storage SET ";
    $query .="store_profit = '{$profit}' ";
    $query .= "WHERE email = '{$_SESSION['email']}' ";
    ?>`

Now this code above works perfectly, but when the user logout out the store_profit would not update anymore simply because this you know would require an active user SESSION to work, so what i am trying to do now is is make it work accordingly even when the user is logged out of the account, if i use the following method

 `<?php
$query = "SELECT * FROM users ";
            $queryResult = mysqli_query($conn,$query);
            while($_row = mysqli_fetch_array($queryResult)){
              $first_name = $_row['first_name'];
            $email_from_loop = $_row['email'];
            $amount  = $_row['amount'];
               
                ?>
                <?php }?>`
    
    
    `
    <?php
    $profit = 10;
    $query = "UPDATE storage SET ";
    $query .="store_profit = '{$profit}' ";
    $query .= "WHERE email = '{$email_from_loop}' ";
    ?>`

it would only return the last user data only, simply because i am outside the loop, now when i try to update inside the loop, it works but it adds 10 to store_profit to the first user and adds 20 to the second user also adds 30 to third user... simply because it inside the loop

now what is want to do is make this work accordingly so i decided to use array, however i am not to familiar with array that why i became confused. An explanation of how i can achieve this purpose would be very much appreciated thanks. however i was thinking of not destroying all the user session maybe i create session for that purpose and not distroy it.. but i don't if that method would be safe or good.

Answer 1

As a result of your loop you have a $data array which you then turn into a string by gluing all its elements using implode function.

You can of course still access every individual element of $data array outside of the loop by addressing to its direct index.

For instance:

echo $data[0]['email'];

will output the email record of first (arrays are 0 based in PHP) element of your array.

Answer 2

You can take a look at below code

    $data = array();
    $query = "SELECT * FROM users ";
    $qeuryResult = mysqli_query($conn,$query);
    while($_row = mysqli_fetch_array($qeuryResult)){
    array_push($data, array(
                            'first_name' => $_row['first_name'],
                            'email' => $_row['email'],
                            'amount' => $_row['amount'],
        ));
       
        }
        //print_r($data[0]['first_name']);
        echo "<pre>";
        print_r($data);