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How to use a While loop in PHP?

开发者 https://www.devze.com 2023-03-11 09:27 出处:网络
I am having trouble with the most simple piece of code. For some reason, the following code only retrieves the first row out of the database. I am trying a while loop and it just is not working for me

I am having trouble with the most simple piece of code. For some reason, the following code only retrieves the first row out of the database. I am trying a while loop and it just is not working for me. Here is my code:

<?php


//Connect t开发者_如何学Co the database
require_once('mysql_connect.php') ;

$query = "SELECT * FROM past_due_students WHERE charged_today = 'No' ORDER BY past_due_id" ;
$result = mysqli_query($dbc, $query) ;
$number_of_students = mysqli_num_rows($result) 

if ($number_of_students >= 1) {

    //Loop through the entire table
    while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {

        $student_id = $row['student_id'] ;
        $number_of_declines = $row['number_of_declines'] - 1;

        //Update the number of declines
        $query = "UPDATE past_due_students SET number_of_declines = $number_of_declines WHERE student_id = $student_id" ;
        $result = mysqli_query($dbc, $query) ;
        $number = mysqli_affected_rows($dbc) ;

        if ($number == 1) {

            echo '<p><b>The number of declines has been successfully updated.</b></p>' ;

        } else {
            echo $query ;
        }

    }//END while loop

}//END if ($number_of_students >= 1) {

?>

It is only grabbing the first row and none of the other rows.


You are overwriting the value of $result:

$query = "SELECT * FROM past_due_students....." ;
$result = mysqli_query($dbc, $query) ;

//Loop through the entire table
 while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {

    $query = "UPDATE past_due_students SET  ...." ;
    $result = mysqli_query($dbc, $query) ; <-- OVERWRITTING HERE.

Use a different variable to hold the inner query result object.


You are destroying the $result resource you got from the SELECT query by assigning the next query resource to the same variable.

Please use a different variable for the UPDATE query.

So change this:

$query = "UPDATE past_due_students SET number_of_declines = $number_of_declines WHERE student_id = $student_id" ;
$result = mysqli_query($dbc, $query) ;

To this:

$query = "UPDATE past_due_students SET number_of_declines = $number_of_declines WHERE student_id = $student_id" ;
$updateResult = mysqli_query($dbc, $query) ;


try to change $result of update query with other variable name


You're overwriting the $result var in your loop it looks like. Also your second mysqli_affected_rows is being passed the database object.

Change

$result = mysqli_query($dbc, $query) ;
$number = mysqli_affected_rows($dbc) ;

to

$result2 = mysqli_query($dbc, $query) ;
$number = mysqli_affected_rows($result2) ;
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