mysql没有从PHP表单更新

Question

I have a very simple PHP form, which shows a checkbox, and will store if it is checked or not in a database. This works for the initial inserting, but not for updating. I have tested cases where $saleid equals $pk and it does not enter the if branch to update...why?

<?php
error_reporting(E_ALL);

if (isset($_GET["cmd"]))
  $cmd = $_GET["cmd"]; 
    else
if (isset($_POST["cmd"]))
  $cmd = $_POST["cmd"]; 
        else die("Invalid URL");

if (isset($_GET["pk"])) { $pk = $_GET["pk"]; }

$checkfield = "";

$checkboxes = (isset($_POST['checkboxes'])? $_POST['checkboxes'] : array());

if (in_array('field', $checkboxes)) $checkfield = 'checked';

$con = mysqli_connect("localhost","user","", "db");
if (!$con) { echo "Can't connect to MySQL Server. Errorcode: %s
". mysqli_connect_error(); exit; }

$con->set_charset("utf8");

$getformdata = $con->query("select saleid, field from STATUS where saleid = '$pk'");
$saleid = "";
while ($row = mysqli_fetch_assoc($getformdata)) {
    $saleid = $row['saleid'];
    $checkfield = $row['field'];
}

if($cmd=="submitinfo") {
    if ($saleid == null) {
       $statusQuery = "INSERT INTO STATUS VALUES (?, ?)";
        if ($statusInfo = $con->prepare($statusQuery)) {
                $statusInfo->bind_param("sssssssssssss", $pk, $checkfield);
                $statusInfo->execute();
                $statusInfo->close();
        } else {
                print_r($con->error);
        }
    } else if ($saleid == $pk) {
        $blah = "what";
        $statusQuery = "UPDATE STATUS SET field = ? WHERE saleid = ?";
        if ($statusInfo = $con->prepare($statusQuery)) {
                $statusInfo->bind_param("ss", $checkfield, $pk);
                $statusInfo->execute();
                $statusInfo->close();
        } else {
                print_r($con->error);
     }  
    }
}
if($cmd=="EditStatusData") {
    echo "<form name=\"statusForm\" action=\"test.php?pk=".$pk."\" method=\"post\" enctype=\"multipart/form-data\">
                <h1>Editing information for Auction No: ".$pk."</h1>
                        <input type=\"checkbox\" name=\"checkboxes[]\" value=\"field\" ".$checkfield." />
                        <label for=\"field\">Test</label>
                        <br />
                        <input type=\"hidden\" name=\"cmd\" value=\"submitinfo\" />
                        <input name=\"Submit\" type=\"submit\" value=\"submit\" />
        </form>";
}
?>

Answer 1

well i created a table and ran your code and it works fine for me

the reason why it doesn't "look" like update is working, is because you are reading $saleid and $checkfield from the database then building an update statement that puts the same two values back into the database

which probably isn't what you are wanting to do

this line here sets $checkfield to 'checked',

 if (in_array('field', $checkboxes)) $checkfield = 'checked';

then you set $checkfield from the database (overwriting the value 'checked' )

while ($row = mysqli_fetch_assoc($getformdata)) {
   $saleid = $row['saleid'];
   $checkfield = $row['field'];

then you write the original value of checkfield back to the database

$statusInfo->bind_param("ss", $checkfield, $pk);