Error reporting in PHP script

Question

I am having minor issue with this script. It is allows the user to upload a CSV and write to a MySQL table along with checking for duplicates against the table for existing records. The script works OK except for message reporting. Please see below.

<?php
 require_once("models/config.php"); //db connection is in this file



function get_file_extension($file_name) {
return end(explode('.',$file_name));
}

function errors($error){
if (!empty($error))
{
        $i = 0;
        while ($i < count($error)){
        $showError.= '<div class="msg-error">'.$error[$i].'</div>';
        $i ++;}
        return $showError;
}// close if empty errors
 } // close function

 if (isset($_POST['upfile'])){

if(get_file_extension($_FILES["uploaded"]["name"])!= 'csv')
{
$error[] = 'Only CSV files accepted!';
}

if (!$error){

$tot = 0;
$handle = fopen($_FILES["uploaded"]["tmp_name"], "r");
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {

for ($c=0; $c < 1; $c++) { 

    //only run if the first column if not equal to firstname
    if($data[0] !='firstname'){         
         $check=mysqli_query($mysqli,"select * from persons where     firstname='$data[0]' and surname='$data[1]'");
  $checkrows=mysqli_num_rows($check);
  if($checkrows>0){echo "$data[0] $data[1] exists in the database. Please remove this     entry before uploading your records.";}  
else{  

        $order = "INSERT INTO persons (firstname, surname, gender, email, address, city, province, postalcode, phone, secondphone, organization, inriding, ethnicity, senior, political_id) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
$stmt = mysqli_prepare($mysqli, $order);
mysqli_stmt_bind_param($stmt, "sssssssssssssss",  $data[0],  $data[1],  $data[2],  $data[3],  $data[4],  $data[5],  $data[6],  $data[7],  $data[8],  $data[9],  $data[10],  $data[11],  $data[12],  $data[13],  $data[14]);
 $result = mysqli_stmt_execute($stmt);          
     }

$tot++;}

}
}
fclose($handle);
$content.= "<div class='success' id='message'> CSV File Imported, $tot records added     </div>";
}// end no error
}//close if isset upfile\\
$er = errors($error);
$content.= <<<EOF
<h3>Import CSV Data</h3>
$er
<form enctype="multipart/form-data" action="" method="post">
File:<input name="uploaded" type="file" maxlength="20" /><input type="submit"     name="upfile" value="Upload File">
</form>
EOF;
echo $content;
?>

When the script finds a duplicate, it echoes out 'Firstname Surname exists in the database. Please remove this entry before uploading your records. CSV File Imported, 1 records added"

Nothing gets written to the table, which is what I want. But I would prefer the message to say 'CSV File Imported, 0 records added' or not have that message appear at all upon finding a duplicate, because that might confuse the user

I know this is a super easy fix (something like a misplaced bracket), but I can't figure out.

Thanks in advance.

jonnu · Accepted Answer · 2014-04-24 17:16:12Z

Move the line that has:

$tot++;

inside the preceding brace. Like so:

//only run if the first column if not equal to firstname
if ($data[0] != 'firstname') {

    $check     = mysqli_query($mysqli, "select * from persons where firstname='$data[0]' and surname='$data[1]'");
    $checkrows = mysqli_num_rows($check);

    if ($checkrows > 0) {
        echo sprintf('%s %s exists in the database. Please remove this entry before uploading your records.', $data[0], $data[1]);
    }
    else {
        $order = "INSERT INTO persons (firstname, surname, gender, email, address, city, province, postalcode, phone, secondphone, organization, inriding, ethnicity, senior, political_id) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)";
        $stmt  = mysqli_prepare($mysqli, $order);
        mysqli_stmt_bind_param($stmt, "sssssssssssssss", $data[0], $data[1], $data[2], $data[3], $data[4], $data[5], $data[6], $data[7], $data[8], $data[9], $data[10], $data[11], $data[12], $data[13], $data[14]);
        $result = mysqli_stmt_execute($stmt);

        $tot++;
    }
}

To be honest, your code is very difficult to read. You may want to re-tab it so you can follow the flow of it better. After re-tabbing your code, the problem was quite easy to see.

Well done on using a statement there, I've lost count of the PHP questions without them - you should add one for your $check query too!

Deele · Accepted Answer · 2014-04-24 17:12:30Z

0

My first guess is that $showError is not defined before you are appending to wt with $showError.=. It should resolve into PHP notification "PHP Notice: Undefined variable: $showError", but it should still output it.

In addition, that function would output null, if there are no errors.

And I would use foreach for that loop.

function errors($errors) {
    $showError = '';
    if (!empty($errors)) {
        foreach ($errors as $error) {
            $showError .= '<div class="msg-error">'.$error.'</div>';
        }
    }
    return $showError;
}

See if that helps to output errors.

answered Apr 24, 2014 at 17:12

Deele

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