CODEWITHSUNDEEP
phpmysql
cala
cala

Reputation: 1421

Update form row in php error

I'm trying to update multiple row/rows in my form. I'm getting the error Notice: Array to string conversion in C:\wamp.....

I'm also getting another error Warning: Cannot modify header information - headers already sent by (output started at C:\wamp\....

Both of these are now fixed.

Form

$ad = "<form action='updaterecords.php' method='post' id='update'> \n";

        $records = $this->query_recs();

        foreach ($records as $record) {
            $ad .= "<p id='id{$record->id}'>";
            $ad .= "<input type='hidden' name='id[]' value='" .$this->render_i18n_data($record->id) . "' />\n";
            $ad .= "<input type='text' name='paname[]' value='". $this->render_i18n_data($record->pa_name) . "' class='paname' />\n";
            $ad .= "<input type='text' name='pcname[]' value='". $this->render_i18n_data($record->pc_name) . "' class='pcname' />\n";
            $ad .= "<input type='text' name='pdname[]' value='". $this->render_i18n_data($record->pd_name) . "' class='pdname' />\n";
            $ad .= "<input type='text' name='pfname[]' value='". $this->render_i18n_data($record->pf_name) . "' class='pfname' />\n";

            $ad .= "<input type='submit' name='update' value='Update' />";
            $ad .= "</p>";
        }

        echo($ad);

PHP

<?php

include 'dbdetails.php';

$con = new mysqli($server, $user, $pass, $db);

// Check connection
if ($con->connect_error) {
    die("Connection failed: " . $con->connect_error);
} 
echo "Connected successfully";


if(isset($_POST['update'])){
    $id         = $_POST['id'];
    $paname = $_POST['paname'];
    $pcname     = $_POST['pcname'];
    $pdname = $_POST['pdname'];
    $pfname    = $_POST['pfname'];

    mysqli_query($con, "UPDATE wp_pbcbc_records
                    SET pa_name = '$paname', pc_name='$pcname', pd_name='$pdname', pf_name='$pfname'
                    WHERE id = '$id' ");   

    header("location: localhost/myp");
    exit;
}   
?>

Update: This has now been solved. Thanks to the people who gave me an answer!

Upvotes: 0

Views: 78

Answers (2)

Logan Wayne
Logan Wayne

Reputation: 5991

Note:

  • You passed values in array. You must run them in a loop before using them in your query.
  • On top of your isset() function, there must have no output of HTML entities, which is the common reason for header() function to fail and cause an error.

Your connection to your database:

include("dbdetails.php");

$con = new mysqli($server, $user, $pass, $db); /* MAKE SURE THAT THE VALUES OF YOUR VARIABLES ARE CORRECT CORRESPONDING TO YOUR DATABASE */

/* CHECK CONNECTION */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

Your updated code:

/* THERE MUST HAVE NO OUTPUT IN THIS SECTION TO FIX THE HEADER ERROR */

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

  $counter = count($_POST["paname"]);

  for($x = 0; $x<=$counter; $x++){

    if(!empty($_POST["paname"][$x])){

      $id = mysqli_real_escape_string($con,$_POST['id'][$x]);
      $paname = mysqli_real_escape_string($con,$_POST['paname'][$x]);
      $pcname = mysqli_real_escape_string($con,$_POST['pcname'][$x]);
      $pdname = mysqli_real_escape_string($con,$_POST['pdname'][$x]);
      $pfname = mysqli_real_escape_string($con,$_POST['pfname'][$x]);

      mysqli_query($con, "UPDATE wp_pbcbc_records
                    SET pa_name = '$paname', pc_name='$pcname', pd_name='$pdname', pf_name='$pfname'
                    WHERE id = '$id' ");   

    } /* END OF IF CHECKING paname */

  } /* END OF FOR LOOP */

  header("location: localhost/myp");
  exit;
} /* END OF ISSET */
?>

On the side note:

  • You must use mysqli_real_escape_string() to sanitize the values of your variables before using them in your query to prevent SQL injections.
  • Better recommendation, is to use mysqli_* prepared statement. It will sanitize the variables automatically and no need to escape the strings per variable.

Your code using mysqli_* prepared statement:

/* THERE MUST HAVE NO OUTPUT IN THIS SECTION TO FIX THE HEADER ERROR */

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

  $counter = count($_POST["paname"]);

  for($x = 0; $x<=$counter; $x++){

    if(!empty($_POST["paname"][$x])){

      if($stmt = $con->prepare("UPDATE wp_pbcbc_records SET pa_name=?, pc_name=?, pd_name=?, pf_name=? WHERE id=?")){

        $stmt->bind_param("ssssi",$_POST["paname"][$x],$_POST["pcname"][$x],$_POST["pdname"][$x],$_POST["pfname"][$x],$_POST["id"][$x]);
        $stmt->execute();
        $stmt->close();

      } /* END OF PREPARED STATEMENT */

    } /* END OF IF CHECKING paname */

  } /* END OF FOR LOOP */

  header("location: localhost/myp");
  exit;
} /* END OF ISSET */
?>

Upvotes: 1

Barmar
Barmar

Reputation: 781300

Since you used names ending with [] in your form, the $_POST variables become arrays, and you have to loop over them.

$stmt = mysqli_prepare("UPDATE wp_pbcbc_records
                SET pa_name = ?, pc_name=?, pd_name=?, pf_name=?
                WHERE id = ? ");  
mysqli_stmt_bind_param($stmt, "ssssi", $cur_paname, $cur_pcname, $cur_pdname $cur_pfname, $cur_id);
for ($i = 0; $i < count($id); $++) {
    $cur_paname = $paname[$i];
    $cur_pcname = $pcname[$i];
    $cur_pdname = $pcname[$i];
    $cur_pfname = $pcname[$i];
    $cur_id = $id[$i];
    mysqli_stmt_execute($stmt);
}

Upvotes: 2

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