Question

我正在尝试更新表单中的多个行/行。我收到错误Notice: Array to string conversion in C:\wamp.....

我还收到了另一个错误Warning: Cannot modify header information - headers already sent by (output started at C:\wamp\....

这两个现在都已修复。

表格

$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;
}   
?>

更新：现在已经解决了。感谢那些给我答案的人！

Answer 1

由于您在表单中使用以[]结尾的名称，$_POST变量将成为数组，您必须循环它们。

$stmt = mysqli_prepare("UPDATE wp_pbcbc_records
                SET pa_name = ?, pc_name=?, pd_name=?, pf_name=?
                WHERE id = ? ");  
mysqli_stmt_bind_param($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);
}

Answer 2

注意：

您在数组中传递了值。在查询中使用它们之前，必须先循环运行它们。
在isset()函数之上，必须没有HTML实体的输出，这是header()函数失败并导致错误的常见原因。

您与数据库的连接：

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();
}

您的更新代码：

/* 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 */
?>

旁注：

在查询中使用变量值之前，您必须使用mysqli_real_escape_string()来清理变量值，以防止SQL injections。
更好的推荐，就是使用mysqli_* prepared statement。它会自动清理变量，不需要为每个变量转义字符串。

您的代码使用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 */
?>

在php错误中更新表单行

2 个答案:

注意：