Question

我已经清理了我的代码，以便将以下内容作为我的表单。但我在发送数据和从新的update.php更新时遇到了问题。该表单可以正常检索数据并显示它。但是在提交时我得到了正确的更新消息，但记录在数据库中没有任何改变。的index.php

<?php
  include 'connectdb.php';
  // include 'query.php';

  $sql = "SELECT id, WeightorMeasure FROM weightsmeasures";
  $result = $conn->query($sql)       
  if ($result->num_rows > 0) {

    // output data of each row
    while($row = $result->fetch_assoc()) {
    echo "<form action=\"update.php\"method=\"post\">";
    echo "<input type=\"text\" name=\"id\" value = ".$row["id"].">";
    echo "<input type=\"text\" name=\"WeightorMeasure\" value = ".$row["WeightorMeasure"] .">";
    echo "<input type=\"submit\" value=\" Submit \" name=\"Update\">";
    }
        echo "</form>";
    } else {
        echo "0 results";
    }  
       $conn->close();
?>

update.php

<?php
  include 'connectdb.php';
  $wm = $_POST['id'];
  $id = $_POST['WeightorMeasure'];
  $sql = "UPDATE weightsmeasures SET WeightorMeasure='$wm' WHERE id='$id'";

    if (mysqli_query($conn, $sql)) {
        echo "Record updated successfully";
    } else {
        echo "Error updating record: " . mysqli_error($conn);
    }

    $conn->close();
?>

已更改为“下方”，现在出现此错误。更新记录时出错：未知栏＆＃39; sdada＆＃39;在＆＃39;字段列表＆＃39;。所以看起来它试图使用表格值$ wm作为表格中的列标题而不是输入值。

$wm = $_POST['WeightorMeasure'];
$id = $_POST['id'];


$sql = "UPDATE weightsmeasures SET WeightorMeasure=$wm WHERE id=$id";

Answer 1

 $wm = $_POST['id'];
 $id = $_POST['WeightorMeasure'];

也许你有这些错误的方式？：d

 $wm = $_POST['WeightorMeasure'];
 $id = $_POST['id'];

顺便说一句，你的查询是针对MySQL注入的，我们请考虑使用预备语句

Answer 2

您是否意识到您已从$ _POST数据中更改了变量分配中的ID和WeightOrMeasure？这会导致更新查询无法找到ID但不会遇到问题。从而告诉你操作成功了

Answer 3

我发现代码中存在以下可能的问题。

String contact_str = "Contact: " + name;
String email_str = "Address: " + email;
String age_str = "Age: " + age;
DisplayAlert(contact_str +"\n"+ email_str +"\n"+ age_str);

和这一行

mysqli_query($conn, $sql); //should be $conn->query($sql);

订单错了。我希望你已经在dpconnect.php文件中创建了$ conn对象。

Answer 4

好的发现问题是上面有$ _POST [＆＃34; WeightorMeasure＆＃34;]的混合物。和$ _POST [＆＃34; id＆＃34;];混淆了，但最重要的因素是我从包含多行发布的表和在_POST上发布到update.php，它不知道如何处理所有不同的行，因为SQL只处理一行。一旦我通过帖子发送单行，它工作正常。现在按照建议学习和添加准备好的语句。

update.php

 <?php
       include 'connectdb.php';
       $wm = $_POST["WeightorMeasure"];
       $id = $_POST["id"];

        echo $wm . "<br>";
        echo $id . "<br"; 

        $sql = "UPDATE weightsmeasures SET WeightorMeasure=\"$wm\", id= 
         $idWHERE id= $id";

        if (mysqli_query($conn, $sql)) {
            echo "Record updated successfully";
        } else {
            echo "Error updating record: " . mysqli_error($conn);
        }
        $conn->close();
    ?>

手动单输入。的index.php

<?php
   include 'connectdb.php';
   // include 'query.php';

   $sql = "SELECT id, WeightorMeasure FROM weightsmeasures WHERE id=11";
       $result = $conn->query($sql);

          if ($result->num_rows > 0) {

           // output data of each row
      while($row = $result->fetch_assoc()) {
          echo "<form action=\"update.php\"method=\"post\">";
          echo "<input type=\"text\" name=\"id\" value = ".$row["id"].">";
          echo "<input type=\"text\" name=\"WeightorMeasure\" value = ".$row["WeightorMeasure"].">";
          echo "<input type=\"submit\" value=\" Submit \" name=\"Update\">";
           }         
              echo "</form>";
             } else {
              echo "0 results";
             }  
       $conn->close();
 ?>

为什么MYSQLi不更新数据库记录但是给出成功消息

4 个答案: